Solutions Manual for Differential Equations Theory, Technique, and Practice (Textbooks in Mathematic by QuentinAxel

CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION? (k) If y 2 = x2 −cx, then 2yy 0 = 2x−c so 2xyy 0 = 2x2 −cx = x2 +x2 −cx = x2 + y 2 . (l) If y = c2 + c/x, then y 0 = −c/x2 so x4 (y 0 )2 = c2 = y − c/x. Use the fact that −c/x = xy 0 to obtain x4 (y 0 )2 = y + xy 0 . 0

(m) If y = cey/x , then y 0 = xyx2−y cey/x = xyyx2−y . Solve for y 0 to obtain y 0 = y 2 /(xy − y 2 ). (n) If y + sin y = x, then y 0 + y 0 cos y = 1 or y 0 = 1/(1 + cos y). Multiply the numerator and denominator of the right side by y to obtain y 0 = y/(y + y cos y). Now use the identity y = x − sin y to obtain y 0 = (x − sin y + y cos y). (o) If x + y = arctan y, then 1 + y 0 = y 0 /(1 + y 2 ). Consequently, (1 + y 0 )(1 + y 2 ) = y 0 . This simplifies to 1 + y 2 + y 2 y 0 = 0. 2. Find the general solution of each of the following differential equations. R (a) If y 0 = e3x − x, then y = xex dx + C = xex − ex + C. R x2 2 2 (b) If y 0 = xex , then y = xex dx + C = e2 + C (substitution u = x2 ). R x R x (c) If (1 + x)y 0 = x, then y 0 = 1+x and y = 1+x dx + C = (1 − 1 1+x )dx + C = x − ln |1 + x| + C. R x x 1 (d) If (1 + x2 )y 0 = x, then y 0 = 1+x 2 and y = 1+x2 dx + C = 2 ln |1 + 2 2 x | + C (substitution u = 1 + x ). R arctan x x (e) If (1 + x2 )y 0 = arctan x, then y 0 = arctan 1+x2 and y = 1+x2 dx + C = 1 x arctan +C (substitution u = arctan x). 2 R (f) y 0 = 1/x, so y = x1 dx + C = ln |x| + C. √ R (g) y = arcsin xdx + C = x arcsin x + 1 − x2 + C (integrate by parts, u = arcsin x). R R R csc x+cot x (h) y 0 = sin1 x so y = sin1 x dx+C = csc xdx+C = csc x csc x+cot x dx+ C. SetR u = csc x + cot x, then du = (− csc x cot x − csc2 x)dx and y = − du u + C = − ln |u| + C = − ln | csc x + cot x| + C. 1 1 R x R −1/3 x 3 x+ 3 (i) y 0 = 1+x 3 , so y = 1+x3 dx = x+1 + x2 −x+1 dx (partial fracR 1 x− 16 R 1/2 tions) = − 31 ln |x + 1| + x23 −x+1 dx + x2 −x+1 dx = − 13 ln |x + 1| + R 1/2 1 2 dx = − 31 ln |x + 1| + 16 ln |x2 − x + 1| + 6 ln |x − x + 1| + (x− 12 )2 + 34 √ R 2/3 √ dx = − 31 ln |x + 1| + 16 ln |x2 − x + 1| + 33 arctan 2x−1 +C √ 1+( 2x−1 )2 3 3

√ ). (substitution u = 2x−1 3 R R −1 x x 2 (j) y 0 = x2 −3x+2 , so y = x2 −3x+2 dx = x−1 + x−2 dx = − ln |x − 1| + 2 ln |x − 2| + C.

3. For each of the following differential equations, find the particular solution that satisfies the given initial condition.

1.3. THE NATURE OF SOLUTIONS

R (a) If y 0 = xex , then y = xex dx + C = (x − 1)ex + C (integrate by parts, u = x). When x = 1, y = C so the particular solution is y(x) = (x − 1)ex + 3. R (b) If y 0 = 2 sin x cos x, then y = 2 sin x cos xdx+C = sin2 x+C. When x = 0, y = C so the particular solution is y(x) = sin2 x + 1. R (c) If y 0 = ln x, then y = ln xdx + C = x ln x − x + C (integrate by parts, u = ln x). When x = e, y = C so the particular solution is y(x) = x ln x − x. R R (d) If y 0 = 1/(x2 − 1), then y = 1/(x2 − 1)dx + C = 1/2 1/(x − 1) − 1/(x + 1)dx + C = 21 ln x−1 x+1 + C (method of partial fractions). When x = 2, y = 12 ln 13 + C = C − ln23 so the particular solution is ln 3 y(x) = 12 ln x−1 x+1 + 2 . R R 1 (e) If y 0 = x(x21−4) , then y = x(x2 −4) dx + C = 1/8 1/(x + 2) + 2

1/(x − 2) − 2/xdx + C = 18 ln |x x−4| + C (method of partial fractions). 2 When x = 1, y = 18 ln 3 + C so the particular solution is y(x) = 2

|x −4| 1 − 18 ln 3 = 18 ln |x3x−4| 2 . 8 ln x2

R 1 R 2x2 +x 2x2 +x 1 (f) If y 0 = (x+1)(x 2 +1) , then y = (x+1)(x2 +1) dx + C = 2 x+1 + 1 3 1 3x−1 2 x2 +1 dx + C = 2 ln(x + 1) + 4 ln(x + 1) − 2 arctan x + C (method of partial fractions). When x = 0, y = C so the particular solution is y(x) = 12 ln(x + 1) + 34 ln(x2 + 1) − 21 arctan x + 1. 2

4. By the Product Rule, the derivative of y = ex 2 2 ex e−x = 2xy + 1.

Rx 0

e−t2 dt is 2xex

Rx 0

e−t2 dt+

5. For the differential equation y 00 − 5y 0 + 4y = 0, carry out the detailed calculations required to verify these assertions. (a) If y = ex , then y 00 − 5y 0 + 4y = ex − 5ex + 4ex ≡ 0. If y = e4x , then y 00 − 5y 0 + 4y = 16e4x − 20e4x + 4e4x ≡ 0. (b) If y = c1 ex +c2 e4x , then y 00 −5y 0 +4y = c1 (ex −5ex +4ex )+c2 (16e4x − 20e4x + 4e4x ≡ 0. 6. Taking a derivative with respect to x on both sides of x2 y = ln y + c, 0 we have 2xy + x2 y 0 = yy . Multiply y to get 2xy 2 + x2 yy 0 = y 0 . Thus 2

2xy y 0 = 1−x 2y .

7. For which values of m will the function y = ym = emx be a solution of the differential equation 2y 000 + y 00 − 5y 0 + 2y = 0?

CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION? Find three such values m. Use the ideas in Exercise 5 to find a solution containing three arbitrary constants c1 , c2 , c3 . Substitute y = emx into the differential equation to obtain 2m3 emx + m2 emx − 5memx + 2emx = 0. Cancel emx in each term (it is never 0) to obtain the equivalent equation 2m3 + m2 − 5m + 2 = 0. Observing that m = m1 = 1 is a solution (and y1 = ex is a solution to the differential equation). Using this we can factor the polynomial–divide by m − 1–to obtain 2m3 + m2 − 5m + 2 = (m − 1)(2m2 + 3m − 2). The quadratic term factors yield two more roots, m2 = −2, m3 = 1/2, and two more solutions y2 = e−2x and y3 = ex/2 . These three solutions can be combined, as in Exercise 5, to produce a solution with three arbitrary constants y = c1 ex + c2 e−2x + c3 ex/2 .

Chapter 2

Solving First-Order Equations 2.1

Separable Equations

1. Use the method of separation of variables to solve each of these ordinary differential equations. dy (a) Write the equation x5 y 0 + y 5 = 0 in Leibnitz form x5 dx + y5 = 0 R R dx dy dy and separate the variables: y5 = − dx x5 . Integrate, y5 = − x5 , to obtain the solution: y −4 /(−4) = x−4 /4+C. This can also be written 4 in the form x4 + y 4 = Cx4 y 4 or y = ( Cxx4 −1 )1/4 . dy (b) Write the equation y 0 = 4xy in Leibnitz form dx = 4xy and separate R R dy dy the variables: y = 4xdx. Integrate, 4xdx, to obtain the y = solution: ln |y| = 2x2 + C. This can also be written in the form 2 y = Ce2x . dy + y tan x = 0 (c) Write the equation y 0 + y tan x = 0 in Leibnitz form dx R dy and separate the variables: y = − tan xdx. Integrate, dy y == R tan xdx, to obtain the solution: ln |y| = ln | cos x| + C. This can also be written in the form y = C cos x.

(d) The equation (1 + xR2 )dy + (1R+ y 2 )dx = 0 can be rearranged and dy dx integrated directly, 1+y 2 + 1+x2 = C. Therefore, the implicit solution is arctan y + arctan x = C. This can also be written in the form y = tan(C − arctan x). (e) Proceed as in part (d). Rearrange − xdy = 0 to the form R dx y lnR ydx dy dy dx − = 0 and integrate: − x y ln y x y ln y = C. This yields the implicit solution ln |x| − ln | ln y| = C which can also be written in the form ln y = Cx or y = eCx . 11

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS dy (f) From Leibnitz form x Rdx = (1 − 4x2 )R tan y we obtain cot ydy = (1/x − 4x)dx. Integrating, cot ydy = 1/x − 4xdx, gives the implicit 2 solution ln | sin y| = ln |x| − 2x2 + C. Consequently, sin y = Cxe−2x 2 so y = arcsin(Cxe−2x ). dy (g) The Leibnitz form dx sin y = x2 separates to sin ydy = x2 dx. Integrate to obtain − cos y = x3 /3 + C or y = arccos(C − x3 /3). dy − y tan x = (h) Write the equation y 0 − y tan x = 0 in Leibnitz form dx R dy 0 and separate the variables: y = tan xdx. Integrate dy y = R tan xdx, to obtain the solution: ln |y| = − ln | cos x| + C. This can also be written in the form y = C/ cos x or y = C sec x. ydy dy = y − 1 we obtain y−1 = dx (i) From Leibnitz form, xy dx x . Write this in y−1+1 dx the form y−1 dy = x and integrate to obtain the implicit solution y + ln |y − 1| = ln |x| + C. dy 2 (j) Leibnitz form xy 2 − dx x = 0 separates to dx x = y−2dy. Integrating yields the implicit solution ln |x| = −1/y + C. The solution can be 1 expressed explicitly as y = C−ln |x| .

2. For each of the following differential equations, find the particular solution that satisfies the additional given property. dy y = x+1 separates to ydy = (x+1)dx. Integrate (a) The Leibnitz form dx 1 2 1 2 to obtain 2 y = 2 x + x + C. When x = 1, y = 3, so 92 = 12 + 1 + C and C = 3. The particular solution is 12 y 2 = 12 x2 + x + 3. dy 2 dx (b) The Leibnitz form dx x = y separates to dy y = x2 . Integrate to obtain ln |y| = − x1 + C. *** dy x 2 (c) The Leibnitz form (1+x 2 )dx = y separates to ydy = x(1 + x )dx. 1 2 1 2 1 4 Integrate to obtain 2 y = 2 x + 4 x + C. When x = 1, y = 3, so 1 2 1 2 1 4 15 C = 15 4 , and the particular solution is 2 y = 2 x + 4 x + 4 . 2

(d) The Leibnitz form ydxdy = x + 2 separates to y 2 dy = (x + 2)dx. Integrate to obtain 31 y 3 = 12 x2 + 2x + C. When x = 0, y = 4, so 1 3 1 2 64 C = 64 3 , and the particular solution is 3 y = 2 x + 2x + 3 . dy 2 = x2 y 2 separates to dy (e) The Leibnitz form dx y 2 = x dx. Integrate to obtain − y1 = 13 x3 + C. When x = −1, y = 2, so C = − 61 , and the particular solution is − y1 = 13 x3 − 61 .

(f) The Leibnitz form (1+y)dy = 1−x2 separates to (1+y)dy = (1−x2 )dx. dx 1 2 Integrate to obtain y + 2 y = x − 13 x3 + C. When x = −1, y = −2, so C = 23 , and the particular solution is y + 21 y 2 = x − 31 x3 + 23 . 0

3. Substituting y 0 = p yields pp = x2 . Separation of variables (or direct 3

integration) gives ln |p| = x3 /3 + C. This implies that p = Cex /3 and

2.2. FIRST-ORDER LINEAR EQUATIONS

R 3 3 so y 0 = Cex /3 . Consequently, y = C ex /3 dx + D. As we expect, the solution contains two arbitrary constants. The integral cannot be evaluated in terms of elementary functions. 4. Substituting y 0 = p yields p0 p = x(1 + x). Separation of variables gives p2 = x2 + 32 x3 + C. Substituting the initial condition p(0) = y 0 (0) = 2 yields C = 4. ***

2.2

First-Order Linear Equations

1. Find the general solution of the following equations. 2

(a) The equation is linear and separable. The integrating factor is e−x /2 2 2 so it simplifies to (e−x /2 y)0 = 0 and e−x /2 y = C. Therefore, y = 2 Cex /2 . (b) This equation is also linear and separable. The integrating factor 2 2 2 is ex /2 so it simplifies to (ex /2y y)0 = xex /2 . Integrate to obtain 2 2 2 ex /2 y = ex /2 + C and y = 1 + Ce−x /2 . (c) The integrating factor is ex so the equation simplifies to (ex y)0 = ex x x 1+e2x . Integrating we obtain e y = arctan e + C. The general −x x −x solution is y = e arctan(e ) + Ce . (d) The integrating factor is ex so the equation simplifies to (ex y)0 = 2x + x2 ex . Integrate to get ex y = x2 + (2 − 2x + x2 )ex + C (integrate x2 ex by parts, twice, or use an integral table). The general solution is y = x2 e−x + 2 − 2x + x2 + Ce−x . (e) Write as xy 0 = 2y − x3 and then y 0 − (2/x)y = −x2 . The integrating factor is x−2 so the equation simplifies to (x−2 y)0 = −1. Integrate to obtain x−2 y = −x + C. The general solution is y = −x3 + Cx2 . 2 2 (f) The integrating factor is ex so the equation simplifies to (ex y)0 = 0. 2 2 Consequently, ex y = C and y = Ce−x . (g) Write as y 0 − (3/x)y = x3 . The integrating factor is x−3 so the equation simplifies to (x−3 y)0 = 1. Integrating we obtain x−3 y = x + C so y = x4 + Cx3 . cot x 2x (h) Express the equation in the form y 0 + 1+x 2 y = 1+x2 . The integrating 2 factor is 1 + x and the equation simplifies to ((1 + x2 )y)0 = cot x. x|+C Consequently, (1 + x2 )y = ln | sin x| + C and y = ln | sin . 1+x2 (i) The integrating factor is sin x and the equation simplifies to (y sin x)0 = 2 2x. Therefore, y sin x = x2 + C and y = xsin+C x . (j) Express the equation in the form y 0 + (1/x + cot x)y = 1. The integrating factor is x sin x so the equation simplifies to (xy sin x)0 = x sin x. Integrate to obtain xy sin x = sin x − x cos x + C (use a table of integrals or integrate x sin x by parts, u = x). Therefore, cos x+C y = sin x−x . x sin x

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS 2. For each of the following differential equations, find the particular solution that satisfies the given initial data. 1

(a) The integrating factor is e− 2 x and the equation simplifies to (e− 2 x y)0 = 1 2 0. Integrate it to obtain e− 2 x y = C. Plug in the initial condition 1 2 1 2 to get C = √3e . Therefore, e− 2 x y = √3e , and y = √3e e 2 x . 2

(b) The integrating factor is e−x and the equation simplifies to (ye−x )0 = 2 2 2 6x. Integrate it to obtain ye−x = 3x2 + C and y = 3x2 ex + Cex . Plug in the initial condition to get C = 1−3e e . Therefore, y = 1−3e x2 2 x2 3x e + e e . (c) The integrating factor is |x|. Given the initial condition, we restrict that x > 0. Thus, the integrating factor becomes x and the equation simplifies to (xy)0 = 3x2 . Integrate it to get xy = x3 + C. Plug in the initial condition to get C = −4. Therefor, xy = x3 − 4 and y = x2 − x4 . (d) The integrating factor is |x|. Given the initial condition, we restrict that x > 0. Thus, the integrating factor becomes x and the equation simplifies to ( x1 y)0 = x. Integrate to get x1 y = 12 x2 + C. Plug in the initial condition to get C = 52 . Therefore, x1 y = 12 x2 + 52 and y = 21 x3 + 52 x. (e) The integrating factor is e4x and the equation simplifies to (e4x y)0 = e3x . Integrate it to get e4x y = 31 e3x + C. Plug in the initial condition to get C = − 13 . Therefore, e4x y = 31 e3x − 13 and y = 31 e−x − 13 e−4x . (f) The integrating factor is |x|. Given the initial condition, we restrict that x > 0. Thus, the integrating factor becomes x and the equation simplifies to (yx)0 = 2. Integrate it to get xy = 2x + C. Plug in the initial condition to get C = −1. Therefore, xy = 2x − 1 and y = 2 − x1 . 3. Bernoulli Equations To verify the technique write the Bernoulli equation in the form y −n y 0 + P y 1−n = Q. The substitution z = y 1−n and 1 z 0 = (1 − n)y −n y 0 yield 1−n z 0 + P z = Q. (a) Bernoulli, n = 3. Substitute z = y −2 , z 0 = −2y −3 y 0 into xy −3 y 0 + y −2 = x4 to obtain (−1/2)xz 0 + z = x4 . This is linear, z 0 − (2/x)z = −2x3 , with integrating factor x−2 . It simplifies to (x−2 z)0 = −2x. COnsequently, x−2 z = −x2 + C and z = −x4 + Cx2 . This means that y −2 = Cx2 − x4 so y 2 = Cx21−x4 . (b) Write the equation in the form y 0 + (1/x)y = y −2 cos x to see that it is Bernoulli, n = −2. Substitute z = y 3 , z 0 = 3y 2 y 0 into the equation y 2 y 0 + (1/x)y 3 = cos x to obtain the linear equation (1/3)z 0 + (1/x)z = cos x. This is z 0 + (3/x)z = 3 cos x, with integrating factor x3 . It simplifies to (x3 z)0 = 3x3 cos x. Consequently, x3 z =

2.2. FIRST-ORDER LINEAR EQUATIONS

3F (x) + C where F (x) is and antiderivative for x3 cos x. Therefore, z = 3x−3 F (x) + x−3 C, and the solution in terms of y can be expressed in the form y 3 x3 = 3F (x) + C. The expression F (x) can be found using multiple integration by parts or, better yet, a table of integrals: F (x) = x3 sin x + 3x2 cos x − 6x sin x − 6 cos x. dy implies that x is the independent variable 4. The usual Leibniz notation dx and y is the dependent variable. In solving a differential equation, it is sometimes useful to reverse the roles of the two variables. Treat each of the following equations by reversing the roles of y and x: dy 2 = y 2 as (ey − 2xy) = dx (a) Rewrite the Leibnitz form (ey − 2xy) dx dy y . If y 2 0 2 y 0 y is the dependent variable, then x y + 2xy = e and x + y x = ye 2 . The integrating factor is y 2 and the equation simplifies to (xy 2 )0 = ey . y Integrate to obtain xy 2 = ey + C, and therefore x = ye 2 + yC2 .

(b) y = y 0 (x + y 2 ey ) yields x0 y = (x + y 2 ey ) and x0 − xy = yey . The integrating factor is y1 . Assume y > 0 for simplicity. Then the equation simplifies to (x y1 )0 = ey . Integrate it to get xy = ey + C, and therefore x = yey + Cy. (c) (x3 y −x3 −x)y 0 = 2 yields x3 y −x3 −x = 2x0 . This is not a first-order linear equation. 0

(y) 3 (d) x0 + 3ff (y) x = ff2(y) (y) . The integrating factor is f (y) if we assume that f (y) > 0 for simplicity. The equation simplifies to [xf 3 (y)]0 = f (y)f 0 (y). Integrate it to obtain xf 3 (y) = 21 f 2 (y) + C and therefore x = 2f1(y) + f 3C(y) .

5. If y = cf (x) + g(x), then g 0 = cf 0 (x) + g 0 (x). Estimate the constant c (multiply the first equation by f 0 (x), the second by f (x), and subtract) to obtain f 0 (x)y − f (x)y 0 = f (x)g(x) − f 0 (x)g 0 (x). This is a first-order linear equation. 0

6. Rewrite the differential equation as yy + p = Q ln y. Plug in z = ln y and z 0 = y1 y 0 to obtain z 0 + P = Qz, which is a first-order linear equation. 0

Rewrite xy 0 = 2x2 y + y ln y as xyy = 2x2 + ln y. Plug in z = ln y and 0

z 0 = yy to obtain xz 0 = 2x2 + z. Write the equation in the form of first1 order linear equation z 0 − xz = 2x. The integrating factor is |x| = x1 if we assume x > 0 for simplicity. The equation simplifies to (z x1 )0 = 2. Integrate it to obtain xz = 2x + C, and z = 2x2 + Cx. Plugging in z = ln y 2 yields ln y = 2x2 + Cx. Therefore, y = e2x +Cx . cos x 7. The equation is linear. Write it in the form y 0 − (2 csc 2x)y = 2sin 2x 2 cos x to find the integrating factor: csc 2x + cot 2x = cot x. Since sin 2x = 2 cos x 2 sin x cos x = csc x, upon multiplying by the integrating factor we obtain

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS x+C (y cot x)0 = csc x cot x. Therefore, y cot x = − csc x + C and y = − csc . cot x π C sin x−1 . As x → the Converting to sines and cosines yields y = cos x 2 denominator approaches 0. To get a finite limit the numerator must also x−1 approach 0 so let C = 1. The solution y = sin cos x has the desired property.

8. The rate of change, x0 (t), equals the rate in: 3 lbs/min, minus the rate out: 4 · Vx(t) (t) lbs/min, where V (t) = 10 − t is the volume of the mixture in 4x the tank at time t. The resulting rate equation is x0 = 3 − 10−t . Rewrite 4 0 it as x + 10−t x = 3 to obtain the integrating factor, (10 − t)−4 . The equation simplifies to [(10 − t)−4 x]0 = 3(10 − t)−4 . Integrate it to obtain (10 − t)−4 x = (10 − t)−3 + C. Consequently, x(t) = (10 − t) + C(10 − t)4 . The initial condition x(0) = 2 determines that C = − 1084 . Therefore, the amount of salt in the tank at time t is given by x(t) = (10 − t) −

8 (10 − t)4 . 104

9. Let x(t) be the amount of salt (pounds) in the tank at time t (minutes). The rate of change, x0 (t), equals the rate in: 6 lbs/min, minus the rate out: 3 · Vx(t) (t) lbs/min, where V (t) = 40 − t is the volume of the mixture in x the tank at time t. The resulting rate equation: x0 = 6 − 3 · 40−t , is linear. 3 0 Write it as x + 40−t x = 6 to obtain the integrating factor, (40 − t)−3 . The equation simplifies to [(40 − t)−3 x]0 = 6(40 − t)−3 implying that (40 − t)−3 x = 3(40 − t)−2 + C. Consequently, x(t) = 3(40 − t) + C(40 − t)3 . The initial condition, x(0) = 0, determines the value of the constant: 3 C = − 1600 . Therefore, the amount of salt in the tank at time t is given by 3 x(t) = 3(40 − t) − (40 − t)3 . 1600 (a) Since V (t) = 40 − t there are 20 gallons of brine in the tank when t = 20. According to the formula derived above, x(20) = 45 pounds of salt. (b) The amount of salt in the tank is maximum when x0 (t) = 0. Solving −3 +

9 (40 − t)2 = 0 1600

40 √ we first obtain (40 − t)2 = 1600 3 , then t = 40 − 3 = 16.905 . . ..

2.3

Exact Equations

All references to M and N refer to the equation in the form M dx + N dy = 0. ∂f ∂x = y ∂f 2 0 = x+ implies that x+φ (y) = x+ y2 ∂y y

∂N 1. M = y, N = x + y2 , so ∂M ∂y = 1 = ∂x . The equation is exact.

implies that f (x, y) = xy +φ(y).

2.3. EXACT EQUATIONS

so φ(y) = 2 ln |y| and f (x, y) = xy + 2 ln |y|. The implicit solution is xy + 2 ln |y| = C. 2 2 2. M = sin x tan y+1, N = cos x sec2 y, so ∂M ∂y = sin x sec y 6= − sin x sec y = ∂N ∂x . The equation is not exact. ∂N 3. M = y − x3 , N = x + y 3 , so ∂M ∂y = 1 = ∂x . The equation is exact. ∂f ∂f 3 4 3 ∂x = y − x implies that f (x, y) = xy − x /4 + φ(y). ∂y = x + y implies 0 3 4 4 4

that x + φ (y) = x + y so φ(y) = y /4 and f (x, y) = xy − x /4 + y /4. The implicit solution is xy − x4 /4 + y 4 /4 = C. ∂N 4. M = 2y 2 − 4x + 5, N = −4 + 2y − 4xy, so ∂M ∂y = 4y 6= −4y = ∂x . The equation is not exact. ∂N 5. M = y + y cos xy, N = x + x cos xy, so ∂M ∂y = 1 − xy sin xy + cos xy = ∂x .

The equation is exact. ∂f ∂x = y+y cos xy implies that f (x, y) = xy+sin xy+ φ(y). ∂f = x + x cos xy implies that x + x cos xy + φ0 (y) = x + x cos xy, so ∂y φ(y) = 0 and f (x, y) = xy+sin xy. The implicit solution is xy+sin xy = C. For a given value of C, the equation t + sin t = C has exactly one solution, t0 . Therefore, the implicit solution for the differential equation also be expressed simply as xy = C. 6. M = cos x cos2 y, N = 2 sin x sin y cos y, so ∂M ∂y = −2 cos x cos y sin y 6= ∂N 2 cos x sin y cos y = ∂x . The equation is not exact. ∂N y 7. M = ey + cos x cos y, N = xey − sin x sin y, so ∂M ∂y = e − cos x sin y = ∂x . y y The equation is exact. ∂f ∂x = e + cos x cos y implies that f (x, y) = xe + ∂f y y sin x cos y + φ(y). ∂y = xe − sin x sin y implies that xe − sin x sin y + φ0 (y) = xey − sin x sin y so φ(y) = 0 and f (x, y) = xey + sin x cos y. The implicit solution is xey + sin x cos y = C. 1 x 1 x x 1 x 8. M = − y1 sin xy , N = yx2 sin xy , so ∂M ∂y = y 2 sin y − y cos y (− y 2 ) 6= y 2 sin y + x x x y 2 cos y (− y 2 ). The equation is not exact. ∂N 9. M = 1 + y, N = 1 − x, so ∂M ∂y = 1 6= −1 = ∂x . The equation is not exact. ∂N 2 10. M = 2xy 3 + y cos x, N = 3x2 y 2 + sin x, so ∂M ∂y = 6xy + cos x = ∂x . 3 2 3 The equation is exact. ∂f ∂x = 2xy + y cos x implies that f (x, y) = x y + 2 2 2 2 0 y sin x + φ(y). ∂f ∂y = 3x y + sin x implies that 3x y + sin x + φ (y) = 2 2 2 3 3x y +sin x so φ(y) = 0 and f (x, y) = x y +y sin x. The implicit solution is x2 y 3 + y sin x = C. 2 2

1+x y ∂N 11. M = 1−xy2 y2 − 1, N = 1−xx2 y2 so ∂M ∂y = (1−x2 y 2 )2 = ∂x . The equation is y exact. ∂f ∂x = 1−x2 y 2 − 1 implies that

f (x, y) =

1 1 ln(xy + 1) − ln(xy − 1) − x + φ(y). 2 2

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS ∂f x 1 x x x 0 ∂y = 1−x2 y 2 implies that 2 ( xy+1 − xy−1 )+φ (y) = 1−x2 y 2 . This simplifies to 1−xx2 y2 + φ0 (y) = 1−xx2 y2 , so φ(y) = 0 and f (x, y) = 12 ln(xy + 1) − 1 1 1 2 ln(xy −1)−x. The implicit solution is 2 ln(xy +1)− 2 ln(xy −1)−x = C.

This equation can be solved for y (exercise) to produce an explicit solution Ce2x +1 y = x(Ce 2x −1) . ∂N 3 12. M = 2xy 4 + sin y, N = 4x2 y 3 + x cos y, so ∂M ∂y = 8xy + cos y = ∂x . 4 2 4 The equation is exact. ∂f ∂x = 2xy + sin y implies that f (x, y) = x y + 2 3 2 3 0 x sin y + φ(y). ∂f ∂y = 4x y + x cos y implies that 4x y + x cos y + φ (y) = 2 3 2 4 4x y + x cos y so φ(y) = 0 and f (x, y) = x y + x sin y. The implicit solution is x2 y 4 + x sin y = C. 2 2

1+x y ∂N 13. M = 1−xy2 y2 + x, N = 1−xx2 y2 , so ∂M ∂y = (1−x2 y 2 )2 = ∂x . The equation is y exact. ∂f ∂x = 1−x2 y 2 + x implies that

f (x, y) =

1 x2 1 ln(xy + 1) − ln(xy − 1) + + φ(y). 2 2 2

∂f x 1 x x x 0 ∂y = 1−x2 y 2 implies that 2 ( xy+1 − xy−1 + φ (y) = 1−x2 y 2 . This simplifies to 1−xx2 y2 + φ0 (y) = 1−xx2 y2 , so φ(y) = 0 and f (x, y) = 12 ln(xy + 1) − 1 x2 1 1 x2 2 ln(xy −1)+ 2 . The implicit solution is 2 ln(xy +1)− 2 ln(xy −1)+ 2 =

C. This equation can be solved for y (exercise) to produce an explicit −x2

Ce +1 . solution y = x(Ce −x2 −1)

p p √−x x2 − y), N = − x2 − y, so ∂M = ∂N ∂y = ∂x . The x2 −y p equation is exact. ∂f x2 − y) implies that f (x, y) = x2 + ∂x = 2x(1 + p 3 2 2 − y) 2 + φ(y) (substitution u = x2 − y). ∂f = − x2 − y implies that 3 (x ∂y p p 3 − x2 − y +φ0 (y) = − x2 − y so φ(y) = 0 and f (x, y) = x2 + 32 (x2 −y) 2 . 3 The implicit solution is x2 + 32 (x2 − y) 2 = C.

14. M = 2x(1 +

y x ∂N 15. M = x ln y + xy, N = y ln x + xy, so ∂M ∂y = y + x 6= x + y = ∂x . The equation is not exact. 2

y 16. M = ey − csc y csc2 x, N = 2xyey − csc y cot y cot x, so ∂M + ∂y = 2ye 2

∂f y csc y cot y csc2 x = ∂N − csc y csc2 x im∂x . The equation is exact. ∂x = e 2 2 y plies that f (x, y) = xey +csc y cot x+φ(y). ∂f ∂y = 2xye −csc y cot y cot x 2

implies that 2xyey − csc y cot y cot x + φ0 (y) = 2xyey − csc y cot y cot x 2 so φ(y) = 0 and f (x, y) = xey + csc y cot x. The implicit solution is 2 xey + csc y cot x = C. 17. M = 1 + y 2 sin 2x, N = −2y cos2 x, so ∂M ∂y = 2y sin 2x = 4y cos x sin x = ∂N ∂x .

The equation is exact.

∂f ∂x

= 1 + y 2 sin 2x implies that f (x, y) =

2.3. EXACT EQUATIONS

2 0 x − 12 y 2 cos 2x + φ(y). ∂f ∂y = −2y cos x implies that −y cos 2x + φ (y) = −2y cos2 x. Since cos2 x = 21 + 12 cos 2x, the last equation is equivalent to −y cos 2x + φ0 (y) = −y − y cos 2x so φ0 (y) = −y, φ(y) = − 12 y 2 , and f (x, y) = x − 21 y 2 cos 2x − 21 y 2 . Therefore, the implicit function is x − 1 2 1 2 2 y cos 2x − 2 y = C. 3

2 2 −2 18. M = x(x2 + y 2 )− 2 , N = y(x2 + y 2 )− 2 , so ∂M = ∂N ∂y = −3xy(x + y ) ∂x . − 23

2 2 The equation is exact. ∂f implies that f (x, y) = −(x2 + ∂x = x(x + y ) 3 ∂f 2 − 12 2 2 − 23 0 y ) +φ(y). ∂y = y(x +y ) +φ (y) implies that y(x2 +y 2 )− 2 +φ0 (y) = 3

y(x2 +y 2 )− 2 so φ(y) = 0 and f (x, y) = −(x2 +y 2 )− 2 . The implicit solution 1 is −(x2 + y 2 )− 2 = C. 3

3x ∂N 19. M = 3x2 (1+ln y), N = xy −2y so ∂M ∂y = y = ∂x . The equation is exact. ∂f ∂f x3 2 3 ∂x = 3x (1 + ln y) implies that f (x, y) = x (1 + ln y) + φ(y). ∂y = y − 2y 3 3 implies that xy + φ0 (y) = xy − 2y, so φ0 (y) = −2y, φ(y) = −y 2 , and 3 2 3 2

f (x, y) = x (1 + ln y) − y . The implicit solution is x (1 + ln y) − y = C.

y x 20. Rewrite the equation as ( (x+y) M = 2 − 1)dx + (1 − (x+y)2 )dy = 0. y (x+y)2 − 1, N

x −y x ∂M ∂N = 1 − (x+y) 2 , so ∂y = (x+y)4 = ∂x . The equation is

exact. y y ∂f (a) ∂f ∂x = (x+y)2 − 1 implies that f (x, y) = − x+y − x + φ(y). ∂y = x x x 0 0 1 − (x+y) 2 implies that 1 − (x+y)2 = phi (y) − (x+y)2 so φ (y) = 1, y φ(y) = y and f (x, y) = − x+y − x + y. The implicit solution is y − x+y − x + y = C. ∂f y x x (b) ∂f ∂y = 1− (x+y)2 implies that f (x, y) = y+ x+y +ϕ(x). ∂x = (x+y)2 −1 y y 0 0 implies that (x+y) 2 + ϕ (x) = (x+y)2 − 1 so ϕ (x) = −1, ϕ(x) = −x x x and f (x, y) = y + x+y − x. The implicit solution is y + x+y − x = D. y Subtracting the left side of the solutions from (a) and (b) yields (− x+y − x x + y) − (y + x+y − x) = −1. Differing by a constant −1, two equations from (a) and (b) represent the same family of equations. 2

4y −2x 8y −x 8xy ∂M ∂N 21. M = 4xy 2 −x3 , N = 4y 3 −x2 y , so ∂y = (x2 −4y 2 )2 = ∂x . The equation is

exact.

∂f ∂x

4y −2x 1 2 2 = 4xy 2 −x3 implies that f (x, y) = ln x + 2 ln(x − 4y ) + φ(y)

(integrate via partial fractions). 2

∂f ∂y

8y −x 4y = 4y 3 −x2 y implies that x2 −4y 2 +

8y −x 1 0 φ0 (y) = 4y Therefore, φ(y) = ln y, and f (x, y) = 3 −x2 y , so φ (y) = y . 1 2 2 ln x + 2 ln(x − 4y ) + ln y. The solution is ln xy + 12 ln(x2 − 4y 2 ) = C or x2 y 2 (x2 − 4y 2 ) = C.

22. For each of the following equations, find the value of n for which the equation is exact. Then solve the equation for that value of n.

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS

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2 ∂N 2 (a) ∂M ∂y = 2xy + nx , ∂x = 3x + 2xy.

∂M ∂N ∂y = ∂x implies that n = 3. ∂f ∂f 1 2 2 2 y 3 2 ∂x = xy +3x implies that f (x, y) = 2 x y +x y+φ(y). ∂y = x y+ x3 +φ0 (y) = x2 y+x3 implies that φ(y) = 0 and f (x, y) = 12 x2 y 2 +x3 y. The implicit solution is 21 x2 y 2 + x3 y = C.

∂N 2xy 2xy (b) ∂M + 2xye2xy , ∂N + nxye2xy . ∂M ∂y = e ∂x = ne ∂y = ∂x implies that 2xy n = 1. ∂f implies that f (x, y) = 21 x2 + 12 e2xy + φ(y). ∂x = x + ye ∂f Then ∂y = xe2xy +φ0 (y) = xe2xy implies that φ(y) = 0 and f (x, y) = 1 2 1 2xy . The implicit solution is 12 x2 + 12 e2xy = C. 2x + 2e

2.4

Orthogonal Trajectories and Families of Curves

1. Sketch the families of curves, find orthogonal trajectories, add them to the sketch. (a) Differentiate xy = c to get xy 0 + y = 0 or y 0 = −y/x. Orthogonal curves are generated by y 0 = x/y, a separable equation. ydy = xdx implies that y 2 /2 = x2 /2 + C or y 2 − x2 = C. See Figure 2.1. (b) Differentiate y = cx2 to get y 0 = 2cx. Divide one by the other, y 0 /y = x ,a 2/x or y 0 = 2y/x. Orthogonal curves are generated by y 0 = − 2y 2 2 separable equation. 2ydy = −xdx implies that y = −x /2 + C or 2y 2 + x2 = C. See Figure 2.2. (c) Differentiate x + y = c to get 1 + y 0 = 0 or y 0 = −1. Orthogonal curves are generated by y 0 = 1 or y = x + C. See Figure 2.3. (d) The curves r = c(1 + cos θ) for c positive are orthogonal to the curves r = c(1 + cos θ) for c negative. See Figure 2.4 where the orthogonal curves on the right correspond to c = −2, −4, −8, −16. To verify this analytically, write the whole family (c positive and negative) in

2.4. ORTHOGONAL TRAJECTORIES AND FAMILIES OF CURVES

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Figure 2.1: The family xy = c and orthogonal trajectories 1.6

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Figure 2.2: The family y = cx2 and orthogonal trajectories

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS

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Figure 2.3: The family x + y = c and orthogonal trajectories

2.4. ORTHOGONAL TRAJECTORIES AND FAMILIES OF CURVES

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Cartesian coordinates: ±

x2 + y 2 = c(1 + √ x2 ±

x +y 2

), and differen-

tiate. Then use the original equation to eliminate c to obtain the following differential equations: x + yy 0 =

y 2 − xyy 0 p . x ± x2 + y 2

The differential equations for the orthogonal trajectories can now be obtained by replacing y 0 with − y10 . Having done this, multiply both sides of the equation by yy 0 , then ”rationalize” p the denominator on the right (multiply top and bottom by x ∓ x2 + y 2 ) and rearrange to y 2 − xyy 0 p = x + yy 0 . x ∓ x2 + y 2 Because these are the same equations, the family is “self-orthogonal”. (e) Differentiate y = cex to get y 0 = cex . Divide one by the other to obtain y 0 /y = 1 or y 0 = y. Orthogonal curves are generated by the separable equation y 0 = −1/y or ydy = −dx. Integrate to obtain y 2 /2 = −x + C. This simplifies to y 2 + 2x = C. See Figure 2.5. (f) Differentiate x−y 2 = c to get 1−2yy 0 = 0 or y 0 = 1/(2y). Orthogonal curves are generated by the separable equation y 0 = −2y or dy/y = −2dx. INtegrate to obtain ln |y| = −2x + C. This simplifies to y = Ce−2x . See Figure 2.6. 2. Differentiate y = cx4 to get y 0 = 4cx3 . Substitute c = xy4 into the equation x 0 to get y 0 = 4y x . Orthogonal curves are generated by y = − 4y . Solve this separable equation to get 2y 2 = − 21 x2 + C or x2 + 4y 2 = C. Differentiate y = cxn to get y 0 = ncxn−1 . Substitute c = xyn into the x 0 equation to get y 0 = ny x Orthogonal curves are generated by y = − ny . 1 Solve this separable equation to get 2 ny 2 = −x2 + C or 2x2 + ny 2 = C.

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS

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Figure 2.4: The family r = c(1 + cos θ), c = 2, 4, 8, 16(left) and orthogonal trajectories 3

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Figure 2.5: The family y = cex and orthogonal trajectories

2.4. ORTHOGONAL TRAJECTORIES AND FAMILIES OF CURVES

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Figure 2.6: The family y = cx4 and orthogonal trajectories

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CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS

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Figure 2.7: The parabolas y 2 = 4c(x + c) 5

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Figure 2.8: The family x − y 2 = c and orthogonal trajectories 3. The sketch is displayed in Figure 2.8. It appears that the curves intersect orthogonally. Differentiate y 2 = 4c(x + c) to get 2yy 0 = 4c. Substitute c = yy 0 /2 into the first equation to obtain y 2 = 2yy 0 (x + yy 0 /2) or y = 2xy 0 + y(y 0 )2 . This is the differential equation defining the parabolas. Replacing y 0 with −1/y 0 yields the differential equation defining the orthogonal trajectories: 1 0 2 0 y = − 2x y 0 + y · (y 0 )2 . It simplifies to y(y ) = −2xy + y which is equivalent to the original, confirming the fact that the parabolas in Figure 2.8 are orthogonal to one another. 4. In each of parts (a) through (f), find the family of curves that satisfy the given geometric condition: (a) Let (x, y) be a point on the curve. The line going through (x, y) with slope y 0 goes through (0, 2y) implies that 2y − y = y 0 (0 − x) so 1 y 0 = − −y x . Solve this separable equation to get ln |y| = ln |x| + C D and y = ± x .

2.5. HOMOGENEOUS EQUATIONS

(b) Let (x, y) be a point on the curve. Assume (X, 0) is the intersection point of the line going through (x, y) with slope − y10 and the x-axis. Then 0 − y = − y10 (X − x) so X = yy 0 + x. The distance of (X, 0) and (x, 0) is 1 implies that |X − x| = |yy 0 | = 1. Thus, yy 0 = ±1. Solve this separable equation to get 21 y 2 = ±x + C. (c) Let (x, y) be a point on the curve. Assume (X, 0) is the intersection point of the line going through (x, y) with slope y 0 and the x-axis. Then 0 − y = y 0 (X − x) so X = − yy0 + x. The distance of (X, 0) and (x, 0) is 1 implies that | − yy0 | = 1 and y 0 = ±y. Solve this separable equation to get ln |y| = ±x + C and y = ±e±x+C . (d) Let (x, y) be a point on the curve. The line going through (x, y) with slope y 0 goes through (0, y2 ) implies that y2 − y = y 0 (0 − x) and 0 2 x = 2yy . Solve this separable equation to get y = ±ex +C . (e) Let (x, y) be a point on the curve. The line going through (x, y) with slope − y10 goes through (0, y2 ) implies that y2 − y = − y10 (0 − x) and 2

−yy 0 = 2x. Solve this separable equation to get − y= x2 + C. (f) Let (x, y) be a point on the curve. The line going through (x, y) with slope − y10 goes through (2x, 0) implies that 0 − y = − y10 (2x − x) and yy 0 = x. Solve this separable equation to get y 2 = x2 + C. Rx 5. Let (x, y) be a point on the curve. The area from 0 to x is 0 y(t)dt. Rx The area of the rectangle is x · y(x) so 0 y(t)dt = x·y(x) 3 . Differentiate 0

with respect to x to obtain y(x) = xy (x)+y(x) or 3y = xy 0 + y. This is a 3 2dx separable equation dy y = x . Integrate to obtain ln y = 2 ln x + C. The equation of the curve is y = Cx2 . 6. Find the differential equation of each of the following one-parameter families of curves: ** 7. Orthogonal trajectories using symbol manipulation software. Neither Maple nor Mathematica can obtain symbolic solutions to the differential equation that defines the orthogonal trajectories in these problems. The following example–problem (a)–shows how to display numerically generated trajectories using Maple. The code for Mathematica is quite similar. (a) The following Maple code has, as its last output, the picture displayed in Figure **.

2.5

Homogeneous Equations

The letters M and N will always refer to the equation M dx + N dy = 0. Note that this equation is equivalent to y 0 = − M N.

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS 1. Verify the equation is homogeneous, solve. (a) M = x2 − 2y 2 and N = xy are homogeneous of degree 2; y 0 = 2 2 y 0 0 − x −2y xy . The substitutions z = x and y = xz + z yield the sep1 0 arable equation xz + z = 2z − z . This equation has solutions z = √ √ ± 1 + Cx2 so the original equation has solutions y = ±x 1 + Cx2 . (b) M = 3xy + 2y 2 and N = −x. M has degree 2 and N has degree 1. This equation is not homogeneous. Note that it is Bernoulli. (c) Divide the equation by x2 to obtain y 0 = 3(1 + ( xy )2 arctan xy + xy , a homogeneous equation. The substitutions z = xy and y 0 = xz 0 + z yield the separable equation xz 0 + z = 3(1 + z 2 ) arctan z + z. This equation has the solution z = tan(Cx3 ) so the original equation has solution y = x tan(Cx3 ). (d) Divide the equation by x to make it homogeneous. The substitutions z = xy and y 0 = xz 0 + z yield the separable equation (xz 0 + z) sinz z = sin z + z1 . This equation has the solution cos z + ln cx = 0 so the original equation has solution cos xy + ln cx = 0. (e) Divide by x to make the equation homogeneous. The substitutions z = xy and y 0 = xz 0 +z yield the separable equation xz 0 +z = z+2e−z . This equation has the solution z = ln(2 ln x + C) so the original equation has solution y = x ln(2 ln x + C). (f) M = x−y and N = −(x+y) are homogeneous of degree 1; y 0 = x−y x+y . The substitutions z = xy and y 0 = xz 0 +z yield the separable equation 2 −2 xz 0 + z = 1−z 1+z . This equation has the solution z + 2z − 1 = Cx 2 2 and the original equation has solution y + 2xy − x = C. (g) This equation is linear and homogeneous. To solve it as a homogey neous equation, divide by x: y 0 = 2 − 6y x , then substitute z = x 0 to obtain the separable equation xz + z = 2 − 6z. The solution is −6 z = 72 +Cx−7 so the original equation has the solution y = 2x . 7 +Cx q 2 (h) Divide by x: y 0 = 1 + xy 2 , then substitute z = xy to obtain the √ √ 2 0 2 2 separable √ equation xz +z = 1 + z . The solution is z +z 1 + z + 2 ln(z+ 1 + z ) = 2 lnp x+C. The solution p to the original equation can be simplified to y 2 +y x2 + y 2 +x2 ln(y+ x2 + y 2 ) = x2 (3 ln x+C). (i) The equation is Bernoulli and homogeneous. To solve it as homoge2 y neous, divide by x2 : y 0 = xy 2 + 2y x , and substitute z = x to obtain x the separable equation xz 0 + z = z 2 + 2z. The solution is z = C−x . 2

x The solution to the original equation is y = C−x . 3

+y (j) M = x3 +y 3 and N = −xy 2 are homogeneous of degree 3; y 0 = x xy 2 . 3

The substitution z = xy yields the separable equation xz 0 + z = 1+z z2 . √ This equation has the solution√z = 3 3 ln x + C and the original equation has the solution y = x 3 3 ln x + C.

2.5. HOMOGENEOUS EQUATIONS

2. The family of all circles tangent to the y-axis is represented by x2 − 2cx + y 2 = 0. Differentiate it to get 2x−2c+2yy 0 = 0. These two equations yield y 2 2x −x2 y 0 = y 2xy . Orthogonal curves are generated by y 0 = − y22xy −x2 = − ( y )2 −1 . 2

−1 dz = dx Set z = xy . Then y 0 = z 0 x + z = − z22z−1 and − z(zz2 −z+2) x . Separate the left to partial fractions so

[

− 3 (2z − 1) − 17 1 dx √ + 24 + √ ]dz = . 7 7 2 2z z −z+2 x ( z− ) +1 2

Integrate it to get √ √ 7 7 1 3 2 2 ln |z| − ln |z − z + 2| − √ arctan( z− ) = ln |x| + C. 2 4 2 4 7 7 Plugging in z = xy yields √ √ 7y 7 1 y 3 y 2 y 2 ln | | − ln |( ) − + 2| − √ arctan( − ) = ln |x| + C. 2 x 4 x x 2 x 4 7 7 ax+by+c dy = F ( dx+ey+f ) 3. Solving dx

(a) The substitutions x = z − h and y = w − k yield the equation az+bw−(ah+bk−c) dy dy dw dz dw dw dz = F ( dz+ew−(dh+ek−f ) ). Note that dx = dw · dz · dx = dz . Since ae 6= bd there are unique numbers h and k such that ah + bk = c and az+bw dh + ek = f . Using these values the new equation is dw dz = F ( dz+ew ) which is homogeneous. (b) If ae = bd, then there is a constant k such that cx + dy = k(ax + by) for all x and y. We are assuming that a and b are not both 0. The ax+by+c dy = F ( k(ax+by)+f ). The substitution equation then has the form dx z+c 0 z = ax + by reduces it to z = a + bF ( kz+f ) which is separable. 4. Solve each of the following differential equations. z+w−h−k+4 (a) Substituting x = z − h, y = w − k yields dw dz = z−w−h+k−6 . h + k = 4 and k − h = 6 have a solution h = −1 and k = 5. When x = z + 1 and y = w − 5, the differential equation is reduced to a homogeneous 1+ w z+w 1+t z . Set t = wz then t0 z + t = 1−t equation dw and dz = z−w = 1− w z R 1 R t R dz 1−t dz dt = . Integrate it to get dt − dt = 1+t2 z 1+t2 1+t2 z and y+5 1 1 2 arctan t − 2 ln(1 + t ) = ln |z| + C. Therefore, arctan x−1 − 2 ln[1 + (y+5)2 ((x−1)2 ] = ln |x − 1| + C. z+4 (b) Substituting z = x + y yields z 0 = 1 + y 0 = 1 + z−6 = 2z−2 and R −5 Rz−6 z−6 1 dz = dx. Integrate it to get ∈ dz + dz = dx and 2z−2 2 2z−2 1 5 1 5 z − ln |z − 1| = ln |x| + C. Therefore, (x + y) − ln |x + y − 1| = 2 2 2 2 ln |x| + C.

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS 2z−2w−2h+2k (c) Substituting x = z − h, y = w − k yields dw . dz = − w−k−1 −2h + 2k = 0 and k + 1 = 0 have a solution h = 1 and k = −1. When x = z − 1 and y = w + 1, the differential equation is reduced 2 w −2 2w−2z = zw . Set t = wz then to a homogeneous equation dw dz = w z

tdt t0 z + t = 2t−2 and −t2 +2t−2 = dz the left to partial t z . Separate R t−1 R R dz 1 fractions and integrate to get − t2 −2t+2 dt− 1+(t−1) 2 dt = z and y−1 2 − 21 ln(t2 −2t+2)−arctan(t−1) = ln |z|+C. Therefore, − 21 ln(( x+1 ) − y−1 y−1 2 x+1 + 2) − arctan( x+1 − 1) = ln |x + 1| + C. z+w−h−k−1 (d) Substituting x = z − h, y = w − k yields dw dz = z+4w−h−4k+2 . h + k = −1 and h+4k = 2 have a solution h = −2 and k = 1. When x = z +2 and y = w − 1, the differential equation is reduced to a homogeneous 1+ w z+w t w 0 z equation dw dz = z+4w = 1+4 w . Set t = z then t z + t = 1+4t and z

1+4t dz 1−4t2 dt = z .

Separate the left to partial fractions and integrate R 32 R R dz 3 1 1 1 to get 1−2t dt + −1/2 1+2t dt = z and − 4 ln |t − 2 | − 4 ln |t + 2 | = y+1 y+1 1 1 1 3 ln |z| + C. Therefore, − 4 ln | x−2 − 2 | − 4 ln | x−2 + 2 | = ln |x − 2| + C. 2z+3w−2h−3k−1 (e) Substituting x = z − h, y = w − k yields dw . dz = 4z−4h+4 2h + 3k = −1 and h = 1 have a solution k = −1 and h = 1. When x = z − 1 and y = w + 1, the differential equation is reduced to 2+ 3w 2z+3w a homogeneous equation dw = 4 z . Set t = wz then dz = 4z 4dt w0 = t0 z + t = 2+3t and 2−t = dz 4 z . Integrate it to get −4 ln |t − 2| = y−1 ln |z| + C. Therefore, −4 ln | x+1 − 2| = ln |x + 1| + C.

5. Observe that if z = xyn then y 0 = xn z 0 + nxn−1 z. −(2n+1)

−z (2n+1) . Setting n = (a) The substitution z = xyn yields z 0 = x 2xz 1 1 0 − 2 this becomes z = 2xz , a separable equation with solution z 2 = ln x + C. Therefore, y 2 = ln x+C . x

(b) The substitution z = xyn yields z 0 = 2x

−(2n+1)

+z 2 (3−4n) . 4xz

Setting

−5/2 n = 34 this becomes z 0 = 2x4xz , a separable equation with solution z 2 = − 25 x−5/2 + C. Therefore, y 2 = Cx3/2 − 2x−1 /5. (The solution

can be also be obtained using n = −1/2.)

n+1

z(n+1) (c) The substitution z = xyn yields z 0 = −z (n−1)x+x . Setting x(x+xn+2 z) 2z 0 n = −1 this becomes z = x(1+z) , a separable equation with solution z = ln z = 2 ln x + C. Therefore, xy + ln y = ln x + C.

6. Every homogeneous equation can be rewritten as y 0 = F ( xy ). Given a straight line through the origin, all points on it share the same xy value. Thus, all tangent lines of the integral curves at intersection points of this straight line and the integral curves share the same slope (y 0 ). Therefore, The angles between these tangent lines and the straight line are the same. x (One exception is when x = 0 and y 0 = ∞. Consider dx dy = G( y ) instead.)

2.6. INTEGRATING FACTORS

7. Solving homogeneous equations using symbol manipulation software. Neither Maple nor Mathematica can obtain complete symbolic solutions to these differential equations. The following example shows Maple’s answer to problem (a). Mathematica’s answer is quite similar. (a) The following Maple code has, as its last output, a partial solution to y 0 = sin xy − cos xy .

2.6

Integrating Factors

The letters M and N will always refer to the equation M dx + N dy = 0. 1. Solve by finding an integrating factor. ∂M

− ∂N

(a) M = −2xy and N = 3x2 − y 2 ; − ∂y M ∂x = − y4 . An integrating factor is µ(y) = y14 , and the solution is x2 − y 2 = Cy 3 . ∂M

− ∂N

(b) M = xy − 1 and N = x2 − xy; ∂y N ∂x = x1 . An integrating factor is µ(x) = x1 , and the solution is 2xy − 2 ln x − y 2 = C. ∂M

− ∂N

∂M

3 4

− ∂N

2+9x y ∂y ∂x (c) M = y and N = −x−3x3 y 4 ; ∂y N ∂x = − x(1+3x = 3 y 4 ) , and − M 3 4

y − 2+9x . Consequently, there are no integrating factors that are y function of x alone or functions of y alone. Try a factor of the form 1 µ = (xy) This is motivated by the observation that the equan. 3 4

3 4

d(xy) y dy 3x y dy tion becomes xdy+ydx = − 3x(xy) or (xy) The left n n = − (xy)n . (xy)n side is directly integrable for any n and, when n = 3, so is the −1/2 3 2 right side: d(xy) (xy)3 = −3ydy. The solution is (xy)2 = − 2 y + C or −2 2 (xy) = 3y + C. ∂M

− ∂N

(d) M = ex and N = ex cot y + 2y csc y; − ∂y M ∂x = cot y. An integrating factor is µ(y) = sin y, and the solution is y 2 + ex sin y = C. ∂M

− ∂N

(e) M = (x + 2) sin y and N = x cos y; ∂y N ∂x = 1 + x1 . An integrating factor is µ(x) = xex , and the solution is x2 ex sin y = C. (f) This equation is similar to the one in part (c). Write it in the form xdy + ydx = 2x2 y 3 dy and divide by (xy)2 to get d(xy) (xy)2 = 2ydy. The 1 2 solution is xy = −y + C. ∂M

− ∂N

∂M

− ∂N

(g) M = x + 3y 2 and N = 2xy; ∂y N ∂x = x2 . An integrating factor is µ(x) = x2 , and the solution is x4 + 4x3 y 2 = C. (h) M = y and N = 2x − yey ; − ∂y M ∂x = y1 . An integrating factor is µ(y) = y and the solution is xy 2 − ey (y 2 − 2y + 2) = C.

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS ∂M

− ∂N

(i) M = y ln y − 2xy and N = x + y; − ∂y M ∂x = − y1 . An integrating factor is µ(y) = y1 and the solution is y − x2 + x ln y = C. (j) M = y 2 + xy + 1 and N = x2 + xy + 1;

∂M ∂N ∂y − ∂x

y−x = x2 +xy+1 , and

∂M ∂N ∂y − ∂x

x−y − M = y2 +xy+1 . Consequently, there are no integrating factors that are functions of x alone or functions of y alone. Let’s see if we can take advantage of the symmetry of the equation. Start by expressing it in the form dx + dy + y(y + x)dx + x(x + y)dy = 0, then divide by x + y: dx+dy x+y + ydx + xdy = 0. This is equivalent

to d(x+y) x+y + d(xy) = 0 and the solution, obtained by integrating is ln(x + y) + xy = C. Note. An integrating factor for this equation is µ = (x + y)−1 . The reader is invited to check that the equation dx+dy x+y + ydx + xdy = 0 is exact and then solve it by the method of exact equations. ∂M

− ∂N

(k) M = x3 + xy 3 and N = 3y 2 ; ∂y N ∂x = x. An integrating factor is 2 2 µ(x) = ex /2 and the solution is ex /2 (y 3 + x2 − 2) = C. ∂M

− ∂N

µ) ∂y ∂x 2. If µ = e g(z)dz , then d(ln = g(z) and µ1 dµ dz dz = g(z) = N y−M x . Rewrite R

dµ ∂M ∂N the equation as µ1 ( dµ dz N y − dz M x) = ∂y − ∂x . Since g(z) is a function dµ dµ dµ ∂µ ∂µ 1 ∂M ∂N of z = xy, dµ dx = dz y and dy = dz x. Thus, µ (N x − M ∂y ) = ∂y − ∂x . ) ) Rewrite it to obtain ∂(µ·M = ∂(µ·N ∂y ∂x . Therefore, µ is an integrating factor. ∂ ∂ (µM ) = ∂x (µN ). Therefore, 3. If µ(M dx + N dy) = 0 is exact, then ∂y ∂µ ∂µ ∂N ∂M ∂N µ ∂M ∂y + ∂y M = µ ∂x + ∂x N . This can be written as µ( ∂y − ∂x ) = ∂µ ∂µ ∂x N − ∂y M .

∂µ 0 If µ = µ(x + y), then ∂µ ∂x = ∂y = µ (x + y) so the last equation can be rearranged to ∂M

∂N

µ0 (x + y) ∂y − ∂x = . µ(x + y) N −M Therefore, there is an integrating factor of the form µ(x + y) if and only ∂M

− ∂N

∂x if ∂y is a function of the form f (x + y). If this is the case, the N −M R integrating factor is µ(z) = e f (z)dz where z = x + y. See Exercise 1 (j) above.

4. Making the substitution z = y/xn yields z 0 xn + nzxn−1 = 2zxn−1 + x3 zxn 0 zxn + x tan x2 . Choosing n = 2 yields a separable equation xz = 1 zdz z + tanRz and 1+z tan z R= dx. Integrate the left side of the equation to cos z obtain 1+z ztan z dz = coszz+z sin z dz = ln | cos z + z sin z| (substitution u = cos z + z sin z. The implicit solution to the original differential equation is ln | cos y/ xn + xyn sin xyn | = x + C.

2.7. REDUCTION OF ORDER

5. The following Maple code has, as its output, the integrating factor for M dx+N dy = 0, provided this factor is either a function of x or a function of y. It is applied to the equation in park (k) of problem 1: (x3 + xy 3 )dx + 3y 3 dy = 0. Only the last output is displayed. **

2.7

Reduction of Order

1. Solve using reduction of order. dp dp to obtain yp dy = −p2 . (a) Make the substitution y 0 = p, y 00 = p dy dy Separate the variables: dp p = − y , and integrate to get ln p = − ln y+ dy C. Equivalently, p = C/y. To finish replace p with y 0 (= dx ) and 2

dy integrate once more: dx = Cy , ydy = Cdx, and y2 = Cx + D. This can be expressed simply as y 2 = Cx + D.

(b) Make the substitution y 0 = p to obtain xp0 = p + p3 , a separable dx equation. Separate to p3dp +p = x and integrate (partial fractions). x The solution is p = ± √C−x2 . Replace p with y 0 and integrate once √ more for the solutions: y = ± C − x2 + D. (c) Proceed as in Example 1.10.2 to obtain y = A sinh(kx) + B cosh(kx). (d) Make the substitution y 0 = p to obtain x2 p0 = 2xp + p2 , a Bernoulli x2 . Replace p with y 0 and equation (n = 2) with solution p = C−x 2

integrate for the solution: y = − x2 − Cx − C 2 ln(x − C) + D. dp dp (e) Make the substitution y 0 = p, y 00 = p dy to obtain 2yp dy = 1 + p2 , √ a separable equation with solution p = ± Cy − 1. Replace p with dy to obtain another separable equation that integrates to Cx = dx √ ±2 Cy − 1 + D. dp dp to obtain yp dy = p2 , a (f) Make the substitution y 0 = p, y 00 = p dy dy separable equation with solution p = Cy. Replace p with dx to Cx obtain another separable equation that integrates to y = De .

(g) Make the substitution y 0 = p to obtain xp0 +p = 4x, a linear equation dy with solution p = 2x + Cx−1 . Replace p with dx and integrate once 2 more for the solution: y = x + C ln x + D. 2. Find the specified particular solution of each of the following equations. (a) Make the substitution y 0 = p, y 00 = p0 to obtain (x2 +2p)p0 +2xp = 0. Solve this exact equation to get x2 p+p2 = C. Initial values determine that C = 0. The equation is x2 p + p2 = 0 or p = −x2 and y 0 = −x2 . Integrate it to obtain y = − 31 x3 + D. Initial values determine that D = 1. Therefore, y = − 13 x3 + 1.

CHAPTER 2. SOLVING FIRST-ORDER EQUATIONS dp dp (b) Make the substitution y 0 = p, y 00 = p dy to obtain yp dy = y 2 p + p2 ∂M

− ∂N

and ydp − (y 2 + p)dy = 0. − ∂y M ∂p = − y2 . The integrating factor is R

thus e − y dy = y12 . Multiple the equation by the integrating factor and solve the exact equation to obtain yp − y = C. Initial values determine that C = − 32 . The equation is yp − y = 23 . Rewrite it as dy (y− 23 )y

= dx. Integrate it by partial fractions to get 23 (ln |y − 32 | − ln |y|) = x + D. Initial values determine that D = 3 ln 2. Therefore, 3 2 3 (ln |y − 2 | − ln |y|) = x + 3 ln 2. dp dp p to obtain dy p = pey and (c) Make the substitution y 0 = p, y 00 = dy dp dy

= ey . Integrate it to get p = ey + C. Initial values determine

that C = 1. The equation is p = ey + 1 or eydy+1 = dx. Integrate it (substitution u = ey ) to obtain ln ey −ln(ey +1) = x+D. Initial values determine that D = − ln 2. Therefore, ln ey − ln(ey + 1) = x − ln 2. 3. Solve using both methods: x missing, y missing. Reconcile the results. (a) y 00 = 1 + (y 0 )2 x missing Substitute y 0 = p, y 00 = p0 to get p0 = 1 + p2 (separable). The solution is p = tan(x + C); that is, y 0 = tan(x + C). Integrate to obtain y = ln(sec(x + C)) + D. dp dp y missing Substitute y 0 = p, y 00 = p dy to get p dy = 1+p2 (separable). √ √ The solution is√p = Ce2y − 1; that is, y 0 = Ce2y − 1.√Integrate to obtain arctan Ce2y − 1 = x+D. This is equivalent to Ce2y − 1 = tan(x + D). To reconcile these solutions note that the second one is equivalent to e2y = C sec2 (x + D). (b) y 00 + (y 0 )2 = 1 x missing Substitute y 0 = p, y 00 = p0 to get p0 = 1 − p2 (separable). The solution is p = tanh(x + C) ; that is, y 0 = tanh(x + C). Integrate to obtain y = ln(cosh(x + C)) + D. dp dp y missing Substitute y 0 = p, y 00 = p dy to get p dy = 1 − p2 (sepa√ √ rable). The solution is p =√ Ce−2y + 1; that is, y 0 = Ce−2y + 1. Integrate to obtain arctanh Ce−2y + 1 = x + D. This is equivalent √ −2y to Ce + 1 = tanh(x + D). To reconcile these solutions note that the second one is equivalent to e2y = C cosh2 (x + D). 4. **

Chapter 3

Some Applications of the First-Order Theory 3.1

The Hanging Chain and Pursuit Curves

1. The statement of the problem is incomplete. The relation T = wy is valid provided the x-axis is placed so that h0 = Tw0 , where T0 denotes the tension at x = 0. This implies that y=

T0 w cosh( x) w T0

q and y 0 = sinh( Tw0 x). Using the identity cosh( Tw0 x) = 1 + sinh2 ( Tw0 x) we p have cosh( Tw0 x) = 1 + (y 0 )2 which allows us to rearrange the displayed p 1 equation to wy = T0 (1 + (y 0 )2 ). But cos θ = 1+(y 0 )2 (see Figure 1.11) so T0 the last equation implies that wy = cos θ = T.

2. 3. The assumption is that the weight supported from the bottom of the cable to the point (x, y) equals L0 x. This means that the displayed equation √ proceeding Equation (1.33) in this section can be changed to y 0 = L0 T0 x where √ T0 denotes the tension at the low point, x = 0. Consequently, y = L0 2T0 x2 + h0 . The cable hangs in the shape of a parabola. 5. Let δ denote the planar weight density of the curtain (weight per unit area). Using this, the weight supported by the cord from the low point, R2 x = 0, to the point (x, y) is equal to δ 0 y(t)dt. Arguing as in this section Rx we are led to y 0 = Tδ0 0 y(t)dt where T0 is the tension in the cord at x = 0. p Differentiate to obtain y 00 = Tδ0 y or y 00 − a2 y = 0 where a = δ/T0 . This second order equation can be solved by reduction of order as in Example 35

36CHAPTER 3. SOME APPLICATIONS OF THE FIRST-ORDER THEORY 1.10.3 (y 0 is missing) to obtain y = c1 eax +c2 e−ax . THe fact that y 0 (0) = 0 implies that c1 = c2 so the solution has the form y = c(eax + e−ax ). The value of the constant c is determined by the height, h0 , of the curtain at x = 0: y = h20 (eax + e−ax ). The shape is similar to that of a catenary (it is not a catenary, why not?) 7. The trajectories orthogonal to the pursuit curve have the differential equation y 0 = √a2x−x2 . Solve to show that the family is defined by x2 +(y−c)2 = a2 .

3.2

Electrical Circuits

1. Since I = ER0 (1 − e−Rt/L ) the theoretical maximum value for I is ER0 . Half this value is attained when 1 − e−Rt/L = 1/2. Solve to obtain t = (L ln 2)/R seconds. 3. The current is controlled by L dI dt + RI = E. (a) The current is maximum or minimum when dI dt = 0, implying that RI = E. (b) Differentiate the controlling equation to obtain E 0 = LI 00 + RI 0 . Since I 0 = 0 when I is a maximum or a minimum, E 0 = LI 00 at such times. If the current is a minimum, then I 00 > 0 so E 0 > 0 and E is increasing. If current is a maximum, then I 00 < 0 so E 0 < 0 and E is decreasing. 1 5. The controlling equations are L dI dt + C Q = 0, Q(0) = Q0 , and I(0) = 0. 1 00 In terms of Q this is LQ + C Q = 0, Q(0) = Q0 , Q0 (0) = 0. Write this 1 as Q00 + LC Q = 0. Since Q0 is missing reduction of order can be used to √ solve for Q (see Example **): Q = A sin ωt + B cos ωt where ω = 1/ LC. Differentiate to find I: I = Q0 = Aω cos ωt−Bω sin ωt. The condition that I(0) = 0 forces A = 0. The √ condition Q(0) = Q0 tells √ us that B = Q0 . 0 Therefore, Q = Q0 cos(t/ LC) and I = − √QLC sin(t/ LC).

38CHAPTER 3. SOME APPLICATIONS OF THE FIRST-ORDER THEORY

Chapter 4

Second-Order Linear Equations 4.1

Second-Order Linear Equations with Constant Coefficients

1. Find the general solution of each of the following differential equations. See Table 4.1. 2. Find the solution of each of the following initial value problems: (a) The associated polynomial r2 − 5r + 6 has roots r = 2, 3. The general solution is y = Ae3x + Be2x . Initial conditions determine that A = 0, B = 1/e. Therefore, 1e e3x . (b) The associated polynomial r2 − 6r + 5 has roots r = 1, 5. The general solution is y = Aex + Be5x . Initial conditions determine that A = 1, B = 2. Therefore, y = ex + 2e5x . (c) The associated polynomial r2 − 6r + 9 has roots r = 3, 3. The general solution is y = Ae3x + Bxe3x . Initial conditions determine that A = 3, B = −1. Therefore, y = 3e3x − xe3x . (d) The associated polynomial r2 + 4r + 5 has roots −2 ± i. The general solution is y = Ae−2 cos x + Be−2 sin x. Initial conditions determine that A = e2 and B = 0. Therefore, y = cos x. √ 2 (e) The associated polynomial r +4r +2 has roots −2± 2. The general √ √ (−2+ 2)x (−2− 2)x solution is y = Ae + Be . Initial conditions √deter√ mine that A = 1, B = −2. Therefore, y = e(−2+ 2)x − 2e(−2− 2)x . (f) The associated polynomial r2 + 8r − 9 has roots r = −9, 1. The general solution is y = Ae−9x + Bex . Initial conditions determine 9 9 9 9 x that A = e5 , B = 5e . Therefore, y = e5 e−9x + 5e e . 39

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r)

Table 4.1: The general solutions for Exercise 1. Assoc Poly Roots General Solution r2 + r − 6 2, −3 Ae2x + Be−3x r2 + 2r + 1 −1,√−1 Ae−x +√Bxe−x √ 2 r +8 ±2 √2i A cos(2 √2x) + B sin(2 2x) √ 2r2 − 4r + 8 1 ± 3i Aex cos( 3x) + Bex sin( 3x) r2 − 4r + 4 2, 2 Ae2x + Bxe2x 2 r − 9r + 20 4, 5 √ Ae4x + Be5x √ √ 5 1 2r2 + 2r + 3 − 2 ± 2 i Ae−x/2 cos( 5x/2) + Be−x/2 sin( 5x/2) 4r2 − 12r + 9 3/2, 3/2 Ae3x/2 + Bxe3x/2 2 r +r −1, 0 Ae−x + B 2 r − 6r + 25 3 ± 4i e3x (A cos 4x + B sin 4x) 2 −5x/2 4r + 20r + 25 ±5/2 √ Ae5x/2 + Be √ √ −x 2 r + 2r + 3 −1 ± 2i Ae cos( 2x) + Be−x sin( 2x) r2 − 4 ±2 √ Ae2x + Be−2x √ √ 3 2 4r − 8r + 7 1± 2 i Aex cos( 3x/2) + Bex sin( 3x/2) 2r2 + r − 1 −1, 1/2 Ae−x + Bex/2 2 16r − 8r + 1 1/4 Aex/4 + Bxex/4 2 r + 4r + 5 −2 ± i Ae−2x cos x + Be−2x sin x 2 r + 4r − 5 −5, 1 Ae−5x + Bex

√ −P ± P 2 −4Q 3. The associated polynomial r2 + P r + Q has roots r = . Sup2 pose that P and Q are both positive. Then P 2 − 4Q ≥ 0 implies that the roots are real and negative so y → 0 as x → ∞ because both exponential terms in the solution have negative exponents. If P 2 − 4Q < 0, then the roots are complex with negative real part. Consequently, the solutions are of the form y = Ae−P x/2 cos ωx + Be−P x/2 sin ωx and will oscillate towards 0 as x → ∞. The other cases are handled similarly. 4. Take a derivative on both sides of y 00 + P y 0 + Qy = 0 to obtain y 000 + P y 00 + Qy 0 = 0. As y 000 + P y 00 + Qy 0 = (y 0 )00 + P (y 0 )0 + Q(y 0 ), y 0 is a solution of the equation. 5. Euler’s equidimensional equation Changing the independent variable dy dz 1 using x = ez is equivalent to z = ln x so y 0 = dx = dy dz · dx = x ẏ where the dot indicates differentiation with respect to the new independent variable, d 1 z. Similarly, y 00 = dx ( x ẏ) = x1 · x1 ÿ − x12 ẏ = x12 (ÿ − ẏ). Making these 2 00 substitutions into x y +pxy 0 +qy = 0 yields x2 · x12 (ÿ− ẏ)+px· x1 ẏ+qy = 0 which simplifies to ÿ + (p − 1)ẏ + qy = 0, an equation with constant coefficients. If y = φ(z) is the general solution to this equation, then y = φ(ln x) will be the general solution to the Euler equidimensional equation. Note that the solution is only valid for x > 0. (a) The z equation is ÿ + 2ẏ + 10y = 0 with solution y = Ae−z cos 3z +

4.2. THE METHOD OF UNDETERMINED COEFFICIENTS

Table 4.2: The differential equations for Exercise 6. Roots Assoc Poly Differential equation (a) 1, −2 r2 + r − 2 y 00 + y 0 − 2y = 0 2 (b) 0, 2 r − 2r y 00 − 2y 0 = 0 2 (c) 3, 5 r − 8r + 15 y 00 − 8y 0 + 15y = 0 2 (d) 1 ± 3i r − 2r + 10 y 00 − 2y 0 + 10y = 0 2 (e) 3, −1 r − 2r − 3 y 00 − 2y 0 − 3y = 0 2 (f) −1, 4 r + 5r + 4 y 00 + 5y 0 + 4y = 0 2 (g) 2, −2 r −4 y 00 − 4y = 0 2 (h) −4 ± i r + 8r + 17 = 0 y 00 + 8y 0 + 17y = 0 Be−x sin(3z). The x equation has the solution y = Ax−1 cos(3 ln x)+ Bx−1 sin(3 ln x). (b) First divide the equation by 2. The z equation is then ÿ +4ẏ +4y = 0 with solution y = Ae−2z + Bze−2z . The x equation has the solution y = Ax−2 + Bx−2 ln x. (c) The z equation is ÿ + ẏ − 12y = 0 with solution y = Ae−4z + Be3z . The x equation has the solution y = Ax−4 + Bx3 . (d) Divide the equation by 4. The z equation is then ÿ − ẏ − 34 y = 0 with solution y = Ae−z/2 + Be3z/2 . The x equation has the solution y = Ax−1/2 + Bx3/2 . (e) The z equation is ÿ − 4ẏ + 4y = 0 with solution Ae2z + Bze2z . The x equation has the solution y = Ax2 + Bx2 ln x. (f) The z equation is ÿ + ẏ − 6y = 0 with solution y = Ae−3z + Be2z . The x equation has the solution y = Ax−3 + Bx2 . (g) The z equation is ÿ + ẏ + 3y = 0 with solution y = Ae−z/2 cos √ Be−z/2 sin 11z 2 . The x equation has the solution √ √ 1 1 11 ln x 11 ln x y = Ax− 2 cos + Bx− 2 sin . 2 2 √

√

11z 2 +

√

(h) The z equation is ÿ − 2y = 0 with solution y = √Ae− 2z + Be 2z . √ The x equation has the solution y = Ae− 2 + Bx 2 . (i) The z equation is ÿ − 16y = 0 with solution y = Ae−4z + Be4z . The x equation as the solution y = Ax−4 + Bx4 . 6. Find the differential equation of each of the following general solution sets. See Table 4.2.

4.2

The Method of Undetermined Coefficients

1. Find the general solution of each of the following equations.

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS (a) The auxiliary roots are −5 and 2 so the homogeneous equation has the solution y = Ae−5x +Be2x . Try y = αe4x as a particular solution. Substitute and simplify to obtain 18αe4x = 6e4x which implies that α = 1/3. The general solution is y = Ae−5x + Be2x + 31 e4x . (b) The auxiliary roots are ±2i so the homogeneous equation has the solution y = A cos 2x + B sin 2x. Try y = α cos x + β sin x as a particular solution. Substitute and simplify to obtain 3α cos x + 3β sin x = 3 sin x which implies that α = 0 and β = 1. The general solution is y = A cos 2x + B sin 2x + sin x. (c) The auxiliary roots are −5, −5 so the homogeneous equation has the solution y = Ae−5x + Bxe−5x . Neither y = αe−5x nor y = αxe−5x can be a particular solution because they are solutions to the homogeneous equation. Try y = αx2 e−5x instead. Substitute and simplify to obtain 2αe−5x = 14e−5x which implies that α = 7. The general solution is y = Ae−5x + Bxe−5x + 7x2 e−5x . (d) The auxiliary roots are 1 ± 2i so the homogeneous equation has the solution y = Aex cos 2x + Bex sin x. Try y = αx2 + βx + γ as a particular solution. Substitute and simplify to obtain 5αx2 + (5β − 4α)x+2α−2β+5γ = 25x2 +12 which implies that α = 5, β = 4, γ = 2. The general solution is y = Aex cos 2x + Bex sin x + 5x2 + 4x + 2. (e) The auxiliary roots are −2, 3 so the homogeneous equation has the solution y = Ae−2x + Be3x . The function y = αe−2x can not be a particular solution because it is a solution to the homogeneous equation. Try y = αxe−2x instead. Substitute and simplify to obtain −5αxe−2x = 20e−2x which implies that α = −4. The general solution is y = Ae−2x + Be3x − 4xe−2x . (f) The auxiliary roots are 1, 2 so the homogeneous equation has the solution y = Aex + Be2x . Try y = α cos 2x + β sin 2x as a particular solution. Substitute and simplify to obtain (−6α−2β) cos 2x+(−2α+ 6β) sin 2x = 14 sin 2x − 18 cos 2x which implies that α = 3 and β = 2. The general solution is y = Aex + Be2x + 3 cos 2x + 2 sin 2x. (g) The auxiliary roots are ±i so the homogeneous equation has the solution y = A cos x + B sin x. The function y = α cos x + β sin x can not be a particular solution because it is a solution to the homogeneous equation. Try y = αx cos x + βx sin x instead. Substitute and simplify to obtain 2β cos x − 2α sin x = 2 cos x which implies that α = 0 and β = 1. The general solution is y = A cos x + B sin x + x sin x. (h) The auxiliary roots are 0, 2 so the homogeneous equation has the solution y = A + Be2x . The function y = αx + β is not a particular solution because part of it is a solution to the homogeneous equation. Try y = αx2 + βx instead. Substitute and simplify to obtain −4αx + 2α − 2β = 12x − 10 which implies that α = −3 and β = 2. The general solution is y = A + Be2x − 3x2 + 2x.

4.2. THE METHOD OF UNDETERMINED COEFFICIENTS

(i) The auxiliary roots are 1, 1 so the homogeneous equation has the solution y = Aex + Bxex . Neither y = αex not y = αxex is a particular solution because they are both solutions to the homogeneous equation. Try y = αx2 ex instead. Substitute and simplify to obtain 2αex = 6ex which implies that α = 3. The general solution is y = Aex + Bxex + 3x2 ex . (j) The auxiliary roots are 1 ± i so the homogeneous equation has the solution y = Aex cos x + Bex sin x. This means that y = αex cos x + βex sin x can not be a particular solution. Try y = αxex cos x + βxex sin x instead. Substitute and simplify to obtain 2βex cos x − 2αex sin x = ex sin x which implies that α = −1/2 and β = 0. The general solution is y = Aex cos x + Bex sin x − 21 xex cos x. (k) The auxiliary roots are −1, 0 so the homogeneous equation has the solution y = Ae−x +B. This means that y = αx4 +βx3 +γx2 +δx+ can not be a particular solution. Try y = αx5 + βx4 + γx3 + δx2 + x instead. Substitute and simplify to obtain 5αx4 + (20α + 4β)x3 + (12β + 3γ)x2 + (6γ + 2δ)x + 2δ + = 10x4 + 2 which implies that α = 2, β = −10, γ = 40, δ = −120 and = 242. The general solution is y = Ae−x + B + 2x5 − 10x4 + 40x3 − 120x3 + 242x. 2. Find the solution of the differential equation that satisfies the given initial conditions: √

(a) The auxiliary roots are 3±2 5 so the homogeneous equation has the √ 3+ 5

3−

√

solution y = Ae 2 x + Be 2 x . Try y = αx + β as a particular solution. Substitute and simplify to obtain −2αx + β = x which implies that α = 1, β = 3. The general solution is y = x√+ 3 + √ √ 3+ 5 3− 5 x x √ 5 and Ae 2 +Be 2 . Initial conditions determine that A = 2− 5 √

√ 5 . The solution that satisfies the given initial conditions B = −2− 5 √

√ √ 3− 5 x √ 3+ 5 2 √ 5 e 2 x + −2− 5 . is y = x + 3 + 2− e 5

√ (b) The auxiliary roots are −2 √ ± 2i so the homogeneous equation has √ the solution y = Ae−2x cos 2x + Be−2x sin 2x. Try y = α cos x + β sin x as a particular solution. Substitute and simplify to obtain 5 cos x(5α+4β)+sin x(5β −4α) = cos x which implies that α = 41 and √ √ 4 −2x −2x β = 41 . The general solution is y = Ae cos 2x+Be sin 2x+ 5 4 5 cos x + sin x. Initial conditions determine that A = − 41 and 41 41 88 √ B = 41 2 . The solution that satisfies the given initial conditions is √ √ 5 −2x √ e−2x sin 2x + 5 cos x + 4 sin x. y = − 41 e cos 2x + 4188 41 41 2 √

i (c) The auxiliary roots are −1± so the homogeneous equation has the √2 √ 3 solution y = Ae−x/2 cos 2 x + Be−x/2 sin 23 x. Try y = α cos x + β sin x as a particular solution. Substitute and simplify to obtain β cos x − α sin x = sin x which implies that α = −1 and β = 0.

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS √

√

The general solution is y = Ae−x/2 cos 23 x + Be−x/2 sin 23 x − cos x. √ √ Initial conditions determine that A = (1 + cos 1) · 3 · cos 23 − √ √ √ √ 3 3 3+cos√1+sin 1 3+cos√1+sin 1 √ · and B = · e · sin 3 · cos 2 2 − e · (1 + 3 3 √

cos 1) · sin 23 . The solution that satisfies the given initial conditions √ √ √ √ 3 −x/2 3 1 √ is y = [(1+cos 1)· 3·cos 23 − 3+cos√1+sin · e·sin ]e cos 2 2 x+ 3 √ √ √ √ √ 1 · 3·cos 23 − e·(1+cos 1)·sin 23 ]e−x/2 sin 23 x−cos x. [ 3+cos√1+sin 3 (d) The auxiliary roots are 1, 2 so the homogeneous equation has the solution y = Aex + Be2x . Initial conditions determine that A = −e and B = e2 . The solution that satisfies the given initial conditions is y = −e · ex + e2 · e2x = −ex+1 + e2x+2 . √

(e) The auxiliary roots are 1±2 3i so the homogeneous equation has the √ √ solution y = Aex/2 cos 23 x + Bex/2 sin 23 x. Try y = αx2 + βx + γ as a particular solution. Substitute and simplify to obtain αx2 + (β − 2α)x − β + 2α = x2 which implies that α = 1, β = √2 and √ γ = 0. The general solution is y = Aex/2 cos 23 x + Bex/2 sin 23 x + x2 + 2x. Initial conditions determine that A = 1 and B = − √53 . The √

solution that satisfies the given initial conditions is y = ex/2 cos 23 x− √ √5 ex/2 sin 3 x + x2 + 2x. 2 3 (f) The auxiliary roots are −1, −1 so the homogeneous equation has the solution y = Ae−x + Bxe−x . Try y = αx + β as a particular solution. Substitute and simplify to obtain αx + 2α + β = 1 which implies that α = 0 and β = 1. The general solution is y = Ae−x + Bxe−x + 1. Initial conditions determine that A = 0 and B = 0. The solution that satisfies the given initial conditions is a constant y = 1. 3. The auxiliary roots are −k, k so the homogeneous equation has the solution y = Aekx + Be−kx . Try y = α cos bx + β sin bx as a particular solution. Substitute and simplify to obtain (k 2 − b2 )α cos bx + (k 2 − b2 )β sin bx = 1 sin bx which implies that α = 0 and β = k2 −b 2 . The general solution is 1 kx −kx y = Ae + Be + k2 −b2 sin bx. 4. Substitute y = y1 + y2 into the left side of the differential equation to get (y1 +y2 )00 +P (y1 +y2 )0 +Q(y1 +y2 ) = y100 +P y10 +Qy1 +y200 +P y20 +Qy2 = R1 +R2 . (a) The auxiliary roots are ±2i so the homogeneous equation has the solution y = A cos 2x + B sin 2x. A particular solution for y 00 + 4y = 4 cos 2x has the form y1 = αx cos 2x + βx sin 2x. Substitute to obtain y1 = x sin 2x. A particular solution for y 00 + 4y = 6 cos x has the form y2 = α cos x + β sin x. Substitute to obtain y2 = 2 cos x. A particular solution for y 00 + 4y = 8x2 − 4x has the form y3 = αx2 + βx + γ. Substitute to obtain y3 = 2x2 − x − 1.

4.3. THE METHOD OF VARIATION OF PARAMETERS

By the superposition principal y = y1 +y2 +y3 is a particular solution to the given equation and y = A cos 2x+B sin 2x+x sin 2x+2 cos x+ 2x2 − x + 1 is a general solution. (b) The auxiliary roots are ±3i so the homogeneous equation has the solution y = A cos 3x + B sin 3x. A particular solution for y 00 + 9y = 2 sin 3x has the form y1 = αx cos 3x + βx sin 3x. Substitute to obtain y1 = − 31 x cos 3x. A particular solution for y 00 + 4y = 4 sin x has the form y2 = α cos x + β sin x. Substitute to obtain y2 = 21 sin x. A particular solution for y 00 + 4y = −26e−2x has the form y2 = αe−2x . Substitute to obtain y4 = 3x3 − 2x. By the superposition principal y = y1 + y2 + y3 + y4 is a particular solution to the given equation and y = A cos 3x+B sin 3x− 31 x cos 3x+ 1 −2x + 3x2 − 2x is a general solution. 2 sin x − 2e 5. **

4.3

The Method of Variation of Parameters

1. Find a particular solution. (a) The homogeneous solution is y = A sin 2x+B cos 2x so the particular solution has the form y = v1 sin 2x + v2 cos 2x where v1 and v2 satisfy the system v10 sin 2x + v20 cos 2x = 0 2v10 cos 2x − 2v20 sin 2x = tan 2x Therefore, v10 = 21 sin x and v20 = − 12 sin 2x tan 2x. Integrate to get v1 = − 14 cos 2x and v2 = 14 sin 2x − 41 ln(sec 2x + tan 2x). Therefore, y = − 14 cos 2x ln(sec 2x + tan 2x). (b) The homogeneous solution is y = Ae−x + Bxe−x so the particular solution has the form y = v1 e−x + v2 xe−x where v1 and v2 satisfy the system v10 e−x + v20 xe−x = 0 −v10 e−x + v20 (e−x − xe−x ) = e−x ln x Therefore, v10 = −x ln x and v20 = ln x. Integrate to get v1 = − 21 x2 ln x+ 14 x2 and v2 = x ln x−x. Therefore, y = 41 x2 e−x (2 ln x−3). (c) The homogeneous solution is y = Ae3x + Be−x so the particular solution has the form y = v1 e3x + v2 e−x where v1 and v2 satisfy the system v10 e3x + v20 e−x = 0 3v10 e−x − v20 e−x = 64xe−x

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS Therefore, v10 = 16xe−4x and v20 = −16x. Integrate to get v1 = −(4x + 1)e−4x and v2 = −8x2 . Therefore, y = −e−x (8x2 + 4x + 1). The last term can be dropped since −e−x is a solution to the homogeneous solution. (d) The homogeneous solution is y = Ae−x sin 2x + Be−x cos 2x so the particular solution has the form y = v1 e−x sin 2x+v2 e−x cos 2x where v1 and v2 satisfy the system v10 e−x sin 2x + v20 e−x cos 2x = 0 v10 (−e−x sin 2x + 2e−x cos 2x) + v20 (e−x cos 2x + 2e−x sin 2x) = e−x sec 2x Therefore, v10 = 21 and v20 = − 21 tan 2x. Integrate to get v1 = 21 x and v2 = 14 ln(cos 2x). Therefore, y = 12 xe−x sin 2x+ 14 e−x cos 2x ln(cos 2x). (e) The homogeneous solution is y = Ae−x/2 + Be−x so the particular solution has the form y = v1 e−x/2 + v2 e−x where v1 and v2 satisfy the system v10 e−x/2 + v20 e−x = 0 1 1 − v10 e−x/2 − v20 e−x = e−3x 2 2 Therefore, v10 = e−5x/2 and v20 = −e−2x . Integrate to get v1 = 1 −3x − 25 e−5x/2 and v2 = 12 e−x . Therefore, y = 10 e . (f) The homogeneous solution is y = Aex + Be2x so the particular solution has the form y = v1 ex + v2 e2x where v1 and v2 satisfy the system v10 ex + v20 e2x = 0 v10 ex + 2v20 e2x = (1 + e−x )−1 −x

−2x

e e 0 Therefore, v10 = − 1+e Integrate to get v1 = −x and v2 = 1+e−x . −x −x −x ln(1 + e ) and v2 = ln(1 + e ) − e . Consequently, the particular solution is y = (ex + e2x ) ln(1 + e−x ) − ex .

2. the homogeneous solution is y = A cos x+B sin x so the particular solution has the form y = v1 cos x + v2 sin x. (a) The functions v1 and v2 satisfy the system v10 cos x + v20 sin x = 0 −v10 sin x + v20 cos x = sec x Therefore, v10 = − tan x and v20 = 1. Integrate to get v1 = ln | sec x| and v2 = x. Therefore, y = (A + ln | sec x|) cos x + (B + x) sin x.

4.3. THE METHOD OF VARIATION OF PARAMETERS

(b) The functions v1 and v2 satisfy the system v10 cos x + v20 sin x = 0 −v10 sin x + v20 cos x = cot2 x 2

cos x x 0 Therefore, v10 = − cos sin x and v2 = − sin2 x . Integrate to get v1 = − ln | csc x − cot x| − cos x and v2 = csc x + sin x. Therefore, y = A cos x+B sin x+(− ln | csc x−cot x|−cos x) cos x+(csc x+sin x) sin x.

(d) The functions v1 and v2 satisfy the system v10 cos x + v20 sin x = 0 −v10 sin x + v20 cos x = x cos x Therefore, v10 = −x sin x cos x and v20 = x cos2 x. Integrate to get v1 = 41 x cos 2x − 18 sin 2x and v2 = 41 x2 + 14 x sin 2x + 18 cos 2x. Therefore, y = A cos x + B sin x + ( 41 x cos 2x − 18 sin 2x) cos x + ( 14 x2 + 1 1 4 x sin 2x + 8 cos 2x) sin x. (e) The functions v1 and v2 satisfy the system v10 cos x + v20 sin x = 0 −v10 sin x + v20 cos x = tan x 2

x 0 Therefore, v10 = − sin cos x and v2 = sin x. Integrate to get v1 = − ln | sec x + tan x| + sin x and v2 = − cos x. Therefore, y = A cos x + B sin x + (− ln | sec x + tan x| + sin x) cos x + (− cos x) sin x.

(f) The functions v1 and v2 satisfy the system v10 cos x + v20 sin x = 0 −v10 sin x + v20 cos x = sec x tan x 2

sin x 0 Therefore, v10 = − cos Integrate to get v1 = 2 x and v2 = tan x. − tan x + x and v2 = ln | sec x|. Therefore, y = A cos x + B sin x + (− tan x + x) cos x + (ln | sec x|) sin x.

(g) The functions v1 and v2 satisfy the system v10 cos x + v20 sin x = 0 −v10 sin x + v20 cos x = sec x csc x

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS 2

sin x 0 Therefore, v10 = − cos Integrate to get v1 = 2 x and v2 = tan x. − tan x + x and v2 = ln | sec x|. Therefore, y = A cos x + B sin x + (− tan x + x) cos x + (ln | sec x|) sin x.

3. By Inspection The auxiliary polynomial is r2 −2r +1 with roots 1, 1 so the homogeneous solution is y = Aex + Bxex . Therefore, there is a particular solution of the form y = αx + β. Substitute to find that y = 2x + 4 is a particular solution. By Variation of Parameters The particular solution has the form y = v1 ex + v2 xex where v1 and v2 satisfy the system v10 ex + v20 xex = 0 v10 ex + v20 (xex + ex ) = 2x Therefore, v10 = −2x2 e−x and v20 = 2xe−x . Integrate to get v1 = (2x2 + 4x + 4)e−x and v2 = −(2x + 2)e−x . Therefore, y = 2x + 4. 4. By undetermined coefficients THe auxiliary polynomial is r− r − 6 with roots −2, 3 so the homogeneous solution is y = Ae−2x + Be3x . Therefore, there is a particular solution of the form y = αe−x . Substitute to find that y = − 41 e−x . By variation of parameters The particular solution has the form y = v1 e−2x + v2 e3x where v1 and v2 satisfy the system v10 e−2x + v20 e3x = 0 −2v10 e−2x + 3v20 e3x = e−x x

−4x

Therefore, v10 = − e5 and v20 = e 5 . Integrate to get v1 = − e5 and v2 = x 1 −4x 1 −4x 3x − 20 e . Therefore, the particular solution is y = − e5 e−2x − 20 e e = 1 −x −4e . 5. The solution yh to the homogeneous equation is given below. Use it to apply the variation of parameters technique to obtain the particular solution, labeled yp . The general solution is y = yh + yp . Warning Write the equation in the form y 00 + P (x)y 0 + Q(x)y = R(x) before applying the variation of parameters algorithm. (a) yh = Ax + B(x2 + 1); yp = x4 /6 − x2 /2. (b) yh = Ax−1 + Bex ; yp = − 31 x2 − x − 1. (c) yh = Ax + Bex ; yp = x2 + x + 1. (d) yh = A(1 + x) + Bex ; yp = 21 e2x (x − 1).

4.4. THE USE OF A KNOWN SOLUTION TO FIND ANOTHER

R −x (e) yh = Ax2 + Bx; yp = −xe−x − (x2 + x) e x dx. To obtain this solution formula you will have to apply integration by R −x parts to ex2 dx. The remaining integral does not evaluate to an elementary function. 6. **

4.4

The Use of a Known Solution to Find Another

1. Find y2 and the general solution, given y1 . R

(a) Since p(x) 0, e− R p(x)dx = e0 = 1 and y2 (x) = sin(x)v(x) where R = 1 v(x) = sin2 x dx = csc2 xdx = − cot x. Therefore, y2 (x) = − cos x. The general solution is y = A sin x + B cos x. R (b) Once more, p(x) = 0R and e− p(x)dx = e0 = 1. Therefore, y2 (x) = 1 ex v(x) where v(x) = e2x dx = − 21 e−2x . Therefore, y2 (x) = ex (− 12 e−2x ) = − 12 e−x . The general solution is y = Aex + Be−x . R

2. Since p(x) = x3 , e− p(x)dx = x−3 and y2 (x) = y1 (x)v(x) = v(x) where R −3 v(x) = x dx = − 21 x−2 . The general solution is y = A + Bx−2 . 3. If y = y1 = x2 , then x2 y 00 + xy 0 − 4y R= x2 · 2 + x · 2x − 4 · x2 = 0. To find y2 observe p(x) = 1/x and e 1/xdx = e− ln x = 1/x. Therefore, R 1 1 that 2 2 −4 2 2 −2 y2 = x . x4 · 4 = x · x4 = −4/x . The general solution is y = Ax +Bx R R −2x 1 1 1 − p(x)dx 2 4. Since p(x) = 1−x = 1−x 2, e 2 . Therefore, y2 = x x4 · 1−x2 dx = 2

1 − 3x − x − log(1 − x) + x2 log(x + 1). The general solution is y = Ax2 + 2 1 B[− 3x − x − log(1 − x) + x2 log(x + 1)].

5. If y = y1 = x−1/2 sin x, then 1 y 0 = x−1/2 cos x − x−3/2 sin x 2 3 y 00 = −x−1/2 sin x − x−3/2 cos x + x−5/2 sin x 4 Substitute carefully into (*). R

To find y2 Robserve that p(x) = 1/x so eR− 1/xdx = 1/x. Therefore, y2 = 1 1 −1/2 sin x csc2 xdx = x−1/2 sin x(− cot x) = x−1/2 sin x x−1 sin 2 x · x dx = x −1/2 −x cos x. Therefore, the general solution is y = Ax−1/2 sin x+Bx−1/2 cos x. 6. Find the general solution given y1 . x (a) Since p(x) = x−1 , e−

p(x)dx

x x−1 dx

= ex−1+ln(x−1) = ex−1 · (x −

x−1 1 x−1 · (x − 1)dx = x · e x = ex−1 . The x2 · e x−1

1). Therefore, y2 = x · general solution is y = Ax + Be

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Table 4.3: Solutions for 1-15. Associated Polynomial General Solution r(r − 1)(r − 2) y = A + Bex + Ce2x 2 x (r − 1)(r − 2r + 2) y = Aex + Bex cos x + Ce √ sin x √ 2 x −x/2 (r − 1)(r + r + 1) y = Ae + e (B cos(√ 3x/2) + C sin(√ 3x/2)) (r + 1)(r2 − r + 1) y = Ae−x + Bex/2 cos 23 x + Cex/2 sin 23 x 3 (r + 1) y = Ae−x + Bxe−x + Cx2 e−x 4 (r + 1) y = Ae−x + Bxe−x + Cx2 e−x + Dx3 e−x 2 2 (r − 1)(r + 1) y = Aex + Be−x + C cos x + D sin x 2 2 (r + 1)(r + 4) y = A cos x + B sin x + C cos 2x + D sin 2x (r − a)2 (r + a)2 y = Aeax + Bxeax + Ce−ax + Dxe−ax 2 2 2 (r + a ) = 0 y = A cos ax + Bx cos ax + C sin ax + Dx sin ax (r + 1)2 (r2 + 1) y = Ae−x + Bxe−x + C cos x + D sin x 2 2 (r − 1) (r + 4r + 5) y = Aex + Bxex + Ce−2x cos 32 x + De−2x sin 32 x (r − 1)(r − 2)(r − 3) y = Aex + Be2x + Ce3x 3 (r − 2)(r + 1) y = Ae2x + Be−x + Cxe−x + Dx2 e−x 2 2 (r − 6)(r − 2) (r + 2) y = Ae6x + Be2x + Cxe2x + De−2x + Exe−2x R

2 − p(x)dx (b) Since = 2x = e−2 ln x = x12 . Therefore, y2 = x2 = x , e R 1 p(x) 1 1 x x2 · x2 dx = x · (− 3x3 ) = − 3x1 2 . The general solution is y = Ax + xB2 .w 2

(c) Since = −x(x+2) , e− p(x)dx = ex+ln x = x2 ex . Therefore, y2 = x2 R 1 p(x) x x2 · ex · x2 dx = xex . The general solution is y = Ax + Bxex . R

7. By inspection, y = y1 = x is one Rsolution. Since p(x) = −xf (x), the R second solution has the form y2 =R x x12 e xf (x)dx dx. The general solution R has the form y = Ax + Bx x12 e xf (x)dx dx. 8. Substitute y1 = y10 R= y100 = ex to get xex −(2x+1)ex +(x+1)ex R= 0. Since 1 − p(x)dx = e2x| ln x = xe2x . Therefore, y2 = ex e2x · e2x · p(x) = − 2x+1 x , e 1 2 x xdx = 2 x e . The general solution is y = Aex + Bx2 ex . 9. If y1 and y2 are linearly dependent, then the function v(x) is a constant R and has a derivative that is identically 0. However, v 0 (x) = y12 e− P (x)dx , 1 which is never 0 (exponentials cannot vanish).

4.5

Higher-Order Equations

1-15. Find the general solution. See Table 4.3. 16. The associated polynomial is r4 so the general solution to the homogeneous equation is yg = A + Bx + Cx2 + Dx3 . Try y = α cos x + β sin x + γx4 as a particular solution. Substitute and simplify to get α cos x + β sin x + γ =

4.5. HIGHER-ORDER EQUATIONS

sin x + 24 which implies that α = 0, β = 1, γ = 24. Therefore, the general solution is y = A + Bx + Cx2 + Dx3 + sin x + 24x4 . 17. The associated polynomial is r(r − 1)(r − 2) so the general solution to the homogeneous equation is yg = A + Bex + Ce2x . Based on the forcing function our fist choice for yp is y = A + Be3x . However, this will not work because y = A is a solution to the homogeneous equation. Try y = Ax+Be3x instead. Substitute this into the forced equation to see that A = 5 and B = 7. The general solution is y = A + Bex + Ce2x + 5x + 7e3x . 18. The associated polynomial is r(r − 1)(r + 1) so the general solution to the homogeneous equation is yg = A + Be−x + Cex . Try yp = αx. Substitute and simply to get α = −1. The general solution is y = A + Be−x + Cex − x. Initial conditions determine that A = 0, B = −1/2, and C = 9/2. Therefore, y = − 21 e−x + 92 ex − x. 19. The Euler Equidimensional Equation (order 3) Using x = ez is equivalent to z = ln x so y 0 = x1 ẏ and y 00 = x12 (ÿ − ẏ). The dot indicates differentiation with respect to the new independent variable, z. See Section 2.1 problem 5. For the third derivative, 1 ... 1 2 d 1 ( 2 (ÿ − ẏ)) = 2 ( y − ẏ) − 3 (ÿ − ẏ) dx x x x x 1 ... ( y − 3ÿ + 2ẏ). x3

y 000 =

Making...these substitutions into x3 y 000 + a2 x2 y 00 + a1 xy 0 + a0 y = 0 yields x3 · x13 ( y − 3ÿ + 2ẏ) + a2 x2 · x12 (ÿ − ẏ) + a1 x · x1 ẏ + a0 y = 0 which simplifies to ... y + (a2 − 3)ÿ + (a1 − a2 + 2)ẏ + a0 y = 0, an equation with constant coefficients. If y = φ(z) is the general solution to the equation, then y = φ(ln x) will be the general solution to the Euler equidimensional equation. Note that this solution is only valid for x > 0. ... (a) The z equation is y − ẏ = 0 with associated polynomial r3 − r = r(r2 − 1). The solution is y = A + Bex + Ce−x so the solution to the original equation is y = A + Bx + Cx−1 . ... (b) The z equation is y − 2ÿ − ẏ + 2y = 0 with associated polynomial r3 −2r2 −r+2 = (r−1)(r+1)(r−2). The solution is y = Aez +Be−z + Ce2z so the solution to the original equation is y = Ax+Bx−1 +Cx2 . ... (c) The z equation is y −ÿ+ẏ−y = 0 with associated polynomial r3 −r2 + 2 r−1 = (r−1)(r +1). The solution is y = Aez +B cos z+C sin z so the solution to the original equation is y = Ax + B cos(ln x) + C sin(ln x). 20. Substituting w = y 0 yields x3 w000 + 8x2 w00 + 8xw0 − 8 = 0. By Problem ... 19, the z equation is w + 5ẅ + 2ẇ − 8w = 0 with associated polynomial (r − 1)(r + 2)(r + 4). The solution is w = Ae−4z + Be−2z + Cez = Ax−4 + Bx−2 + Cx. Integrate to get y = − 31 Ax−3 − Bx−1 + C2 x2 .

CHAPTER 4. SECOND-ORDER LINEAR EQUATIONS

21. The equation is m1 m2 22. **

d2 x1 d4 x1 +(m (k +k )+m (k +k )) +(k1 k2 +k1 k3 +k2 k3 )x1 = 0. 1 2 3 2 1 3 dt4 dt2

Chapter 5

Applications of the Second-Order Theory 5.1

Vibrations and Oscillations

1. The amplitude A = √

F0 (k−ω 2 M )2 +ω 2 c2

attains its maximum at the ω value

that minimizes the polynomial φ(ω) = (k − ω 2 M )2 + ω 2 c2q . A simple 2

k c calculation will show that φ0 (ω) = 0 when ω = 0 or ω = ± M − 2M 2. √ 2 c k Thus if M ≤ 2M 2 , i.e. c ≥ 2kM , then there is no resonance frequency and as ω increases from 0, the√amplitude A will steadily decrease to 0. On the other hand, if 0 < c < 2kM , thenqA will increase as ω increases c2 k M − 2M 2

reaching its maximum value at the ω ∗ =

and decrease to 0

1 k thereafter. The resonance frequency is 2π − 2M 2 . This frequency is qM k 1 clearly less than the natural frequency 2π M.

3. Let b denote the density of the buoy (weight per unit volume) and ω the density of water. The volume of the buoy is V = 43 πr3 . The volume of a slice of the buoy from its center to a point y units from center is πr2 y − 13 y 3 (exercise). Since the buoy floats half-submerged, b = ω/2. As it bobs up and down let y be the distance from its center to the surface of the water (up is positive). If y > 0, then the net force on the buoy is negative given by the difference between the upward buoyant force of the water: V 1 − πr2 y + πy 3 ), 2 3 and the downward weight of the buoy b · V . Subtracting, the net force is ω(−πr2 y + 31 πy 3 ). Newton’s law (ma = F ) applied to the sphere, at w·(

CHAPTER 5. APPLICATIONS OF THE SECOND-ORDER THEORY its center of mass, yields the following equation (g is the gravitational constant) b · V 00 1 y = −πωr2 y + ωπy 3 . g 3 This is a second-order non-linear differential equation. However, if the buoy is only “slightly” depressed, then the linearized version (ignore the y 3 term) provides an excellent model for the q motion. The linearized equation simplifies to y 0 + a2 y = 0 where a = q 2r 2π 3g seconds.

3g 2r .

The period of the motion is

4. Similar to the solution of Exercise 3. 5. Recall, Section 1.10 problem 4, that inside the Earth the force of gravity on an object is proportional to its distance from the center. Let x be the distance from the train to the center of a tunnel of length 2L. Draw a picture to see that the distance from the train to the center of the Earth √ is x2 + R2 − L2 where R is the radius of the Earth. The magnitude of the force √ on the train, in the direction of the center of the Earth, is then Fc = k x2 + R2 − L2 . The value of k can be found from this equation when the train is at the surface of the Earth: mg = kR, so k = mg/R. The magnitude of the force on the train parallel to the tracks is the comx ponent of Fc in that direction: Fc · cos θ = Fc · √x2 +R = kx. When x 2 −L2 is positive, the force is negative. Applying Newton’s Second Law we have g x00 + R x = 0, and the period of motion is mx00 = −kx = − mg R x. Thus q R independent of L: T = 2π g seconds; this is approximately 90 minutes. The equation of motion for a particular L value is found fromp the initial g conditions: x(0) = L and x0 (0) = 0. This yields x(t) = L cos R t. The pg 0 greatest speed is |x (T /4)| = R L ≈ 4.43L miles per hour. 6. Adjusting the weight with a gravitational constant to be a mass, the differential equation becomes 4

d2 x + 64x = 32 sin 4t . dt2

(∗)

Solving the homogeneous equation as usual, we find that x(t) is either cos 4t or sin 4t. We solve the inhomogeneous equation by undetermined coefficients. Since the obvious guess is already a solution of the homogeneous equation, we instead guess xp (t) = At cos 4t + Bt sin 4t . Substituting this guess into (∗), and solving for the coefficients, we find that 8 xp (t) = − t cos 4t . 5

5.2. NEWTON’S LAW OF GRAVITATION AND KEPLER’S LAWS

Thus the general solution of (∗) is 8 x(t) = − t cos 4t + A cos 4t + B sin 4t . 5 The presence of the t factor guarantees that x will take arbitrarily large values.

5.2

Newton’s Law of Gravitation and Kepler’s Laws

1. Kepler’s Third (a) In astronomy the semi-major axis of the orbit is called the mean distance to the Sun because it is the average of the least and greatest values of r. Let au and Tu denote the semi-major axis and period of Uranus. These are known from Example 2.6.1. According to T2 T2 Kepler’s Third Law, a3u = a3m where am and Tm are Mercury’s semiu m major axis and period. Consequently, being careful with the units–see Example 2.6.1–we have am = (

88 · (3.16 × 107 ) 2/3 Tm 2/3 ) · (2.87 × 1014 ) ) · au = ( 365 Tu 26.516 × 108 = 5.800 × 1012 centimeters.

This is 5.800 × 1012 centimeters. (b) When distance is measured in astronomical units and time in years, 4π 2 then GM = 1 (verify). Therefore, in this system of units, T 2 = a3 . For example, the value of am calculated above can also be found 2/3 88 2/3 (in astronomical units) using am = Tm = ( 365 ) = 0.3874 au. Multiply by 93, 000, 000 to obtain am = 36, 000, 000 miles. 3/2

Regarding Saturn, Ts = as

= (9.54)3/2 = 29.5 years.

3. According to Exercise 2, in the instant after the explosion, the motion of every particle that moves into an elliptical orbit about the Sun obeys the equation v 2 = GM ( 2r − a1 ). Consequently all of these particles move in an orbit with the same semi-major axis, a astronomical units, and (according to Kepler’s Third Law) the same period, T = a3/2 years. This means that T years later all of them will return to their original positions. 4. The formula derived in the text is r=

kCk2 1 · . GM 1 + kKk cos θ

If we let h = kCk and k = GM then we get the desired formula representing Kepler’s first law.

CHAPTER 5. APPLICATIONS OF THE SECOND-ORDER THEORY Formula (22) in the text is R × R0 = C . Taking lengths gives the formula in the exercise. 5. See Exercise 1, part (b). (a) T = 23/2 = 2.83 years. (b) T = 33/2 = 5.20 years. (c) T = 253/2 = 125 years.

Chapter 6

Power Series Solutions and Special Functions 6.1

Introduction and Review of Power Series

j+1 P∞ = limj→∞ (j + 1)|x| = 1. For the series (1): j=1 j! · xj , limj→∞ (j+1)!x j!xj ∞ when x 6= 0. The series converges only when x = 0, R = 0. P∞ j xj+1 /(j+1)! |x| For the series (2): = limj→∞ j+1 = 0 j=0 x /j!, limj→∞ xj /j! for all x. The series converges for all x, R = ∞. P∞ j+1 For the series (3): j=0 xj , limj→∞ | xxj | = limj→∞ |x| = |x|. The series converges when |x| < 1 and diverges when |x| > 1, R = 1.

2. Apply the ratio test to get limj→∞ p−j j+1 · x = |x|. The series converges when |x| < 1 and diverges when |x| > 1. j−1 2j−1 P∞ x ; 3. sin x = j=1 (−1)(2j−1)! x2(j+1)−1 /(2(j + 1) − 1)! |x2 | = lim = 0. 2j−1 j→∞ j→∞ (2j + 1)2j x /(2j − 1)! lim

cos x =

P∞

j=0

(−1)j x2j (2j)! ;

x2(j+1) /(2(j + 1))! |x|2 = lim = 0. j→∞ j→∞ (2j + 2)(2j + 1) x2j /(2j)! lim

4. sin x =

P∞

j=1

(−1)j−1 x2j−1 ; (2j−1)!

d2j d2j−1 j j−1 cos x, so a2j = 0 and a2j−1 = dx2j sin = (−1) sin x, dx2j−1 sin x = (−1) j−1 2n−1 (2n−1) (−1) (−1) cos(ξ) 2n−1 f 2n−1 = x . cos(ξ) is bounded (2j−1)! . R2n+1 (x) = (2n−1)! x (2n−1)!

by 1 and the series goes to 0. 57

CHAPTER 6. POWER SERIES SOLUTIONS cos x =

P∞

j=0

(−1)j x2j (2j)! ;

d2j d2j−1 j j j dx2j cos x = (−1) cos x, dx2j−1 cos x = (−1) sin x, so a2j = (−1) and (2n) 2n a2j−1 = 0. R2n (x) = f (2n)!(ξ) x2n = (−1)(2n)!cos ξ x2n . cos(ξ) is bounded by 1

and the series goes to 0. 5. If |x| < 1, then limn→∞ xn+1 = 0 so n X 1 − xn+1 1 lim xj = lim = . n→∞ n→∞ 1 − x 1 − x j=0 1 = 1 − x + x2 − x3 + · · · . Replace x with −x to verify that 1+x 2

1 Integrate 1+x = 1−x+x2 −x3 +· · · to get ln(1+x) = x− x2 + x3 − x4 +· · · . 3

Then replace x with x2 and integrate to obtain arctan x = x − x3 + x5 − x7 7 − ···. 1 = 1 + x + x2 + x3 + · · · ; 6. (a) 1−x

1 = (1 + x + x2 + x3 + · · · )(1 + x + x2 + x3 + · · · ) (1 − x)2 = 1 + (x + x) + (x2 + x · x + x2 ) + · · · = 1 + 2x + 3x2 + 4x3 + · · · (b) 1 = (1 − x)2

1 1−x

= (1 + x + x2 + x3 + · · · )0 = 1 + 2x + 3x2 + 4x3 + · · · 2

7. (a) If y = 1 − x2! + x4! − x6! + · · · , then y 0 = −x + x3! − x5! + · · · and y 00 = −1 + 2(j+1)

(b) limj→∞ x

x2 x4 − + · · · = −y. 2! 4!

/(22 ·42 ···((2(j+1))2 ) x2j /(22 ·42 ···(2j)2 )

Starting with y =

(−1)j x2j j=0 22 ·42 ···(2j)2

P∞

xy = y0 = xy 00 =

|x | = limj→∞ (2j+2) 2 = 0.

we have

∞ X

(−1)j x2j+1 22 · 42 · · · (2j)2 j=0

∞ X (−1)j · 2j · x2j−1 j=1 ∞ X

22 · 42 · · · (2j)2

(−1)j · 2j · (2j − 1) · x2j−1 22 · 42 · · · (2j)2 j=1

6.2. SERIES SOLUTIONS FOR FIRST-ORDER EQUATIONS

j 2j−1 P∞ x Observe that xy 00 +y 0 = j=1 22 ·4(−1) 2 ···(2(j−1))2 . If the sum is reindexed by replacing j with j + 1, then it its the negative of the sum for xy.

6.2

Series Solution of First-Order Differential Equations

P j 1. Find a power series solution of the form j aj x . Then solve directly using methods from earlier parts of the book. All of the details for the first solution are given. The calculations for equations (b)-(f) given in Pare ∞ less detail. In all cases, the first step is substitution of y = j=0 aj xj . P∞ P∞ j−1 j+1 (a) Since y 0 = and 2xy = . Writing the j=1 jaj x j=0 2aj x 0 differential equation in the form y − 2xy = 0 we have ∞ X

jaj x

j−1

−

∞ X

2aj xj+1 = 0.

j=0

j=1

Reindex the second sum, j → j − 2: ∞ X

jaj xj−1 −

j=1

∞ X

2aj−2 xj−1 = 0,

j=2

then split off the first term from the first sum and move it to the right side ∞ X X jaj xj−1 − xj−1 = a1 . j=2

Equivalently,

∞ X

j=2

(jaj − 2aj−2 )xj−1 = a1 .

j=2

It follows that a1 = 0 and jaj − 2aj−2 ) = 0 for j ≥ 2. This is the recursion relation; a0 can be chosen arbitrarily. Written in the form aj = 2j aj−2 the recursion relation implies that aj = 0 if j is odd. For the even coefficients let a0 = A. Then 1 a2 = 22 A = A, a4 = 42 a2 = − 12 A, a6 = 26 a4 = 2·3 A, a8 = 28 a6 = 1 1 2·3·4 A, and, in general, a2j = j! A. The power series solution is P∞ 2j 2 y = A j=0 xj! . This can be recognized as y = Aex , the same solution that is obtained by separating variables and integrating. P∞ P∞ (b) Substitute to obtain j=1 jaj xj−1 + j=0 aj xj = 1. Reindex the P∞ second sum, j → j−1, and combine to get j=1 (jaj +aj−1 )xj−1 = 1. It follows that a1 + a0 = 1 and jaj + aj−1 = 0 for j ≥ 2; a0 can be chosen arbitrarily. To obtain a solution formula, let a0 = A so a1 = 1 − A. Then write the recursion equation in the form aj =

CHAPTER 6. POWER SERIES SOLUTIONS 1 − 1j aj−1 to get a2 = − 21 a1 = − 12 (1 − A), a3 = − 31 a2 = 2·3 (1 − A), j

1 (1−A) and, in general, aj = − (−1) a4 = − 14 a3 = − 2·3·4 j! (1−A), j ≥ 1. P∞ j The power series solution is y = A − (1 − A) j=1 (−1)j xj! . This can be recognized as y = 1 − (1 − A)e−x , which is equivalent to the solution obtained by solving the equation as first order linear.

(c,d,e) Problems (c), (d), (e) are very similar to (b). P∞ P∞ (f) Substitute to obtain j=1 jaj xj−1 − j=0 aj xj = x2 . Reindex the second and combine the two sums to get the equation, P∞ sum, j → j −1, j−1 (ja −a )x = x2 . It follows that a1 −a0 = 2, 2a2 −a1 = 0, j j−1 j=1 3a3 − a2 = 1, and jaj − aj−1 = 0 for j ≥ 4; a0 can be chosen arbitrarily. To obtain a solution formula, let a0 = A so a1 = a0 = A, a2 = 1 1 1 A 1 1 1 1 2 a1 = 2 A, and a3 = 3 (1 + a2 ) = 3 (1 + 2 ) = 3 + 3·2 A = 3·2 (2 + A). Now write the last recursion equation in the form aj = 1j aj−1 to 1 1 get a4 = 41 a3 = 4·3·2 (2 + A), a5 = 15 a4 = 5·4·3·2 (2 + A) and, in 1 general, aj = j! (2 + A) when j ≥ 3. The power series solution is P∞ j y = A + Ax + 21 Ax2 + (2 + A) j=3 xj! . This can be rearranged into y = Aex + 2(ex − 1 − x − 21 x2 ) or y = Cex − 2 − 2x − x2 , where C = A+2. This is also the solution that is obtained using the method of undetermined coefficients. 2. Find a power series solution and then solve the equation directly. P∞ P∞ (a) Substitute to obtain j=1 jaj xj = j=0 aj xj . Combine these two P∞ sums to obtain j=1 (jaj − aj )xj = a0 . It follows that a0 = 0 and (j −1)aj = 0, j ≥ 1. So aj = 0, j ≥ 1, j 6= 1 and a1 could be arbitrary. The solution is y = a1 x. P−1 P∞ P−1 (b) Assume y = −∞ aj xj + j=0 aj xj . Then y 0 = j=−∞ jaj xj−1 + P∞ P P∞ −1 j−1 . Substitute to obtain j=−∞ jaj xj+1 + j=1 jaj xj+1 − j=1 jaj x P−1 P∞ j j −∞ aj x − j=0 aj x = 0. Reindex the first and second sums, P0 P∞ j → j − 1, to obtain j=−∞ (j − 1)aj−1 xj + j=2 (j − 1)aj−1 xj − P P−1 ∞ j j j=−∞ aj x − j=0 aj x = 0. Combine sums and simplify to obtain P∞ P−1 j j j=−∞ [(j−1)aj−1 +aj ]x + j=2 [(j−1)aj−1 −aj ]x = a−1 +a0 +a1 x. It follows that (j−1)aj−1 −aj = 0, j ≤ −1, (j−1)aj−1 −aj = 0, j ≥ 2, a−1 + a0 = 0 and a1 = 0; a0 can be arbitrary. To obtain a solution formula, let a0 = A so a−1 = −A. Then write the recursion equation aj A A to get a−2 = A2 , a−3 = − 3·2 , a−4 = 4·3·2 , in the form aj−1 = j−1 A jA a−5 = − 5·4·3·2 and, in general, a−j = (−1) j! , j ≥ 1. THe power sej P−1 A j ries solution is y = j=−∞ (−1) x + A, which can be recognized as j! y = Ae−1/x , which is equivalent to the solution obtained by solving the equation as a separable equation.

6.2. SERIES SOLUTIONS FOR FIRST-ORDER EQUATIONS

P∞ P∞ (c) Substitute to obtain j=1 jaj xj−1 − x1 · j=1 aj xj−1 = x2 . Simplify P∞ P ∞ to get j=1 jaj xj−1 − ax0 − j=1 aj xj−1 = x2 . Combine sums to P j−1 get = x2 + ax0 . It follows that a0 = 0, (j − j=1 (j − 1)aj x 1)aj = 0, j ≥ 1, j 6= 3, and (3 − 1)a3 = 1. Consequently, a1 can be arbitrary, a3 = 1/2 and aj = 0, j 6= 1, 3. The powers series solution is y = 21 x3 + c1 x, which is equivalent to the solution obtained by solving the equation as first order linear. P∞ (d) Similar to (c), substitute and simplify to obtain j=1 (j −1)aj xj−1 = x+ ax0 . It follows that a0 = 0, (j−1)aj = 0, j ≥ 1, j 6= 2, (2−1)a2 = 1. Consequently, a0 = 0, a1 can be arbitrary, a2 = 1 and aj = 0, j ≥ 3. The power series solution is y = x2 + a1 x, which is equivalent to the solution obtained by solving the equation as a separable equation. 3. Solve y 0 = (1 − x2 )−1/2 in two different ways. Method 1. Using the binomial series: (1 + x)p = 1 + px + =1+

p(p − 1) 2 p(p − 1) · · · (p − j + 1) j x + ··· + x + ··· 1·2 j!

∞ X p(p − 1) · · · (p − j + 1)

j=1

we have (1 − x2 )−1/2 = 1 +

=1+

∞ X (− 1 ) · (− 3 ) · · · (− 2j−1 2

j=1 ∞ X

(−x2 )j

1 · 3 · · · (2j − 1) 2j x . 2j j! j=1

Integrate y 0 = (1 − x2 )−1/2 term by term to obtain the general solution in P∞ 2j+1 power series from: y = x+ j=1 1·3···(2j−1) +C. The arcsine function 2j j!(2j+1) x is obtained by setting C = 0 to obtain the solution to the differential equation satisfying y(0) = 0: arcsin x = x +

∞ X 1 · 3 · · · (2j − 1) j=1

2j j!(2j + 1)

x2j+1 , |x| < 1.

The formula for π/6 is obtained by substituting x = 1/2. Note that 2j j! = 2 · 4 · · · 2j. Method 2. Using the power series method to solve y 0 = (1 − x2 )−1/2 we P∞ could start off by substituting y = j=0 aj xj into the equation y0 = 1 +

∞ X 1 · 3 · · · (2j − 1) j=1

2j j!

x2j

CHAPTER 6. POWER SERIES SOLUTIONS to get

∞ X

jaj xj−1 = 1 +

j=1

∞ X 1 · 3 · · · (2j − 1) j=1

2j j!

x2j .

The reader is invited to finish. Hint. First observe that a0 can be chosen arbitrarily, let a0 = A. Then let bj denote the j th coefficient for the series on the right side. The equation (j + 1)aj+1 = bj , j ≥ 0, serve as the recursion relations. 4. Find the coefficients aj in two ways: P∞ P∞ (a) Since y = j=0 aj xj , y 0 = j=1 jaj xj−1 and   ∞ m X X  y2 = aj am−j  xm . m=0

j−1 = 1+ j=1 jaj x

P∞

m j=0 aj am−j

xm . m=0 P∞ Reindex the first sum, j → m + 1, to obtain m=0 (m + 1)am+1 xm = P∞ Pm m 2 1 + m=0 j=0 aj am−j x . It follows that a1 = 1 + a0 , (m + Pm 1)am+1 = so a = 1. j=0 aj am−j . Since tan(0) = 0, a0 = 0 Pm 1 1 Write the recursion equation in the form am+1 = m+1 j=0 aj am−j to get a2 = 0, a3 = 13 (a0 a2 + a21 + a2 a0 ) = 1/3, a4 = 41 (a0 a3 + 2 a1 a2 + a2 a1 + a3 a0 ) = 0, a5 = 15 . Thus, the power series of tan x is 1 3 2 5 x + 3 x + 15 x + · · · . Since y(0) = 0, y 0 (0) = 1 + 02 = 1 and a1 = 0. y 00 = 2yy 0 implies y 00 (0) = 2y(0)y 0 (0) = 0 and a2 = 0. Similarly, y 000 = 2yy 00 + 2(y 0 )2 2 implies y 000 (0) = 2 and a3 = 3! = 13 . y (4) = 2y 0 y 00 + 2yy 000 + 4y 0 y 00 (4) (5) 00 2 0 000 0 000 (−4) 00 00 Substitute to obtain

(b)

P∞

j=0

implies y (0) = 0. y = 2(y ) +2y y +2y y +2yy +4y y + 2 4y 0 y 000 implies y (5) (0) = 16 and a5 = 16 = . Therefore, the power 5! 15 2 5 series for tan x is x + 13 x3 + 15 x + ···. P∞ 5. First solve y 0 = x − y, y(0) = 0 by substituting y = j=0 aj xj . Note that y(0) =P0. Substituting, rearranging, and reindexing produces P∞ a0 = j−1 P∞ ∞ j−1 j−1 ja x + a x = x, so (ja = x. Conj j + aj−1 )x j=1 j=1 j−1 j=1 sequently, a1 + a0 = 0 so a1 = 0 also. For j = 2, 2a2 + a1 = 1, so a2 = 21 . Now use jaj + aj−1 = 0, j ≥ 3, in the form aj = − 1j aj−1 j

1 1 to obtain a3 = − 3·2 , a4 = 4·3·2 , · · · , aj = (−1) j! , · · · . The solution is P∞ j x y = j=2 (−1)j j! = e−x − 1 + x. The algorithm for first order linear equation yields the same solution.

Now solve the IVP using the method of repeated differentiation as described in problem 4(b). We know that y(0) = 0. Since y 0 (0) = 0 − y(0), y 0 (0) = 0 also. For the higher order terms, repeated differentiation yields y 0 = x − y, y 00 = 1 − y 0 , y 000 = −y 00 , and y (j) = −y (j−1) , j ≥ 3.

6.3. ORDINARY POINTS

Therefore, y 00 (0) = 1 − y 0 (0) = 1, y 000 (0) = −y 00 (0) = −1, y (4) (0) = −y (3) (0) = 1, and so on. Consequently, a0 = y(0) = 0, a1 = y 0 (0) = 0, a2 = y 00 (0)/2! = 1/2, and aj = (−1)j y (j) (0)/j! = (−1)j /j! for j ≥ 3, as above. 6. **

6.3

Ordinary Points

P∞ 1. Make sure p and q are real analytic at 0, then substitute y = j=0 aj xj . We work the first problem in detail and summarize the results for the remaining five. In each case a0 = A and a1 = B. P∞ (a) p(x) = x and q(x) = 1. Substitution yields j=2 j(j − 1)aj xj−2 + P∞ P ∞ j xj = 0 which, after reindexing the first sum, j=0 jaj x + j=0 ajP ∞ can be rearranged to j=0 [(j + 2)(j + 1)aj+2 + (j + 1)aj ]xj = 0. Consequently, (j + 2)(j + 1)aj+1 + (j + 1)aj = 0 for all j ≥ 0. 1 aj , and startUsing the recursion relation in the form aj+2 = − j+2 1 1 ing with a0 = A we have a2 = − 2 A, a4 = 2·4 A, and in general, j 2j P∞ (−1)j x a2j = 2·4···(2j) A. One solution is y1 = A(1 + j=1 (−1) 2·4···(2j) ) = j 2j P∞ x 1 1 A j=0 (−1) 2j j! . Starting with a1 = B we get a3 = − 3 B, a5 = 3·5 B, j−1

(−1) B. Another solution is y2 = and in general a2j−1 = 1·3···(2j−1) P∞ (−1)j−1 x2j−1 B j=1 1·3···(2j−1) and the complete general solution, in power sej−1 2j−1 j 2j P∞ P∞ x x + B j=1 (−1) ries form, is y = A j=0 (−1) 2j j! 1·3···(2j−1) .

(b) p(x) = −1, q(x) = x. Substitute and rearrange to ∞ X

[j(j − 1)aj − (j − 1)aj−1 + aj−3 ]xj−2 = a1 − 2a2 .

j=3

Consequently, a2 = 12 B, and the remaining coefficients can be ob(j−1)aj−1 −aj−3 tained using the three term relation aj = , j ≥ 3. j(j−1) Solution: 1 1 1 1 y = A(1 − x3 − x4 − · · · ) + B(x + x2 + x3 + · · · ). 6 24 2 6 (c) p(x) = 2x, q(x) = −1. Substitute and rearrange to ∞ X [(j + 2)(j + 1)aj+2 + (2j − 1)aj ]xj = a0 − 2a2 + x. j=1

Therefore, a2 = 21 A and 6a3 + a1 = 1, implying that a3 = 16 (1 − B). 2j−1 When j ≥ 2, aj+2 = − (j+2)(j+1) aj . Solution: 1 1 1 1 y = A(1 + x2 + · · · ) + B(x − x3 + · · · ) + x3 − x5 + · · · . 2 6 6 24

CHAPTER 6. POWER SERIES SOLUTIONS (d) p(x) = 1, q(x) = −x2 . Substitute and rearrange to ∞ X [j(j −1)aj +(j −1)aj−1 −aj−4 ]xj−2 = −2a2 −a1 +1−(6a3 +2a1 )x. j=4

Consequently, a2 = 21 (1 − B) and a3 = − 13 B. When j ≥ 4, aj = aj−4 −(j−1)aj−1 . Solution: j(j−1) y = A(1 +

1 4 1 1 1 x + · · · ) + B(x − x2 + · · · ) + x2 − x3 + · · · . 12 2 2 6

(e) p(x) = x/(1 + x2 ) and q(x) = 1/(1 + x2 ). Substitute and rearrange to ∞ X [(j + 2)(j + 1)aj+2 + (j 2 + 1)aj ]xj = −(2a2 + a0 ) − (6a3 + 2a1 )x. j=2

Consequently, a2 = − 21 A and a3 = − 13 B. If j ≥ 2, then aj+2 = 2

j +1 − (j+2)(j+1) aj . Solution:

5 1 1 1 y = A(1 − x2 + x4 + · · · ) + B(x − x3 + x5 + · · · ). 2 24 3 6 (f) p(x) = 1 + x, q(x) = −1. Substitute and rearrange to ∞ X

[(j + 2)(j + 1)aj+2 + (j + 1)aj+1 + (j − 1)aj ]xj = a0 − a1 + 2a2 .

j=1

Therefore, a2 = 21 (A−B), and when j ≥ 1, aj+2 = − Solution:

(j+1)aj+1 +(j−1)aj . (j+2)(j+1)

1 1 1 1 y = A(1 + x2 − x3 + · · · ) + B(x − x2 + x3 + · · · ). 2 6 2 6 2. Substitute and rearrange to ∞ X

[(j + 2)(j + 1)aj+2 + (j + 2)(j − 1)aj ] xj + (2a2 − 2a0 ) + 6a3 x = 0.

j=2

If a0 = A and a1 = B, then a2 = A, ai = 0 for all odd i ≥ 3, a4 = − 4−1 4+1 = − 35 A, a6 = 37 A, a8 = − 93 A. The solution is: 3 3 y = A(1 + x2 − x4 + x6 − · · · ) + Bx. 5 7 3. Consider y 00 + xy 0 + y = 0.

6.3. ORDINARY POINTS (a) Substitute y = ∞ X

P∞

j j=0 aj x , reindex

(j + 2)(j + 1)aj+2 xj +

j=0

∞ X

jaj xj +

j=1

∞ X

aj xj = 0.

j=0

and rearrange to ∞ X [(j + 2)(j + 1)aj+2 + (j + 1)aj ]xj = −(2a2 + a0 ). j=1 1 aj for j ≥ 1. If a0 = 1 Consequently, a2 = − 21 a0 and aj+2 = − j+2 1 and a1 = 0, then aj = 0 if j is odd. Also a4 = − 41 a2 = 2·4 , j

(−1) 1 , and, in general, a2j = 2·4···(2j) . Thus, one a6 = − 16 a4 = − 2·4·6 P∞ (−1)j x2j solution is y1 (x) = j=0 2j j! . A similar calculation, starting P∞ (−1)j−1 x2j−1 with a0 = 0 and a1 = 1, yields y2 (x) = j=1 1·3·5···(2j−1) as a second, linearly independent, solution.

(b) The ratio test applied to the series for y1 (x): x2(j+1) /(2j+1 (j + 1)!) x2 = lim = 0, j→∞ j→∞ 2(j + 1) x2j /(2j j!) lim

shows that the radius of convergence is ∞. A similar calculation will show that the series for y2 (x) also converges for all x. P∞ (−x2 /2)j (c) Write the series defining y1 as to see that y1 (x) = j=0 j! 2

e−x /2 . According to the method developed in Section 2.4 there is a solution y3 of the form Z Z R 1 1 −x2 /2 − p(x)dx −x2 /2 y1 (x) e dx = e e dx. 2 −x y1 (x) e 2 R 2 2 Thus y3 (x) = e−x /2 ex /2 dx. Choosing the antiderivative so its Rx 2 value is 0 at x = 0, say 0 et /2 dt, will make y3 the solution of the differential equation satisfying the initial conditions y(0) = 0 and y 0 (0) = 1. Since y2 satisfies the same conditions, y2 = y3 . 4. Substitute and rearrange to ∞ X

[(j + 2)(j + 1)aj+2 + (j + 1)aj+1 − aj−1 ] xj + 2a2 + a1 = 0.

j=1 B−A If a0 = A, a1 = B, then a2 = − B2 , a3 = A+B 6 , a4 = 24 , and so on. The solution is

1 1 1 1 y = A(1 + x3 − x4 + · · · ) + B(x − x2 + x3 + · · · ). 6 24 2 6

CHAPTER 6. POWER SERIES SOLUTIONS The initial condition y1 (0) = 1 implies that A = 1. The initial condition y20 (0) = 1 implies that B = 1. Therefore, 1 1 1 1 y = (1 + x3 − x4 + · · · ) + (x − x2 + x3 + · · · ). 6 24 2 6 5. Investigate power series solution to y 00 + (p + 1/2 − x2 /4)y = 0. (a) Substitute y =

P∞

j j=0 aj x . Then reindex and combine to obtain

∞ X 1 1 (j + 2)(j + 1)aj+2 + (p + )aj − aj−2 xj = 0, 2 4 j=0 where, by convention, a−2 = a−1 = 0. 2

(b) If y = we−x /4 , then 2

y 0 = w0 e−x /4 −

x −x2 /4 we 2

2 2 2 2 1 x2 y 00 = w00 e−x /4 − xw0 e−x /4 − we−x /4 + we−x /4 . 2 4

Substitute and simplify, eventually canceling the exponential terms, to obtain the desired equation: w00 − xw0 + pw = 0. P∞ (c) Substitute w = j=0 bj xj . Reindex and rearrange to ∞ X [(j + 2)(j + 1)bj+2 + (p − j)bj ]xj = 0. j=0 p−j The two-term recursion formula is bj+2 = − (j+2)(j+1) bj , j ≥ 0. p(p−2) p Starting with b0 = A, b2 = − 2·1 A, b4 = − p−2 4·3 b2 = 4! A, b6 = p(p−2)(p−4) p−4 A, and so on. One solution is − 6·5 b4 = − 6!

p p(p − 2) 4 y1 = A 1 − x2 + x − ··· . 2! 4! Starting with b1 = B, the second solution is p − 1 3 (p − 1)(p − 3) 5 x + x − ··· . y2 = B x − 3! 5! 6. Airy’s equation y 00 + xy = 0. R∞ R∞ (a) q(x) = x. When x > 0, q(x) > 0 and 1 q(x)dx = 1 xdx = +∞. By Theorem 3.3.3, y has infinitely many positive zeros. If x < 0, q(x) < 0. By Theorem 3.3.2, y has at most one negative zero.

6.3. ORDINARY POINTS

(b) Substitute and rearrange to ∞ X

[j(j − 1)aj + aj−2 ]xj−2 + 2a2 = 0.

j=3 B , a5 = 0, If a0 = A and a1 = B, then a2 = 0, a3 = − A6 , a4 = − 12 A a6 = − 180 , and so on. The solution is:

1 1 5 1 y = A(1 − x3 − x − · · · ) + B(x − x4 − · · · ). 6 180 12 (c) If y is the solution to y 00 +xy = 0, then −y is the solution of y 00 −xy = 0 (verify). The solution is 1 5 1 1 x − · · · ) − B(x − x4 − · · · ). y = −A(1 − x3 − 6 180 12 7. Chebyshev’s equation: (1 − x2 )y 00 − xy 0 + p2 y = 0. P∞ (a) Substitute y = j=0 aj xj . Reindex and rearrange to ∞ X

[(j + 2)(j + 1)aj+2 + (p2 − j 2 )aj ]xj = 0.

j=0 2

p −j aj the general soluUsing the recursion relation aj+2 = − (j+2)(j+1) tion is p2 (p2 − 22 ) 4 p2 x − ··· + y =A 1 − x2 + 2! 4! 2 p − 1 3 (p2 − 1)(p2 − 32 ) 5 B x− x + x − ··· . 3! 5!

(b) If p is an even integer, then the first series terminates in a polynomial of degree p. If p is an odd integer, then the second series terminates in a polynomial of degree p. 8. Hermite’s equation y 00 − 2xy 0 + 2py = 0 where p is a constant. (a) Substitute and rearrange to ∞ X

[j(j − 1)aj − 2(j − 2)aj−2 + 2paj−2 ]xj−2 + 2a2 + 2pa0 = 0.

j=3 2(p−1) If a0 = A and a1 = B, then a2 = − 2pA 2 , a3 = − 3! B, a4 = ( 22 p(p−2)A , a5 = 2 p−3)(p−1) B, and so on. 4! 5!

the same as stated in the problem.

Therefore, y1 and y2 are

CHAPTER 6. POWER SERIES SOLUTIONS (b) If p = 0, y1 (x) = 1. If p = 1, y2 (x) = x. If p = 2, y1 (x) = 1 − 2x2 . If p = 3, y2 (x) = x − 32 x3 . If p = 4, y1 (x) = 1 − 4x2 + 34 x4 . If p = 5, 4 5 y2 (x) = x − 43 x3 + 15 x . (c) H0 (x) = 1 H1 (x) = 2x H2 (x) = −2(1 − 2x2 ) = 4x2 − 2 H3 (x) = −12(x − 2x3 /3) = 3x3 − 12x H4 (x) = 12(1 − 4x2 + 4x3 /3) = 16x4 − 48x2 + 12 H5 (x) = 120(x − 4x3 /3 + 4x5 /15) = 32x5 − 160x3 + 120x (d) 2

H0 (x) = (−1)0 ex e−x = 1 2 2 d e−x = 2x H1 (x) = (−1)ex dx 2 2 x2 d H2 (x) = (−1) e [e−x (−2x)] = 4x2 − 2 dx 2 2 3 x2 d H3 (x) = (−1) e (4x2 e−x − 2e−x ) = 8x3 − 12x dx 2 2 4 x2 d (12e−x − 8x3 e−x ) = 12 − 48x2 + 16x4 H4 (x) = (−1) e dx 2 2 2 5 x2 d (12e−x − 48x2 e−x + 16x4 e−x = 32x5 − 160x3 + 120x H5 (x) = (−1) e dx 9. **

6.4

Regular Singular Points

1. Locate and classify the singular points. 3 (a) p(x) = − x23 , q(x) = x2 (x−1) . The singular points are 0 and 1, 1 is regular and 0 is not. 1 (b) p(x) = x(x+1) , q(x) = x2 (x22 −1) . The singular points are 0 and ±1, all three are regular.

(c) p(x) = − x−2 x2 , q(x) = 0. The singular point is 0 which is not regular. x+1 2 (d) p(x) = − x(3x+1) , q(x) = x(3x+1) . The singular points are 0 and −1/3, both are regular.

2. Determine the nature of the point x = 0 for each of the following differential equations.

6.4. REGULAR SINGULAR POINTS

(a) p(x) = 0, q(x) = sin x so x = 0 is a regular point. (b) p(x) = 0, q(x) = sinx x , xp(x) = 0, x2 q(x) = x sin x so x = 0 is a regular singular point. x 2 (c) p(x) = 0, q(x) = sin x2 , xp(x) = 0, x q(x) = sin x so x = 0 is a regular singular point. x sin x 2 (d) p(x) = 0, q(x) = sin so x = 0 is a x3 , xp(x) = 0, x q(x) = x irregular singular point. sin x x 2 (e) p(x) = 0, q(x) = sin x4 , xp(x) = 0, x q(x) = x2 so x = 0 is a irregular singular point.

3. Find the indicial equation and its roots. Note that if the equation is in q(x) 0 the form y 00 + p(x) x y + x2 y = 0, then the indicial equation is m(m − 1) + p(0)m + q(0) = 0. (a) p(x) = cos x2x−1 , q(x) = 2, p(0) = −2 (use p’s Taylor series) and 2 q(0) = 2 so the indicial equation is m(m − 1) − 2m + 2 = 0. The roots are m1,2 = 1, 2. 3

(b) p(x) = 2x 4−5 , q(x) = 3x 4+2 . p(0) = −5/4 and q(0) = 1/2 so the indicial equation is m(m−1)− 54 m+ 12 = 0. The roots are m1,2 = 14 , 2. (c) p(x) = 3, q(x) = 4x. p(0) = 3 and q(0) = 0 so the indicial √ equation is m(m − 1) − 4m + 3 = 0. The roots are m1,2 = 5/2 ± 13/2. 4. Verify that the origin is a regular singular point and calculate two independent Frobenius series solutions: 3 (a) p(x) = 4x3 2 , xp(x) = 4x so x = 0 is an irregular singular point. x 2 (b) xp(x) = 3−x 0 is a regular singular point. 2 , x q(x) = − 2 so x =P ∞ Substitute, reindex, and rearrange to j=1 [(m + j)(2m + 2j + 1)aj − m+j−1 m−1 (m + j)aj−1 ]x + a0 x [2m(m − 1) + 3m] = 0. The indicial equation is 2m(m − 1) + 3m = 0 with solutions m1,2 = − 12 , 0. The 1 · aj−1 , j ≥ 1. recursion equation is aj = 2m+2j+1 1 1 When m = − 21 , aj = 2j aj−1 . If a0 = A, then a1 = 12 A, a2 = 2!2 A, 1 a3 = 23 3! A, etc. 1 1 When m = 0, aj = 2j+1 · aj−1 . If a0 = B, then a1 = 31 B, a2 = 5·3 B, 1 a3 = 7·5·3 B, etc.

The solution is y = Ax−1/2 (1 + 21 x + 2212! x2 + 2313! x3 + · · · ) + B(1 + 1 1 2 1 3 3 x + 5·3 x + 7·5·3 x + · · · ). 3x 2 (c) xp(x) = x+1 0 is a regular singular point. 2 , x q(x) = 2 so x = P ∞ Substitute, reindex, and rearrange to j=1 [(m + j)(2m + 2j − 1)aj + (m + j + 2)aj−1 ]xm+j−1 + a0 xm−1 [2m(m − 1) + m] = 0. The indicial equation is 2m(m − 1) + m = 0 with solutions m1,2 = 0, 12 . The m+j+2 recursion equation is aj = − (m+j)(2m+2j−1) aj−1 .

CHAPTER 6. POWER SERIES SOLUTIONS j+2 When m = 0, aj = − j(2j−1) aj−1 . If a0 = A, then a1 = −3A, 2 a2 = 2A, a3 = − 3 A, etc. 2j+5 7 aj−1 . If a0 = B, then a1 = − 3! B, When m = 12 , aj = − 2j(2j+1) 9·7 11·9·7 a2 = 5! B, a3 = − 7! B, etc. 1

7 The solution is y = A(1 − 3x + 2x2 − 23 x3 + · · · ) + Bx 2 (1 − 3! x+ 9·7 2 11·9·7 3 x − x + · · · ). 5! 7!

so x = 0 is a regular singular point. (d) xp(x) = 1/2, x2 q(x) = − x+1 2 P[ Substitute, reindex, and rearrange to j=0 [2(m + j)2 − (m + j) + 1]aj − aj−1 ]xm+j + a0 xm [2m(m − 1) + m] = 0. The indicial equation is 2m(m − 1) + m = 0 with solutions m1,2 = 0, 1/2. The recursion 1 equation is aj = 2(m+j)2 −(m+j)+1 aj−1 . 1 1 aj−1 . If a0 = A, then a1 = 21 A, a2 = 14 A, When m = 0, aj = 2j 2 −j+1 1 1 a3 = 16 · 14 · A, etc. 1 aj−1 . If a0 = B, then a1 = 14 B, When m = 1/2, aj = 2j 2 +j+1 1 1 a2 = 11·4 B, a3 = 22·11·4 B, etc. 1

1 2 1 The solutions is y = A(1 + 12 x + 14 x + 16·14 x3 + · · · ) + Bx 2 (1 + 1 1 1 2 3 4 x + 11·4 x + 22·11·4 x + · · · ). 2

5. Write the equation in the form y 00 + x1 y 0 + xx2 y = 0 to see that the indicial equation is m(m − 1) + m = 0 so the indicial roots are m1,2 =P0, 0. This ∞ suggests that there is an ordinary power series solution: y = j=0 aj xj . P∞ 2 j Substitute, reindex, and rearrange to j=1 [j aj + aj−2 ]x = 0 where we use the convention that a−1 = 0. Consequently, a0 can be chosen arbitrarily, a1 = 0, and aj = − j12 aj−2 for all j ≥ 2. Start with a0 = 1 to obtain the series displayed in the text. 6. Consider the differential equation y 00 + x12 y 0 − x13 y =. (a) xp(x) = x1 , x2 q(x) = − x1 so x = 0 is an irregular singular points. hR i R 1 1 − x12 dx (b) By Section 2.4, y2 (x) = e dx · x = −xe x . 2 x 1

1 1 −n (c) y2 (x) = −xe x = −x(1 + x1 + 2!x term for 2 + 3!x3 + · · · ). It has x any n > 0 and thus cannot be written as a Frobenius series.

7. Frobenius solutions for y 00 + xpb y 0 + xqc = 0, p and q nonzero real numbers. q 3 00 0 (a) Consider y 00 + xp2 y 0 +P x3 = 0. Write it in the form x y +pxy +qy = 0 ∞ and substitute y = j=0 aj xm+j . Since

qy = q(a0 xm + a1 xm+1 + · · · ) pxy 0 = p(a0 mxm + a1 (m + 1)xm+1 + · · · ) x3 y 00 = a0 m(m − 1)xm−1 + a1 (m + 1)mxm+2 + · · ·

6.5. MORE ON REGULAR SINGULAR POINTS

we obtain (pm + q)a0 xm + [m(m − 1)a0 + (p(m + 1) + q)a1 ]xm+1 + · · · = 0. Therefore, a0 can be chosen arbitrarily provided pm + q = 0. This is the indicial equation for this case and there is one possible value for the exponent, m = −q/p. (b) Consider now the general case y 00 + px−b y 0 + qx−c y = 0. THe same substitution that was made in part (a) now entails qx−c y = q(a0 xm−c + a1 xm+1−c + · · · ) px−b y 0 = p(a0 mxm−1−b + a1 (m + 1)xm−b + · · · ) y 00 = a0 m(m − 1)xm−2 + a1 (m + 1)mxm−1 + · · · Add these up, and cancel xm from each term, to obtain the following equation qa0 x−c + pma0 x−b−1 + a0 m(m − 1)x−2 + qa1 x1−c + p(m + 1)a1 x−b + a1 (m + 1)mx−1 + · · · = 0.

(6.1)

If b = 1 and c ≤ 2, then after multiplying Equation 6.1 by x2 it can be rearranged to [m(m − 1) + pm]a0 + qa0 x2−c + higher order terms = 0. Thus we can hope for two values of m for Frobenius solutions. If b = 1 and c > 2, then it is clear from the first term in Equation 6.1 that a0 = 0 and there is no hope for a Frobenius solution. If b 6= 1 and c = b + 1, then there will be one value of m that might yield a Frobenius solution, as in part (a). Otherwise, and still assuming b 6= 1, there will be no Frobenius solutions because the three leading terms in Equation 6.1 force a0 = 0. P∞ 8. Substitute, reindex, and rearrange to j=1 [(m+j)2 aj−1 −(m+j)aj ]xm+j−1 − ma0 xm−1 = 0. It implies that m = 0 and aj = jaj−1 . If a0 = A, then a1 =PA, a2 = 2!A, a3 = 3!A, and in general aj = j!xj . Therefore, ∞ y = j=0 j!xj .

6.5

More on Regular Singular Points

1. Since p(x) = −3 and q(x) = 4x + 4, the indicial equation is m(m − 1) − 3m + 4 = 0. P The exponents are m1,2 = 2, 2 and there is a solution of the ∞ form y = x2 j=0 aj xj . Substitute, reindex, and rearrange to ∞ X (j 2 aj + 4aj−1 )xj = 0. j=0

CHAPTER 6. POWER SERIES SOLUTIONS We are using the convention that a−1 = 0. Consequently, a0 can be chosen arbitrarily, say a0 = 1, and the rest of the coefficients are found using the relation aj = − j42 aj−1 , j ≥ 1. The Frobenius solution is y = P∞ 4n j x2 j=0 (−1)n (n!) 2x . 2

(4x +1)/4 2. Since p(x) = −2x , the indicial equation is m(m − x and q(x) = x2 1 1) + 4 = 0. The exponents are m1,2 = 12 and there is a solution of the 1 P∞ form y = x 2 j=0 aj xj . Substitute, reindex, and rearrange to ∞ X 1 [4j 2 aj − 8(j − aj−1 + 4aj−2 ]xm+j + 4(a1 − a0 )xm+1 = 0. 2 j=2

Consequently, a0 can be chosen arbitrarily, say a0 = 1. Then a1 = a0 = 1 (2j−1)aj−1 −aj−2 and aj = , j ≥ 3. Thus, a2 = 21 A, a3 = 16 A, etc. The j2 1

Frobenius solution is y = x 2 (1 + x + 12 x2 + 61 x3 + · · · ). 3. Find two independent Frobenius solutions. (a) p(x) = 2 and q(x) = x2 . The indicial equation is M 2 +P m = 0 so the ∞ exponents are m1,2 = 0, −1. We will substitute y = j=0 aj xm+j with the hope that both solutions can be obtained using m2 = −1. IF that is not the case, then we will at least have the solution y + 1 corresponding to m1 = 0 and can build an independent solution using the integral formula for a second independent solution. Substitute, reindex, and simplify to obtain the following sum X [(m + j)(m + j + 1)aj + aj−2 ]xj−1 = 0. j=0

We are using the convention that a−1 and a−2 are both zero. The recursion relations start with j = 0 =⇒ m(m + 1)a0 = 0 j = 1 =⇒ (m + 1)(m + 2)a1 = 0

(6.2) (6.3)

Equation 6.2 confirms that m = 0 and m = −1 are the exponents. Either value for m will allow us to choose a0 arbitrarily. Moreover, if we choose m = −1, then Equation 6.3 implies that a1 can also be chosen arbitrarily. We get two independent Frobenius series solutions using m = −1. We got lucky. This is pure luck. Sometimes it happens, sometimes it does not. Unfortunately, there is no way to tell in advance that it will happen. In view of what we have just seen, we let m = −1 and use the 1 aj−2 , j ≥ 2, to determine the rest of recursion relation aj = − j(j−1)

6.5. MORE ON REGULAR SINGULAR POINTS

the coefficients. Independent solutions are generated by first starting out with a0 = 1 and a1 = 0, an then starting out with a0 = 0 and a1 = 1. a0 = 1, a1 = 0 j

1 1 , a4 = 4·3·2·1 , and, in general, a2j = (−1) In this case, a2 = − 2·1 (2j)! . j P ∞ 2j = cosx x . The solution is y1 = x−1 j=0 (−1) (2j)! x a0 = 0, a1 = 1 j

(−1) 1 1 In this case, a3 = − 3·2 , a5 = 5·4·3·2 and, in general, a2j+1 = (2j+1)! . J P ∞ (−1) x2j = sinx x . The solution is y2 = x−1 j=0 (2j+1)!

(b) p(x) = −x, q(x) = x2 − 2. The indicial equation is m(m − 1) − 2 = 0 and m1,2 = 2, −1. Substitute and simplify to obtain ∞ X

[(m + j − 2)(m + j + 1)aj − (m + j − 1)aj−1 + aj−2 ]xj = 0.

j=0

This is correct under the assumption that a−1 and a−2 are both zero. The exponents are obtained from the coefficients when j = 0: (m−2)(m+1)a0 = 0. THis confirms that the exponents are 2 and −1 as expected. Let’s investigate what happens when we set m = −1. The recursion relation in this is j(j − 3)aj − (j − 2)aj−1 + aj−2 = 0, and the coefficient calculations for j = 0, 1, 2, 3 go like this: j = 0 =⇒ 0 · a0 = 0, j = 1 =⇒ −2 · a1 + a0 = 0, j = 2 =⇒ −2 · a2 + a0 = 0, j = 3 =⇒ 0 · a3 − a2 + a1 = 0,

so a0 is arbitrary 1 so a1 = a0 2 1 so a2 = a0 2 so a3 is arbitrary

We got lucky again. Observe that once j > 3 the recursion relation (j−2)aj−1 −aj−2 can be rearranged to aj = allowing us to calculate the j(j−3) remaining coefficients in terms of a0 and a3 . Independent solutions y1 and y2 arise from a0 = 1, a3 = 0 and a0 = 0, a3 = 1 respectively. x x2 + + ···) 2 2 x5 x4 y2 = x−1 (x3 + + + · · · ). 2 20

y1 = x−1 (1 +

(c) p(x) = −1, q(x) = 4x4 . The indicial equation is m(m − 1) − m = 0 so the exponents are m1,2 = 0, 2. Substitute and simplify to obtain ∞ X j=0

[(m + j)(m + j − 2)aj + 4aj−4 ]xj−1 = 0.

CHAPTER 6. POWER SERIES SOLUTIONS The coefficients with negative indices are set to 0. The exponents are confirmed using the j = 0 relation: m(m − 2)a0 = 0. They are 0 and 2 as we expect. Let’s try out lick again. Set m = 0 and investigate the first few recursion relations. We already know a0 is arbitrary and the recursion relations are derived from the equation j(j − 2)aj + 4aj−4 = 0. j = 1 =⇒ −a1 = 0,

so a1 = 0

j = 2 =⇒ 0 · a2 = 0,

soa2 is arbitrary

j = 3 =⇒ aj = −

4 , j(j − 2)aj−4

to calculate the rest

Once more we have two arbitrary constants. Starting from a0 = 1 and a2 = 0, a1 = 0 and a3 = 0 also, and the recursion relation implies that all coefficients that are not multiples of 4 must be 0. j 1 a4j−4 , it is easily seen that a4j = (−1) Since a4j = − 2j(2j−1) (2j)! , and P∞ x4j = cos(x2 ). one solution is y1 = j=0 (−1)j (2j)! The solution derived from a0 = 0 and a2 = 1 is y2 = sin(x2 ) (verify). 4. (x − 1)p(x) = −3, (x − 1)2 q(x) = 2 so x = 1 is a regular singular point for dy dt the equation. Change the variable by t = x + 1. Then y 0 = dx = dy dt · dx = dy 00 2 dt = ẏ and similarly y = ÿ. The differential equation becomes t ÿ −3tẏ + 2y = 0. This is an Euler equidimensional equation, the substitution y = tm works just as well. √ The simplified result is the equation m(m−1)−3m+2 = so m1,2 = 2 ± 2. The general solution to the original equation is √ √ y = (x − 1)2 c1 (x − 1) 2 + c2 (x − 1)− 2 . 1 1 5. The equation y 00 − 3(x+1) y 0 − 3(x+1) 2 y = 0 and x = −1 is clearly a singular point. Since (x + 1)p(x) = −1/3 and (x + 1)2 q(x) = −1/3, −1 is a regular singular point and the differential equation will have at least one solution P∞ of the form y = (x + 1)m j=0 aj (x + 1)j , a0 arbitrary.

To facilitate the computation of the coefficients it is convenient to make dy = a simple change of independent variable: t = x + 1. Then y 0 = dx dy dt dy 00 · = = ẏ, and a similar computation will show y = ÿ. Therefore, dt dx dt the differential becomes 3t2 ÿ − tẏ − y = 0 and we may substiP∞equation m j tute y = t j=0 aj t . Because it is an Euler equidimensional equation, the substitution y = tm works just as well. The simplified result is √ the 2 2 equation 3m(m − 1) − m − 1 = 0 or 3m − 4m − 1 = 0 so m1,2 = 3 ± 37 . Consequently, the general solution to the original equation is √ √ y = (x + 1)2/3 c1 (x + 1) 7/3 + c2 (x + 1)− 7/3 .

6.5. MORE ON REGULAR SINGULAR POINTS

6. The indicial equation is m(m−1)+m−1 P∞ = 0 with solutions m1,2 = 1, −1 so m1 −m2 = 2. Substituting y = x j=0 aj xj , reindexing and rearranging to P∞ 2 j+1 + 3a1 x2 = 0. a0 could be chosen arbitrarily, j=2 [(j + 2j)aj + aj−2 ]x 1 say a0 = 1. Then aj = 0 for odd j, and a2 = − 81 , a4 = 24 , etc. The Frobenius series solution is 1 1 y = x(1 − x2 + x3 − · · · ). 8 24

7. Since p(x) = 1 and q(x) = x2 − 1/4, the indicial equation is m(m − 1) + m − 1/4 = 0 and the exponents are m1,2 = ±1/2. Thus m1 − m2 = 1. P∞ Make the substitution y = xm j=0 aj xj into the equation and simplify to ∞ X

1 [((m + j)2 − )aj + aj−2 ]xj = 0. 4 j=0

As usual, a−1 = 0 and a−2 = 0. Let m = −1/2 to obtain the recursion relation j(j − 1)aj + aj−2 = 0, j ≥ 0. Then 0 · a0 = 0 and 0 · a1 = 0 imply that both a0 and a1 can be chosen arbitrarily. The remaining coefficients are found using the relation aj = 1 − j(j−1) aj−2 , j ≥ 2. x4 x2 + − · · · ) = x−1/2 cos x 2! 4! x3 x5 a0 = 0, a1 = 1 =⇒ y2 = x−1/2 (x − + − · · · ) = x−1/2 sin x. 3! 5! a0 = 1, a1 = 0 =⇒ y1 = x−1/2 (1 −

P∞ 8. Substituting y2 = y1 ln x + xm2 j=0 cj (m2 + j)xm2 +j−1 to the differen2 00 0 2 2 00 tial equation x + xy10 ln x + x2 y1 ln x + P∞ x y + xy + x y = 0 yieldsm2x+jy1 lnP ∞ 0 + j=0 cj (m2 + j)xm2 +j + 2xy1 + j=2 cj (m2 + j)(m2 + j − 1)x j P∞ P∞ 4j 2j m2 +j+2 = 0. Substitute y1 , reindex, and rearrange to j=1 (−1) j=0 cj x 22j (j!)2 x + P∞ 2 m2 +j +c0 m22 xm2 +c1 (m2 +1)2 xm2 +1 = 0. Choose j=2 [cj (m2 +j) +cj−2 ]x P∞ (−1)j 4j 2j m2 = 0 and a0 could be chosen arbitrarily. Then + j=1 22j (j!)2 x P∞ 2 j 2 [c j + c ]x + c (m + 1) x = 0. Consequently, c = 0, and j−2 1 2 1 j=2 j j

j+1

4j (−1) 2j−2 1 c2j (2j)2 + c2j−2 + (−1) 22j (j!)2 = 0 or c2j = − 4j 2 + 22j (j!)2 · j .

CHAPTER 6. POWER SERIES SOLUTIONS Choose c0 = 0. It follows that (−1)1+1 1 · 22·1 (1!)2 1 (−1)2+1 1 (−1)1 + 1 · 1 + 2·2 · c4 = − 2 2·1 2 4 · 2 · 2 (1!) 2 (2!)2 2 (−1)2+1 1 = 2·2 (1 + ) 2 2 (2!) 2 (−1)2+1 1 1 (−1)3+1 1 · 2·2 (1 + ) + 2·3 · c6 = − 2 2 4 · 3 2 (2!) 2 2 (3!)2 3 1 1 (−1)3+1 (1 + + ) = 2·3 2 (3!)2 2 3 .. . c2 =

Therefore, y2 (x) is as displayed in the text.

6.6

Gauss’s Hypergeometric Equation

1. Verify the identity. (a) F (−p, b, b, −x) = 1 +

P∞

j=1

F (−p, b, b, −x) = 1 +

−p(−p+1)···(−p+j−1) (−x)j . Therefore, j!

∞ X p(p − 1) · · · (p − j + 1)

j=1

xj = (1 + x)p .

(b) See the last displayed equation in this section. P∞ 1 ( 1 +1)···( 21 +j−1) 12 ( 12 +1)···( 21 +j−1) 2j+1 (c) xF ( 21 , 12 , 33 , x2 ) = x+ j=1 2 2 x . Canj! 23 ( 32 +1)···( 32 +j−1) cel terms and simplify to obtain ∞ X 1 1 3 1 · 3 · 5 · · · (2j − 1) 2j+1 xF ( , , , x2 ) = x + x = arcsin x. 2 2 2 2j j!(2j + 1) j=1

(d) xF ( 12 , 1, 32 , −x2 ) = x + x terms and simplify to

P∞

j=1

1 1 1 2 ( 2 +1)···( 2 +j−1)1·2···j j! 23 ( 32 +1)···( 32 +j−1)

(−x2 )j . Cancel

∞ X 1 3 (−1)j 2j+1 xF ( , 1, , −x2 ) = x + x = arctan x. 2 2 2j + 1 j=1

P∞ P∞ 1(1+ 1b )···(1+ j−1 j b ) j · xbj = 1+ j=1 x . (e) F (a, b, a, xb = 1+ j=1 b(b+1)···(b+j−1) j! j! Allow b → ∞, term by term, to obtain ∞ X x xj F (a, b, a, ) = 1 + = ex . b j! j=1

6.6. GAUSS’S HYPERGEOMETRIC EQUATION 2

(f) xF (a, a, 23 , −x 4a2 ) = x + x

a2 (a+1)2 ···(a+j−1)2 (−x2 )j j=1 j! 32 ( 23 +1)···( 23 +j−1) · 4j a2j . Distribute

P∞

a2j into the numerator and simplify the denominator to obtain

∞ 2 X 1(1 + a1 )2 · · · (1 + j−1 3 −x2 a ) xF (a, a, , 2 = x + (−1)j x2j+1 . j · 3 · 5 · · · (2j + 1) 2 4a j! · 2 j=1

Now let a → ∞ term by term, and use j! · 2j = 2 · 4 · · · (2j), to end up with ∞ X 3 −x2 (−1)j 2j+1 x xF (a, a, , 2 ) = x + = sin x. 2 4a ) (2j + 1)! j=1

(g) Just like (f). 2. Find the general solution of each of the following differential equations near the indicated singular point. (a) The differential equation is hypergeometric with a = 2, b = −1, c = 3/2. The general solution is y = AF (2, −1, 3/2, x)+Bx−1/2 F (3/2, −3/2, 1/2, x). dt (b) Make the substitution t = −x. This implies that y 0 = dy dt · dx = 5 00 −ẏ and y = ÿ. The equation becomes t(1 − t)ÿ + (1/2 − 2 t)ẏ − 1 2 y = 0. This is hypergeometric with a = 1/2, b = 1, c = 1/2. The general solution is y = AF (1/2, 1, 1/2, t) + Bt1/2 F (1, 3/2, 3/2, t) = AF (1/2, 1, 1/2, −x) + B(−x)1/2 F (1, 3/2, 3/2, −x). 1 0 (c) Make the substitution t = x+1 2 . This implies that y = 2 ẏ and 1 1 00 y = 4 ÿ. The equation becomes t(1 − t)ÿ + ( 2 − 5t)ẏ − 4y = 0. This is hypergeometric with a = 2, b = 2, c = 21 . The general solution is y = AF (2, 2, 12 , t) + Bt1/2 F ( 25 , 52 , 32 , t) = AF (2, 2, 12 , x+1 2 )+ 5 5 3 x+1 1/2 B( x+1 ) F ( , , , ). 2 2 2 2 2 1 0 (d) Make the substitution t = − x−3 5 . This implies that y = − 5 ẏ and 1 14 00 y = 25 ÿ. THe equation becomes t(1−t)ÿ +( 5 −3t)ẏ −y = 0, which is hypergeometric with a = 1, b = 1, c = 14 5 . The general solution is 4 4 4 14 x−3 −9/5 y = AF (1, 1, 14 , t)+Bt F (− , − , − , 5 5 5 5 t) = AF (1, 1, 5 , − 5 )+ x−3 −3/5 4 4 4 x−3 B(− 5 ) F (− 5 , − 5 , − 5 , − 5 ). dy dy dt 1 0 3. Make the substitution t = 1−x 2 . This implies that y = dx = dt · dx = − 2 ẏ. 1 00 Similarly, y = 4 ÿ, and the differential equation is converted into

4t(1 − t) ·

1 1 · ÿ + (2t − 1) · (− ) · ẏ + p2 y = 0. 4 2

Simplify to t(1 − t)ÿ + ( 21 − t)ẏ + p2 y = 0, which is easily recognized as hypergeometric with c = 21 , a = p, and b = −p. Consequently, the general solution, in terms of t, is y = c1 F (p, −p, 1/2, t) + c2 t1/2 F (p + 1/2, −p + 1/2, 3/2, t). The solution in terms of x is obtained by replacing t with (1 − x)/2.

CHAPTER 6. POWER SERIES SOLUTIONS 4. Consider the differential equation x(1 − x)y 00 + [p − (p − 2)x]y 0 − py = 0, where p is a constant. (a) The equation is hypergeometric with a = p, b = 1, c = p. The general solution is y = AF (p, 1, p, x) + Bx1−c F (1, 2 − p, 2 − p, x). (b) F (p, 1, p, x) ∞ X p(p + 1) · · · (p + j − 1)1 · 2 · · · · j j x =1+ j!p(p + 1) · · · (p + j − 1) j=1 =1+

∞ X

xj =

j=1

1 , 1−x

F (1, 2 − p, 2 − p, x) ∞ X 1 · 2 · · · j(2 − p)(2 − p + 1) · · · (2 − p + j − 1) j =1+ x j!(2 − p)(2 − p + 1) · · · (2 − p + j − 1) j=1 =1+

∞ X j=1

xj =

1 . 1−x

1 1 Therefore, y = A · 1−x + Bx1−c 1−x . 1 (c) y1 (x) = 1−x . When p = 1, the differential equation becomes x(1 − 00 x)y + (1 − 3x)y 0 − y = 0 (correction) so Z R 1−3x 1 y2 (x) = [ (1 − x)2 e− x(1−x) dx dx] 1−x Z 1 1 = [ (1 − x)2 (1 − x)2 dx] x 1−x 4 1 1 = (ln x − 4x + 3x2 − x3 + x4 ) 3 4 1−x A +B(ln x−4x+3x2 − 43 x3 + Therefore, the general solution is y = 1−x 1 4 1 4 x ) 1−x .

5. If t = ex , then y 0 = tẏ and y 00 = t2 ÿ + tẏ. After substitution and simplification, the differential equation for y(t) is hypergeometric of the form t(1 − t)ÿ + (3/2 − t)ẏ + y = 0. Clearly c = 3/2 and a = −b = 1. The formula for the general solution near t = 1 yields y = c1 F (1, −1, −1/2, 1 − t) + c2 (1 − t)3/2 F (5/2, 1/2, 5/2, 1 − 5). Replace t with ex to obtain the general solution of the original equation near x = 0.

6.6. GAUSS’S HYPERGEOMETRIC EQUATION

6. (a) F 0 (a, b, c, x) ∞ X a(a + 1) · · · (a + j − 1)b(b + 1) · · · (b + j − 1) j−1 ·x = j· j!c(c + 1) · · · (c + j − 1) j=1 =

∞ ab X a(a + 1) · · · (a + j − 1)b(b + 1) · · · (b + j − 1 j−1 + x c (j − 1)!c(c + 1) · · · (c + j − 1) j=2

∞ ab X a(a + 1) · · · (a + j)b(b + 1) · · · (b + j) + c j!c(c + 1) · · · (c + j) j=1

ab ab (a + 1)(a + 2) · · · [(a + 1) + j − 1](b + 1)(b + 2) · · · [(b + 1) + j − 1] + c c j!(c + 1)(c + 2) · · · [(c + 1) + j − 1] ab = F (a + 1, b + 1, c + 1, x) c =

(b) By (a), y10 = −2p2q F (p + 1, −p + 1, 3/2, (1 − x)/2), which is bounded

2(p+1/2)(−p+1/2) near x = 1. y20 = 1−x F (p+3/2, −p+3/2, 5/2, (1− 2 · 3 x)/2) which is unbounded near x = 1. y = C · F (p, −p, 1/2, (1 − P∞ j x)/2) = C·[1+ j=1 p(p+1)···(p+j−1)(−p)(−p+1)···(−p+j−1) ( 1−x 2 ) ], which j! 12 ·( 12 +1)···( 21 +j−1) is a polynomial only when p is an integer.

Chapter 7

Fourier Series: Basic Concepts 7.1

Fourier Coefficients

R π/2 R 1. a0 = π1 −π πdx = 3π/2; aj = π1 −π π/2π cos(jx)dx = 1j sin( jπ 2 ); bj = R jπ 1 π/2 1 j π −π π sin(jx)dx = j ((−1) − cos( 2 )). 2. a0 = π1

R π/2 0

1 dx = 12 ;

  0 1/(jπ) cos jx dx =  0 −1/(jπ)  Z  1/(jπ) 1 π/2 2/(jπ) bj = sin jx dx =  π 0 0

1 aj = π

Z π/2

if j is even, if j is 1 plus an even multiple of 2, if j is 1 plus an odd multiple of 2. if j is odd, if j is an odd multiple of 2, if j is an even multiple of 2.

Rπ Rπ j 3. a0 = π1 0 sin xdx = π2 ; aj = π1 0 sin x cos jxdx = 1+(−1) π(j 2 −1) , j > 1; a1 = 0; bj = π1 sin x sin jxdx = 0, j > 1; b1 = 1/2. Rπ Rπ Rπ 4. a0 = π1 0 cos x dx = 0; aj = π1 0 cos x cos jx dx = π1 0 cos((1+j)x)+cos((1−j)x) dx = 2 R R −j 1 π 1 π sin((j+1)x)−sin((j−1)x) 0; bj = π 0 cos x sin jx dx = π 0 dx = π(1−j 2 ) · [1 + 2 (−1)j ]. 5. Consider alternative lines of thought. In each case the terms in the Fourier series are the same terms that define the function. 7. Let f be the function in Exercise 2 and g and h be the functions in Exercise 1 and Example 6.1.2 respectively. Draw their graphs and observe that f (x) = π1 (g(x) − (π − h(x)) = π1 (g(x) + h(x)) − 1. It follows that the 81

CHAPTER 7. FOURIER SERIES: BASIC CONCEPTS Fourier coefficients satisfy similar relationships. For example, if f0 , g0 , and h0 denote the constants in the three Fourier series, then f0 = π1 (g0 +h0 )−1. On the other hand, f1 , g1 , and h1 denote the first cosine coefficient of each of the three Fourier series, then f1 = π1 (g1 + h1 ), and so on.

7.2

Some Remarks about Convergence

1. Exercise. Rπ x2 dx = π 2 /3; aj = π1 x2 cos jxdx = (−1)j j22 ; bj = π1 0 x2 sin jxdx = 0 j −1 2 · (−1) + (−1)j πj (integration by parts, or use a table of integrals). π·j 3

3. (a) a0 = π1

Rπ

(b) Exercise. P∞

(−1)j j2 .

P∞

(−1)2j j2 .

j=1

Consequently, 1 − 212 + 312 − 412 + · · · = π12 . 2

If x = π, then the series converges to π 2 /2 so π2 = π6 +2

j=1

Consequently, 1 + 212 + 312 + 412 + · · · = π6 . (d) Use Exercise 2 and the first sum in (c) to conclude that 2

π 8

1 j (2j)2 =

− π12 = π24 .

4. According to Theorem 6.2.7, the series converges to the average of the left and right limits at 0, which is 0. One can also calculate the Fourier series explicitly to be X 2 − [(−1)j − 1] sin jx jπ j and observe that the sum at 0 is 0. Rπ π −π π π 5. (a) a0 = π1 −π ex dx = e −e = 2 sinh ; aj = π1 ex cos jxdx = 2 sinh · π π π R j π (−1) j (−1)j 1 2 sinh π x · j 2 +1 (integration by parts, j 2 +1 ; bj = π −π e sin jxdx = − π or use a table of integrals). (b) Exercise. π

−π

(c) At x = π the series converges to e +e = cosh π. Therefore, 2 P∞ (−1)2j sinh x 2 sinh π cosh x = π + π j=1 j 2 +1 , and a little algebra will produce the first formula. P∞ (−1)j 2 sinh π At x = 0 the series converges to 1 so 1 = sinh j=1 j 2 +1 , π + π and a little more algebra will produce the second formula. 6. By taking the derivative, one can easily verify that f and g satisfy the Dirichlet conditions. However the sum function 2 x sin(1/x) if x 6= 0 (f + g)(x) = 0 if x = 0

7.3. EVEN AND ODD FUNCTIONS

is not monotone away from 0 (just take the derivative), so does not satisfy the Dirichlet conditions. 7. Let x be an interior point where f is not continuous. Since f is piecewise monotone, it muse be monotone to the left of x, for example monotone increasing is an open interval (a, x). If s is the least upper bound of the set S = {f (t) : a < t < x}, then limt→x− f (t) = s.

7.3

Even and Odd Functions

1. Determine whether the each function is odd or even or neither. Functions x5 sin 2x x2 sin 2x ex (sin x)3 sin x2 cos(x + x3 ) x + x2 + x3 ln 1−x 1+x

Parity even odd neither odd even even neither odd + even odd

Reason odd×odd even×even oddodd f (even) feven (odd) f (−x) = ln 1−x 1+x = −f (x)

2. Write f (x) =

f (x) + f (−x) f (x) − f (−x) + ≡ fe (x) + fo (x) . 2 2

Note that fe (−x) =

f (−x) + f (x) = f (x) , 2

so fe is even and fo (−x) =

f (−x) − f (x) f (x) − f (−x) =− = −fo (x) , 2 2

so fo is odd. Ra R0 Ra 3. Assume f is even. Then −a f (x)dx = −a f (x)dx + 0 f (x)dx. Make the change-of-variable x = −u in the first integral to obtain Z a Z 0 Z a f (x)dx = f (−u)(−du) + f (x)dx −a a 0 Z a Z a f (u)du + f (x)dx = a 0 Z a =2 f (x)dx. 0

The verification of formula (4) is quite similar.

CHAPTER 7. FOURIER SERIES: BASIC CONCEPTS 4. We calculate that 2 bj = π

Z π 0

π 1 dx = · [1 − (−1)j ] . 4 2j

Thus the sine series written explicitly is sin 3x sin 5x + + ··· . 3 5

sin x + At x = π/2 the series is

1−

1 1 + − +··· . 3 5

The cosine series for the same constant function is just π . 4 5. The function is even. a0 = π2

Rπ 0

j+1

cos x/2dx = π4 ; aj = π2

Rπ 0

cos x2 cos jxdx =

4 (−1) π · 4j 2 −1 .

6. The sine series for sin x is just sin x. The cosine series has coefficients calculated by Z Z 2 π 2 π sin((1 + j)x) + sin((1 − j)x) aj = sin x cos jx dx = dx π 0 π 0 2 =

2 [1 + (−1)j ] . π(1 − j 2 )

7. (a) Direct computation. The function is even and a0 = 0. aj = π2 j+1

π 2 1+(−1) 2 ) cos jxdx = π · j2

f (x) =

4 π

Rπ 0

(−x+

. Note that the even coefficients are 0.

Z ∞

cos(2j − 1)x , −π ≤ x ≤ π. (2j − 1)2 j=1

(b) Observe that f (x) = π2 − |x|. Rπ Rπ 8. (a) a0 = π1 −π π − x dx = π1 [πx − x2 /2]π−π = 2π; aj = π1 −π (π − Rπ Rπ x) cos jx dx = 0; bj = π1 −π (π − x) sin jx dx = −π sin jx dx − R 2(−1)j 1 π π −π x sin jx dx = j R Rπ 1 π (b) a0 = π 0 π − x dx = π; aj = π1 0 (π − x) cos jx dx = πj2 2 [1 − (−1)j ]. (c) Similar to (b).

7.3. EVEN AND ODD FUNCTIONS

9. The average value of f is clearly equal to π/4 (so a0 = π/2). Moreover, R π/2 Rπ aj = π2 ( 0 x cos jxdx+ π/2 (π −x) cos jxdx). These coefficients evaluate as follows.   0, j = 1, 3, 5, · · · 2 jπ − πj8 2 , j = 2, 6, 10, · · · aj = 2 · 2 cos( − (1 + (−1)j =  πj 2 0, j = 4, 8, 12, · · · Note that the indices for the nonzero coefficients are of the form 2(2j − 1) 2 1 8 and a2(2j−1) = − π(2(2j−1)) 2 = − π · (2j−1)2 . 11. (a) a0 = π2

Rπ 0

x3 dx = π 3 /2; aj = π2

Rπ 0

1+(−1) x3 cos jxdx = 12 π · j4

j+1

j 6π (−1) j2 .

(b)

i. Substitute x = 0 and x = π/2 into the Fourier series for x3 to obtain the following two equations. 0=

∞ ∞ X π3 (−1)j 1 24 X + 6π + 2 4 j π (2j − 1)4 j=1 j=1

∞ ∞ X X (−1)j π3 π3 +6π = + 6π 8 4 (2j)2 j=1 j=1

(7.1)

(7.2)

j P∞ Use Equation 7.2 to calculate j=1 (−1) j 2 . Then substitute the value into Equation 7.1 to obtain the desired result. P∞ P∞ P∞ 1 1 1 ii. Let S = j=1 j14 . Then S = j=1 (2j) 4 + j=1 (2j−1)4 = 16 S +

π4 96 . Solve for S.

13. The identity sin2 x = 12 − 12 cos 2x can be used to evaluate the integrals for the Fourier series coefficients for sin2 x. Of course, they will evaluate to the coefficients that appear in the identity so the identity and the Fourier series are one in the same. 15. These identities can be established by appealing to the complex formulas 1 for the sine and cosine functions: sin x = 2i (eix − e−ix ) and cos x = 1 ix −ix ). FOr example, 2 (e + e

1 3ix (e − 3e2ix e−ix + 3eix e−2ix − e−3ix ) 8i 1 = − (e3ix − e−3ix − 3eix + 3e−ix ) 8i 3 1 = sin x − sin 3x. 4 4

sin3 x = −

CHAPTER 7. FOURIER SERIES: BASIC CONCEPTS

7.4

Fourier Series on Arbitrary Intervals

1. Calculate the Fourier series for the given function. R1 2 (a) L = 1. The function is odd so aj = 0, bj = 1−0 x sin nπxdx = 0 j+1 P ∞ (−1) sin jπx 2 ; x = π2 j=1 (−1)j+1 j , −1 < x < 1. π · j R2 (b) L = 2. The function is odd so aj = 0. bj = 22 0 sin x sin(jπx/2)dx = 2π sin 2 · (−1)j+1 π2 jj2 −4 . sin x = 2π sin 2

∞ X

(−1)j+1

j=1

j π2 j 2 − 4

sin(

jπx ), −2 < x < 2. 2

R3 ex dx = 23 sinh(3); aj = 13 −3 ex cos(jπx/3)dx = R j j j 1 3 6 sinh(3) · π(−1) ex sin(jπx/3)dx = −2π sinh(3) · π(−1) 2 j 2 +9 ; bj = 3 2 j 2 +9 . −3

ex =

−3

∞ X sinh 3 (−1)j jπx + 6 sinh 3 cos( ) 2 2+9 3 π j 3 j=1

− 2π sinh 3

∞ X (−1)j j j=1

π2 j 2 + 9

sin(

jπx ), −3 < x < 3 3

R1 R1 (d) L = 1. The function is even. a0 = 2 0 x2 dx = 2/3; aj = 2 0 x2 cos(jπx)dx = j

(−1) 4 π2 · j 2

. ∞ 1 4 X cos(jπx) x = + 2 (−1)j , −1 ≤ x ≤ 1. 3 π j=1 j2 2

√ R π/3 (e) L = π/3. THe function is even. a0 = π6 0 cos 2xdx = 32π3 ; aj = √ R j+1 6 π/3 cos(2x) cos(3jx)dx = 6 π 3 · (−1) π 0 9j 2 −4 .

√ ∞ √ 3 3 6 3 X (−1)j+1 π π cos 2x = + cos(3jx), − ≤ x ≤ . 4π π j=1 9j 2 − 4 3 3 √ R1 3 sin(2x − π/3)dx = − 2 sin 2; aj = −1 sin(2x − −1 √ R1 j π/3) cos(jπx)dx = 2 3 sin 2· π(−1) sin(2x−π/3) sin(jπx)dx = 2 j 2 −4 ; bj = −1

(f) L = 1. a0 =

j −π sin 2 · π(−1) 2 j 2 −4 .

√ sin(2x−π/3) = −

∞ X √ 3 (−1)j sin 2+2 3 sin 2 sin(jπx), −1 < x < 1. 2 4 π j2 − 4 j=1

7.5. ORTHOGONAL FUNCTIONS

2. The Fourier series for g is ∞ 3 X πx −6 . + sin (2j − 1) 2 j=0 (2j − 1)π 2

The Fourier series for f is ∞

3 X πx −6 − + . sin (2j − 1) 2 j=0 (2j − 1)π 2 3. Find the Fourier series. (a) The function is even with average value 1/2 so a0 = 1 and aj = R1 j+1 2 0 (1 − x) cos(jπx)dx = π22 · 1+(−1) . j2 ∞ 4 X cos(2j − 1)πx 1 , −1 ≤ x ≤ 1. f (x) = + 2 2 π j=1 (2j − 1)2

(b) The function is even with average value 1 so a0 = 2 and aj = R2 j −1 2 0 x cos(jπx/2)dx = π42 · (−1) . j2 ∞

|x| = 1 −

8 X cos((2j − 1)πx/2) , −2 ≤ x ≤ 2. π 2 j=1 (2j − 1)2

R1 R1 5. L = 1. a0 = 2 1 x2 − x + 61 dx = 0. aj = 2 0 (x2 − x + 16 ) cos jπxdx = P∞ 1+(−1)j 2 . Therefore, x2 − x + 61 = π12 j=1 cosj2jπx , 0 ≤ x ≤ 1. 2 π2 · j2 R1

2 cos jπx 2 dx = 0;  Z 1 if j is odd, 2 1 jπx 4  2 if j is twice an odd integer, bj = 2 sin dx = · 2 0 2 jπ  0 if j is twice an even integer.

6. a0 = 22

2 dx = 2; aj = 22

7. Since f has period 2, the Fourier series is f (x) = cos πx.

7.5

Orthogonal Functions

1. Verify the functions are orthogonal on the given interval. π Rπ 1 (a) −π sin 2x cos 3xdx = 21 cos x − 10 cos 5 −π = 0. π Rπ 1 (b) 0 sin 2x sin 4xdx = 14 sin 2x − 12 sin 6x 0 = 0. R1 (c) −1 x2 · x3 dx = [x6 /6]1−1 = 0.

CHAPTER 7. FOURIER SERIES: BASIC CONCEPTS (d)

R2 −2

x cos 2xdx =

2 1 2 cos 2x + 2 x sin 2x −2 = 0.

2. Now kf + gk2 + kf − gk2

hf + g, f + gi + hf − g, f − gi

hf, f i + hf, gi + hg, f i + hg, gi

2hf, f i + 2hg, gi

2kf k2 + 2kgk2 .

+hf, f i − hf, gi − hg, f i + hg, gi

3. Observe that kf − gk2 = hf − g, f − gi = hf, f i − 2hf, gi + hg, gi = kf k2 − 2hf, gi + kgk2 It follows that kf − gk2 = kf k2 + kgk2 if and only if hf, gi = 0. 4. If f is not zero for all x then there is a nontrivial interval I and an > 0 so that |f | > on I. Then Z Z kf kL2 = |f (x)|2 dx1/2 ≥ 2 dx1/2 = · |I|1/2 I

and that is a contradiction. 5. A calculation like the one in Exercise 3 will verify that φ(λ) = kf + λgk2 = kf k2 + 2λhf, gi + λ2 kgk2 . Since φ0 (λ) = 2hf, gi + 2λkgk2 , the function φ is minimized when λ has the value λ0 = − hf,gi kgk2 . Because φ(λ) ≥ 0 for all λ, φ(λ0 ) ≥ 0 as well, so kf k2 − 2 ·

hf, gi · hf, gi + kgk2

hf, gi kgk2

kgk2 ≥ 0.

This simplifies to kf k2 kgk2 ≥ hf, gi2 which, in turn, implies the desired inequality: kf kkgk ≥ |hf, gi|.

Chapter 8

Laplace Transforms 8.1

Introduction

1. Evaluate the integrals in the Laplace transform table. Unless otherwise indicated it is assumed that p > 0. Z ∞ L[1] =

e 0

Z ∞ L[x] = 0

L[xn ] = Z ∞

L[eax ] =

L[sin ax] =

x→∞ a p sin ax + a cos ax −px e = 2 . sin axdx = − p2 + a2 p + a2 x=0

−px

x→∞ p cos ax − a sin ax −px p cos axdx = − e = 2 . 2 2 p +a p + a2 x=0

e 0

1 2

−(p−a)x x→∞ e 1 = , p > a. −(p − a) x=0 p−a

−px

Z ∞

L[sinh ax] =

. See the calculation in this section.

eax e−px dx =

L[cos ax] =

−px x→∞ 1 e = . dx = − p x=0 p

x→∞ 1 x 1 = 2. xe−px dx = −( 2 + )e−px p p p x=0

pn+1

Z ∞

−px

Z ∞

(eax −e−ax )e−px dx = −

x→∞ 1 e−(p−a) e−(p+a) − = 2 p−1 p + a x=0 a , p > |a|. p2 − a2

CHAPTER 8. LAPLACE TRANSFORMS

L[cosh ax] =

1 2

Z ∞

(eax +e−ax )e−px dx = −

x→∞ 1 e−(p−a) e−(p+a) + = 2 p−1 p + a x=0 a , p > |a|. p2 − a2

2. (a) Now ax e − e−ax 1 1 −ax L[sinh ax = L = L [eax ] − e 2 2 2 =

1 1 a 1 1 + = 2 . 2p−a 2p+a p − a2

(b) Now L[cosh ax] = L =

ax e + e−ax 1 −ax 1 e = L [eax ] + 2 2 2

1 1 1 1 p + = 2 . 2p−a 2p+a p − a2

3. Since sin2 ax = 12 (1 − cos 2ax), and the Laplace transform is a linear operator, 1 (L[1] − L[cos 2ax]) 2 p 2a2 1 1 − 2 = . = 2 p p + 4a2 p(p2 + 4a2 )

L[sin2 ax] =

2a +p Similarly, L[cos2 ax] = 12 (L[1] + L[cos 2ax]) = p(p 2 +4a2 ) .

Since sin2 ax + cos2 ax = 1, the two Laplace transforms should add up to 1/p (verify). 4. (a) L[10] = 10L[1] = 10 p 1 5 (c) L[2e3x − sin 5x] = 2L[e3x ] − L[sin 5x] = 2 · p−3 + p2 +25

(e) L[x6 sin2 3x + x6 cos2 3x] = L[x6 ] = p6!7 5. Find the function whose Laplace transform is given. 30 3 (a) Since L[x3 ] = p64 , the function f (x) = 30 6 x transforms to p4 . 1 2 (b) Since L[e−3x ] = p+3 , the function f (x) = 2e−3x transforms to p+3 .

8.2. APPLICATIONS TO DIFFERENTIAL EQUATIONS

1 (d) Using the technique of partial fractions, p21+p = p1 − p+1 . Therefore, 1 −x the function f (x) = 1 − e will transform to p2 +p . 1 1 1 (e) Using partial fractions, p4 +p 2 = p2 − p2 +1 . Therefore, the function 1 f (x) = x − sin x will transform to p4 +p2 .

7. Male is unable to calculate any of the transforms; Mathematica can calculate the transforms for part (a) and part (c). The following code shows how Mathematica calculates the Laplace transform for part (c). LaplaceTransform[Sin[Log[x]],x,p] 1 −1−i 2i (p Gamma[−i] + Gamma[i]) 2p

8.2

Applications to Differential Equations

1. The Laplace transforms are displayed in Table **. The transforms for (d) and (g) were obtained using a property that is introduced in the next dn section: L[xn f (x)] = (−1)n dp n L[f (x)]. Table 8.1: Some Laplace transforms Function Transform 120 (a) x5e−2x (p+2)6 1 2 (b) (1 − x2 )e−x p+1 − (p+1) 3 1 −x (c) e sin x (p+1)2 +1 6p (d) x sin 3x (p2 +9)2 p−3 3x (e) e cos 2x (p−3)2 +4 1 (f) xex (p−1)2

2. (a) L−1

p 4p2 +1

(f ) L−1

6 (p−1)3

(g) (h)

x2 cos x sin x cos x

= 2e−2x sin 3x

6 (p+2)2 +9

= 12 cos x/2

= 3 · ex · x2

2p(p2 −3) (p2 +1)3 1 p2 +4

3. Use the Laplace transform to solve the given IVPs. Each solution begins with the transform of the IVP. We let Y = L[y]. 1 1 (a) pY −0+Y = p−2 implies that Y = (p+1)(p+2) . Using partial fractions, 1 1 Y = 13 p−2 − p+1 . Therefore, the solution is y = 13 (e2x − e−x ).

(b) p2 Y −p·0−3−4·(pY −0)+4Y = 0 implies that (p2 −4p+1)Y −3 = 0. 3 3 2x Therefore, Y = p2 −4p+4 = (p−2) . 2 and the solution is y = 3xe

CHAPTER 8. LAPLACE TRANSFORMS (c) p2 Y − p · 0 + 2 · (pY − 0) + 2Y = p2 implies that (p2 + 2p + 2)Y − 1 = p2 . p+2 Therefore, y = p21+2/p +2p+2 = p(p2 +2p+2) . Use partial fractions and p+1 p+1 complete the square to obtain Y = p1 − p2 +2p+2 = p1 − (p+1) 2 +1 . −x Therefore, the solution is y = 1 − e cos x.

(d) p2 Y − p · 0 − 1 + pY − 0 = p63 implies that (p2 + p)Y − 1 = p63 . 3

p +6 Therefore, Y = 1+6/p p2 +p = p3 (p2 +p) . Using partial fractions (the corB C D E rect decomposition is of the form A p + p2 + p3 + p4 + p+1 , we obtain the solution y = −5 + 6x − 3x2 + x3 + 5e−x . 3 (e) p2 Y − p · 0 − 3 + 2 · (pY − 0) + 5Y = (p+1) 2 +1 implies that Y sat3 . Therefore, Y = isfies (p2 + 2p + 5)Y − 3 = p2 +2p+5

3 3+ p2 +2p+2

p2 +2p+2 = 3p2 +6p+9 1 (p2 +2p+2)(p2 +2p+5) . The partial fraction decomposition is Y = p2 +2p+2 + 2 1 2 −x (sin x + sin 2x). p2 +2p+5 = (p+1)2 +1 + (p+1)2 +4 . Therefore, y = e

4. According to the text, L[y] =

(p − 2a)y0 + y00 p2 − 2ap + a2

hence y=L

−1

p 1 −1 0 y0 +L (y0 − 2ay0 ) · = y0 ·[eax x]0 +(y00 −2ay0 )eax ·x . (p − a)2 (p − a)2

Rx 5. Let y(x) = 0 f (t)dt and F (p) = L[f (x)]. By the Fundamental theorem, Rx y 0 (x) = f (x) so F (p) = L[y 0 (x)] = pL[y(x)] − y(0) = pL[ 0 f (t)dt]. ConRx purpose of finding an sequently, L[ 0 f (t)dt] = F (p) p , as desired. For the h i R x 1 inverse transform, this can be expressed as L−1 p F (p) = 0 f (t)dt. This inverse transform can be used to avoid partial fractions. For i formula h R x −t 1 −1 −t x −x example, L . p(p+1) = 0 e dt = [−e ]0 = 1 − e

8.3

Derivatives and Integrals of Laplace Transforms

d d L[cos ax] = − dp 1. L[x cos ax] = −L[−x cos ax] = − dp 2

p p2 +a2

−a = (pp2 +a 2 )2 . 2

p +a −2a 1 2a 1 To invert (p2 +a 2 2 note that L[x cos ax] = (p2 +a2 )2 = p2 +a2 − (p2 +a2 )2 . ) 2 Therefore, L x cos ax − a1 sin ax = − (p22a +a2 )2 . It follows that 1 1 1 L−1 = − 2 (x cos ax − sin ax) (p2 + a2 )2 2a a 1 = 3 (sin ax − ax cos ax). 2a

8.3. DERIVATIVES AND INTEGRALS OF LAPLACE TRANSFORMS 93 d2 d2 a 2. (a) L x2 sin ax (p) = dp 2 L[sin ax](p) = dp2 p2 +a2 p d d (c) L[x cos ax](p) = − dp L[cos ax](p) = − dp p2 +a2

3. Start with the Laplace transform. Let y(0) = A. d d [p2 Y − Ap] − 3 dp [pY ] − (pY − A) + 4 dY (a) − dp dp − 9Y = 0. This evaluates 2 0 to the equation (−p − 3p + 4)Y − (3p + 12)Y = −2A which, in turn, simplifies to (p − 1)Y 0 + 3Y = 2A/(p + 4). Consequently, 2 ln(p+4) 1 2 x Y = A · p −12p+50 + B · (p−1) 3 and, setting A = 0, y = Bx e . (p−1)3 d d 3 (b) − dp [p2 Y −Ap]−2 dp [pY ]+3(pY −A)− dY dp +3Y = p+1 . This evaluates 3 2 0 to the equation (−p − 2p − 1)Y + (p + 1)Y = 2A + p+1 which, in 2A 3 0 turn, simplifies to (p + 1)Y − Y = − p+1 − (p+1)2 . Consequently, A 1 −x Y = p+1 − (p+1) . 2 + B(p + 1) and, setting B = 0, y = (A + x)e

4. (a) Let A be the Laplace transform of 1. Then Z ∞ 0

e−ax − e−bx dx = x

Z ∞

e−ax dx − x

Z ∞ A(s) ds −

Z ∞

e−bx dx x

1 ds − s

A(s) ds = b

Z ∞

Z ∞ b

1 ds s

(b) Let B be the Laplace transform of sin bx. Then Z ∞ 0

e−ax sin bx dx = x

Z ∞

5. (a) Let p = 0 in the equation √ 12

Z ∞ B(s) ds =

p +1

= L[J0 (x)] =

b ds s2 + b2

R∞ 0

J0 (x)e−px dx.

Rπ R π P∞ (−1)j 2j (b) Observe that π1 0 cos(x cos t)dt = π1 0 j=0 (2j)! (x cos t) dt. Move the integral across the sum and proceed as follows. To justify the last (2j)! step use 2j j = (j!)2 . 1 = π

Z π cos(x cos t)dt = 0

Z ∞ X (−1)j 1 π j=0 ∞ X

(2j)!

cos2j tdt x2j

(−1) 1 2j = · 2j · x2j (see below) (2j)! 2 j j=0 = J0 (x) (exercise)

CHAPTER 8. LAPLACE TRANSFORMS The verification that π1 Z π

Rπ 0

2j 1 it (e + e−it dt 2 0 Z πX 2j 2j 1 (eit )2j−k (e−it )k dt = 2j k 2 0 k=0 2j Z π 1 X 2j e2(j−k)it dt = 2j k 2 0 k=0 1 2j = 2j · · π. j 2

cos2j tdt =

cos2j tdt = 212j 2j j goes like this:

Z π

R∞ P∞ R (j+1)·a −px 7. (a) Start with F (p) = 0 e−px f (x)dx = j=0 j·a e f (x)dx, make the change of variable x = u + j · a in the j th integral, and proceed as follows: ∞ Z a X F (p) = e−p(u+ja) f (u + ja)du

j=0 ∞ X

−ap j

Z a

)

j=0

Z a =

e−pu f (u)du

e−pu f (u)du ·

1 1 − e−ap

∞ X (e−ap )j j=0

Z a

e−px f (x)dx

R2 (b) The function f has period a = 2 so F (p) = 1−e1−2p 0 e−px f (x)dx = h i R 1 −px −px x=1 1 1−e−p 1 = p1 · 1−e dx = 1−e1−2p · e−p −2p = p(1+e−p . 1−e−2p 0 e x=0

8.4

Convolutions

i h i h Rx 1 1 1. Since L−1 p2 +a = a1 sin ax, L−1 (p2 +a = a12 0 sin a(x − t) sin atdt. 2 2 )2 Therefore, Z x Z x 1 1 L−1 2 = sin ax cos at sin atdt − cos ax sin at sin atdt p + a2 a2 0 0 1 1 1 1 2 = 2 sin ax · sin ax − cos ax · (ax − sin 2ax) . a 2a 2a 2 This simplifies to the solution found in Exercise 1 of the last section (verify).

8.4. CONVOLUTIONS

2. (a) Taking the Laplace transform of both sides, we see that L[y] = L[1] − L[x] · L[y] so that L[y] = or L[y] =

L[1] 1 + L[x]

p 1/p = 2 . 2 1 + 1/p p +1

Hence y = cos x . (c) Taking the Laplace transform of both sides, we see that p 1 =Y +2 2 ·Y , p+1 p +1 where we have let Y denote the Laplace transform of Y . Solving for Y gives p2 1 p2 + 1 = + . Y = (p + 1)3 (p + 1)3 (p + 1)3 It follows that y=

1 −x 2 00 (e x ) + e−x x2 . 2

q L[T (y)] 2g 3. Divide both sides of ** by p to obtain L[fp(y)] = π · p1/2 . Take the inverse transform of both sides, using convolution, to get the equation √ Ry T (t) 2g R y √ f (y)dy = π 0 y−t dt. Then differentiate with respect to y to obtain 0 √

f (y) =

T (t) 2g d √ π dy y−t dt. √

If T (y) = T0 , a constant, then f (y) = √

Therefore, f (y) =

2g d R y √T0 dt = π dy 0 y−t

√

2g d 1/2 ). π dy (2T0 y

2g T0 √ π · y , as in **.

4. We know that

r F = L[f ] =

2g 1/2 · p · L[T (y)] . π

√ If T (y) = k y, then the equation becomes r F =

2g 1/2 √ · p · L[k y] . π

But we have a formula for the Laplace transform of the square root function. So we have r r 2g 1/2 1 π F = ·p ·k· . π 2p p

CHAPTER 8. LAPLACE TRANSFORMS This simplifies to

√ gk F = √ 2p

hence

√ gk f= √ . 2

5. Take the Laplace transform of the IVP to obtain p2 Y + a2 Y = F (p) where 1 F is the Laplace transform of f . Therefore Y = F (p) · p2 +a 2 so, using R 1 x convolution on the right side, y(x) = a 0 f (t) sin a(x − t)dt.

8.5

The Unit Step and Impulse Functions

1. Find the convolution of the following pairs of functions. t Rt (a) a ∗ sin at = 0 sin aτ dτ = − cosaaτ 0 = a1 (1 − cos at). h (b−a)t it Rt Rt (b) eat ∗ ebt = 0 ea(t−τ ) ebτ dτ = eat 0 e(b−a)τ dτ = eat e b−a = 0

ebt −eat b−a .

R t a(t−τ ) Rt at . e τ dτ = eat 0 τ e−aτ dτ = e −1−at a2 0 h i a b −1 (d) sin at ∗ sin bt = L p2 +a2 · p2 +b2 . Using partial fractions, 1 ab ab −1 sin at ∗ sin bt = 2 · L − a − b2 p2 + b2 p2 + a2 a sin bt − b sin at . = a2 − b2 (c) t ∗ eat =

2. (a) The convolution of the two given functions is Z x − cos ax 1 1 · sin at dt = + . a a 0 The Laplace transform of this convolution is −1 p 1 1 a + · = . · a p2 + a2 a p p(p2 + a2 ) This is clearly the product of the individual Laplace transforms. (c) The convolution of the two given functions is Z x −x eax 1 (x − t)eat dt = + 2 − 2. a a a 0 The Laplace transform of this function is −

1 1 1 1 1 + 2· − = 2 . ap2 a p − a a2 p p (p − a)

This is clearly the product of the individual Laplace transforms.

8.5. THE UNIT STEP AND IMPULSE FUNCTIONS

3. These are solved using the impulse response function h(t) = i h problems Rt 1 −1 L z(p) . The solution is y(t) = h(t) ∗ f (t) = 0 h(t − τ )f (τ )dτ . The details of the integration are not shown. h i h i 1 1 1 (a) h(t) = L−1 (p+2)(p+3) = L−1 p+2 − p+3 = e−2t − e−3t . ThereR t 3τ −2(t−τ ) fore, y(t) = 0 5e · [e − e−3(t−τ ) ]dτ = 61 (5e−3t − 6e−2t + e3t ). h i h i 1 1 1 (b) h(t) = L−1 (p−2)(p+3) = 15 L−1 p−2 − p−3 = 15 (e2t −e−3t ). ThereR t 1 2(t−τ ) −3(t−τ ) 1 fore, y(t) = 0 τ · t [e −e ]dτ = − 180 (t+30t−9e2t +4e−3t ). i h i h 1 1 = L−1 p−1 − p1 = et − 1. Therefore, y(t) = (c) h(t) = L−1 p(p−1) Rt 2 τ · [et−τ − 1]dτ = 2et − 13 t3 − t2 − 2t − 2. 0 5. Make the substitution σ = t − τ to obtain f ∗ g = R0 Rt f (σ)g(t − σ)(−dσ) = 0 g(t − σ)f (σ)dσ = g ∗ f . t

Rt 0

f (t − τ )g(τ )dτ =

Regarding associativity, interchange the order of integration, then make the change of variable τ − σ = µ. Z τ Z t f ∗ [g ∗ h] = f (t − τ ) g(τ − σ)h(σ)dσ dτ 0 0 Z t Z t = f (t − τ )g(τ − σ)dτ h(σ)dσ 0 σ Z Z t t−σ = f (t − σ − µ)g(µ)dµ h(σ)dσ 0

= [f ∗ g] ∗ h. 6. We see that z(p) = M p2 + k . So L[A] =

1 1/k −M p/k 1 = = + . p · z(p) p(M p2 + k) p M p2 + k

Using partial fractions, we find that we can write L[A] =

1 1 1 p · − · 2 . k p k p + k/M

As a result, 1 1 A = − cos k k

k t. M

Similarly, L[h] =

1 1 = . z(p) M p2 + k

CHAPTER 8. LAPLACE TRANSFORMS It follows that 1 h= · sin M Therefore Z x x(t) = 0

k t. M

" r # k 1 sin (t − τ ) f (τ ) dτ . M M

i h 1 = L1 e−Rt/L and the 7. The impulse response function is h(t) = L−1 Lp+R output current is Z Z 1 −Rt/L t 1 t −R(t−τ )/L E(τ ) · e dτ = e I(t) = E(τ ) · eRτ /L dτ. L 0 L 0 Rt (a) I(t) = EL0 e−Rt/L 0 u(τ ) · eRτ /L dτ = ER0 (1 − e−Rt/L ). (b) I(t) = EL0 e−Rt/L . Rt (c) I(t) = EL0 e−Rt/L 0 sin(ωτ ) · eRτ /L dτ . This evaluates to I(t) =

E0 [ωLe−Rt/L + R sin ωt − ωL cos ωt]. 2 R + ω 2 L2

Chapter 9

The Calculus of Variations 9.1

Introductory Remarks

9.2

Euler’s Equation

1. Find the extremals for I =

R x2 x1

f (x, y, y 0 )dx.

∂f 0 (a) Since x is missing, the extremals satisfy ∂y 0 · y − f = C. This is √ p 0 2 0 2 ) √(y ) 0 2 − 1+(y = C which becomes −1 = Cy 1 + (y 0 )2 . This y y

1+(y )

in turn, is equivalent to y 2 (1 + (y 0 )2 ) = A2 where A is positive. 2 2 = dx and Consequently, (y 0 )2 = A y−y . This separates to √ ydy 2 A2 −y 2 p A2 − y 2 = x + B. Equivalently, (x + B)2 + y 2 = A2 . ∂f 0 (b) Since x is missing, ∂y 0 · y − f = C. Therefore, the extremals are solutions to −2(y 0 )2 − (y 2 − (y 0 )2 ) = C or −(y 0 )2 − y 2p= C. This is equivalent to (y 0 )2 + y 2 = A2 , A ≥ 0. Hence y 0 = A2 − y 2 so y √ dy = dx. Integrate to arcsin A = x + B so y = A sin(x + B). 2 2 A −y

(c) Both x and y are missing so the extremals are solutions to the equation fy0 y0 · y 00 = 0. This is 2y 00 = 0 so y = Ax + B. √ √ (d) Make the change of independent variable t = 2x. Then y 0 = 2ẏ and the function f (x, y, y 0 ) = 2y 2 − (y 0 )2 becomes g(t, y, ẏ) = 2(y 2 − (ẏ)2 ). The calculations in part (b) show that the extremals for g are y = A√ sin(t + B). Therefore, the extremals for f are the curves y = A sin( 2x + B). 2. This problem falls under Case II in the text. So the Euler equation becomes d ∂f = 0. dx ∂y 0 99

100

CHAPTER 9. THE CALCULUS OF VARIATIONS Now we calculate that

∂f = x − 2y 0 . ∂y 0

Hence 0=

d dx

∂f ∂y 0

= 1 − 2y 00 .

We conclude that y 00 = 1/2 or y 0 = x/2 + C or y = x2 /4 + Cx + D . Plugging in the two boundary conditions yields that C = −1/4 and D = 0. So y = x2 /4 − x/4 . 3. Since fy0 = 2a(x)y 0 + 2b(x)y, we have fy0 y0 = 2a(x), fy0 y = 2b(x), and fy0 x = 2a0 (x) + 2b0 (x)y. Moreover, fy = 2b(x)y 0 + 2c(x)y. Therefore, the extremals satisfy the equation a(x)y 00 + a0 (x)y 0 + [b0 (x) − c(x)]y = 0. 4. Since the variable θ is missing, we are in Case III. According to the text, Euler’s equation then becomes ∂f · r0 − f = C . ∂r0 Here we use r0 to denote dr/dθ. Since f = ((r0 )2 + r2 )1/2 , we see that 1 0 2 ((r ) + r2 )−1/2 · 2r0 · r0 − ((r0 )2 + r2 )1/2 = C 2 or −r2 = C((r0 )2 + r2 )1/2 . It is easy to check directly, for instance, that r = C cos θ satisfies this equation. That is the equation of a line. q 5. Observe that f (φ, θ, θ0 ) = 1 + (θ0 )2 sin2 φ. Since θ is missing, the ex0

∂f φ tremals satisfy the equation ∂θ That is, √ θ sin 0 = C. 2 0 2

1+(θ ) sin φ

= C. If

C = 0, then θ0 = 0 and θ = θ0 , a constant. This is the geodesic if the points P and Q happen to lie on a great circle through the North and South poles. This actually describes all of the geodesics since any two points will lie on such a curve after a suitable rotation of the sphere.

9.3. ISOPERIMETRIC PROBLEMS AND THE LIKE

101

7. The arc length integral for this parametrization is Z Q L= P

dx dx

dy dz

dz dz

2 dz.

dy 0 0 cos θ · θ0 + g 0 (z) sin θ, and Since dx dz = −g(z) sin θ · θ + g (z) cos θ, dz R= g(z) p Q dz integral is L = P g(z)2 (θ0 )2 + [g 0 (z)]2 + 1dz. dz = 1, the arc lengthp 0 Therefore, f (z, θ, θ ) = g(z)2 (θ0 )2 + [g 0 (z)]2 + 1. Observe that θ is miss∂f ing so the extremals are solutions for the equation ∂θ 0 = C. Consequently, θ 0 g(z)2 √ = C. Solve for θ0 to obtain θ(z) is the solution to g(z)2 (θ 0 )+ [g 0 (z)]2 +1 √ R √1+[g0 (z)]2 C 1+[g 0 (z)]2 √ 2 2 . Integration yields θ = C √ 2 2 dz + D. θ0 = g(z)

9.3

g(z) −C

g(z)

g(z) −C

Isoperimetric Problems and the Like

1. Solve by the method of Lagrange multipliers.

(a) Let F (x, y, z, λ) = x2 + y 2 + z 2 + λ(ax + by + cz − d). The extreme values must satisfy the following equations. ∂F = 2x + λa = 0 ∂x ∂F = 2y + λb = 0 ∂y ∂F = 2z + λc = 0 ∂z ∂F = ax + by + cz − d = 0 ∂λ Solve the first 3 equations for x, y, z respectively and substitute into the fourth to obtain − λ2 · (a2 + b2 + c2 ) = d which implies that λ = − a2 +b2d2 +c2 . Consequently, x=

bd cd ad ,y = 2 ,z = 2 . a2 + b2 + c2 a + b2 + c2 a + b2 + c 2

(b) We maximize f (x, y, z) = s(s − x)(s − y)(s − z) subject to the constraint g(x, y, z) = (x + y + z)/2 − s = 0. Let F (x, y, z) =

102

CHAPTER 9. THE CALCULUS OF VARIATIONS f (x, y, z) − λg(x, y, z) and ∂F = −s(s − y)(s − z) + λ/2 = 0 ∂x ∂F = −s(s − z)(s − z) + λ/2 = 0 ∂y ∂F = −s(s − x)(s − y) + λ/2 = 0 ∂z ∂F = (x + y + z)/2 − s = 0 ∂λ Multiply the first equation by s − x, the second by s − y, and the third by s − z, to see that (s − x) ·

λ λ λ = (s − y) · = (s − z) · . 2 2 2

Since λ 6= 0, it follows that x = y = z. (c) We wish to maximize f (x1 , x2 , . . . , xn ) = x1 · x2 · · · xn subject to the constraint g(x1 , x2 , . . . , xn ) = x1 + x2 + · · · + xn − s = 0. Let F (x1 , x2 , . . . , xn ) = f (x1 , x2 , . . . , xn ) + λg(x1 , x2 , . . . , xn ). The extrema must satisfy ∂F = x2 · x3 · · · · · xn + λ = 0 ∂x1 ∂F = x1 · x3 · · · · · xn + λ = 0 ∂x2 .. . ∂F = x1 · x2 · · · · · xn−1 + λ = 0 ∂xn ∂F = x1 + x2 + . . . + xn − s = 0. ∂λ Multiply the j th equation by xj to see that x1 · λ = x2 · λ = · · · = xn · λ. Since λ 6= 0 it follows that x1 = x2 = · · · = xn = ns and the maximum n value of the produce is ns . Consequently, n x1 + x2 + . . . + xn x1 · x2 · · · · · xn ≤ n and

√ n

x1 · x2 · · · · · xn ≤

x1 + x2 + . . . + xn . n

2. Set up the obvious extremal problem as in the text and solve it.

9.3. ISOPERIMETRIC PROBLEMS AND THE LIKE

103

3. Let the chain have length L and hang from (a, A) to (b, B) where a < b and A and B are positive. We may assume that if y = y(x) is the curve that describes the shape of the chain, then y(x) > 0. The chain’s potential p Rb energy is proportional to I(y) = a y(x) 1 + y 0 (x)2 dx. We will minimize Rbp this integral subject to J(y) = a 1 + y 0 (x)2 dx = L. p p 2 + λ 1 + y 0 (x)2 and the extremals Define F (x, y, y 0 ) = y(x) 1 + y 0 (x) d ∂F ∂F must satisfy the Euler equation dx ∂y 0 − ∂y = 0. Calculate the deriva00

y 1 tives and simplify to obtain the equation 1+(y 0 )2 = y+λ . Since the variable dy dv dv dv = dy · dx = v dy x is missing we make the substitution y 0 = v, y 00 = dx dy vdv 1 2 to obtain 1+v 2 = y+λ . This can be integrated to yield 2 ln(1 + v ) = 0 2 2 2 ln(y + λ) + C which implies that 1 + (y ) = C (y + λ) where the constant C is now positive. It will be convenient √ to let y +λ = u/C for a moment to obtain 1 + (u0 )2 /C 2 = u2 or du = C u2 − 1. This equation separates to dx du √ = Cdx implying that arccosh(u) = Cx + D so u = cosh(Cx + D). u2 −1 Finally, since y = u/C − λ, the chain must lie along the curve determined by the function y(x) = C1 cosh(Cx + D) − λ. This is a catenary. d 5. Since the surface is defined by g(x, z) = 0, gy = 0 implying that dt (ẏ/f ) = 0. Therefore, there is a constant C such that

ẏ(t) p

ẋ(t)2 + ẏ(t)2 + ż(t)2

for any geodesic r(t) = (x(t), y(t), z(t)) and for all t. If the geodesic intersects the y-axis at t = t0 , the angle between the curve and the positive ẏ(t),ż(t)) is y-axis is the dot product T (t0 ) · (0, 1, 0) where T (t) = √ (ẋ(t), 2 2 2 ẋ(t) +ẏ(t) +ż(t)

the unit tangent vector. According to the equation displayed above, this angle is C for any geodesic. 6. Of course the result follows immediately from Green’s theorem or Stokes’s theorem. But one can also note that the first integral gives the area below the upper boundary curve and the second integral gives the area below the lower boundary curve. The difference is the area that we seek. R 2π dy R 2π 2 7. 12 0 x dt − y dx = 12 0 (a2 cos2 t + a2 sin2 )dt = a2 · 2π = πa2 . dt

104

CHAPTER 9. THE CALCULUS OF VARIATIONS

Chapter 10

Systems of First-Order Equations 10.1

Introductory Remarks

1. Replace the differential equations with an equivalent system of first-order equations. (a) y 00 − xy 0 − xy = 0; let y0 = y, y1 = y 0 to obtain y00 = y1 y10 = xy0 + xy1 . (b) y 000 = y 00 − x2 (y 0 )2 ; let y0 = y, y1 = y 0 , y2 = y 00 to obtain y00 = y1 y10 = y2 y20 = −x2 y12 + y2 . (c) xy 00 − x2 y 0 − x3 y = 0; let y0 = y, y1 = y 0 to obtain y00 = y1 y10 = x2 y0 + xy1 . (d) y (4) − xy 000 + x2 y 00 − x3 y = 1; let y0 = y, y1 = y 0 , y2 = y 00 , y3 = y 000 to obtain y00 = y1 y10 = y2 y20 = y3 y30 = x3 y0 − x2 y2 + xy3 + 1. 105

106

CHAPTER 10. SYSTEMS OF FIRST-ORDER EQUATIONS

2. We have

10.2

x0 = x

y0 = y

x1 = x00 mx01 = f (t, x, y)

y1 = y00 my10 = g(t, x, y)

Linear Systems

1. The Wronskian for (*) is x1 (t)x02 (t)−x01 (t)x2 (t). Since y1 = x01 and y2 = x02 this is also the Wronskian of the two solutions to (**): x (t) x2 (t) x (t) x2 (t) W (x) = det 1 = det 10 . y1 (t) y2 (t) x1 (t) x02 (t) 2. (a) For the first solution, dx/dt = 4e4t and x + 3y = 4e4t . Also dy/dt = 4e4 t and 3x + y = 4e4t . So both equations are satisfied. For the second solution, dx/dt = −2e−2t and x + 3y = −2e−2t . Also dy/dt = 2e−2t and 3x + y = 2e−2t . So both equations are satisfied. (b) If it there were constants c1 and c2 so that c1 e4t + c2 e−2t ≡ 0 then, taking derivatives, we would also have 4c1 e4t − 2c2 e−2t ≡ 0. The only simultaneous solutions of these two equations are c1 = c2 = 0. So the solutions are linearly independent. As another approach, if c1 e4t + c2 e−2t = 0 then e6t = −c2 /c1 (assuming that c1 6= 0). But the right side of this equation is constant while the left side is not. That is a contradiction unless c1 = c2 = 0. So the two solutions are linearly independent. (c) The general solution is x(t)

Ae4t + Be−2t

y(t)

Ae4t − Be−2t

Substituting in the initial conditions, we find that 5

= A+B

= A−B

The solution of this system is A = 3 , B = 2. So the particular solution that we seek is x(t) = 3e4t + 2e−2t , y(t) = 3e4t − 2e−2t .

10.2. LINEAR SYSTEMS

107

3. (a) If x = 2e4t and y = 3e4t , then x0 = 8e4t and y 0 = 12e4t . Since x + 2y = 8e4t and 3x + 2y = 12e4t , the first pair of functions do form a solution to the given homogeneous system. Verification for the second pair can be made with a similar calculation. (b) The two solutions are clearly linearly independent. One pair is not a scalar multiple of the other. This can formally be verified by calcu2e4t e−t lating their Wronskian: W (t) = det = −5e3t 6= 0. 3e4t −e−t (c) The verification is straightforward. Simply substitute and simplify. Since the two solutions in part (a) are linearly independent there is a general solution of the form ( x(t) = 2c1 e4t + c2 e−t + 3t − 2 . y(t) = 3c1 e4t − c2 e−t − 2t + 3 4. (a) Differentiating the first equation in t gives dx dy d2 x = +2 . dt2 dt dt Now subtracting the two differential equations gives dx dy = + 2x . dt dt Substituting this last equation into the first equation yields d2 x dx −3 − 4x = 0 . 2 dt dt This is second order linear, and we solve it as usual to obtain x(t) = e4t , x(t) = e−t . We can use the first differential equation to see that x = e4t corresponds to y = (3/2)e4t . We can use the first differential equation also to see that x = e−t corresponds to y = −e−t . That accounts for our two linearly independent solutions. (b) Differentiating the second equation in t gives d2 y dy dx +2 . =3 2 dt dt dt Now subtracting three times the first differential equation from the second gives dx 1 dy 4 = + y. dt 3 dt 3

108

CHAPTER 10. SYSTEMS OF FIRST-ORDER EQUATIONS Substituting this last equation into the first equation yields dy d2 y −3 − 4y = 0 . dt2 dt This is second order linear, and we solve it as usual to obtain y(t) = e4t , y(t) = e−t . We can use the first differential equation to see that y = e4t corresponds to x = (2/3)e4t . We can use the first differential equation also to see that y = e−t corresponds to x = −e−t . That accounts for our two linearly independent solutions.

5. Differentiate the first equation: x00 = x0 + y 0 , and use the second equation to replace y 0 with y: x00 = x0 +y. Now use the first equation to eliminate y: x00 = x0 + x0 − x. This second order equation simplifies to x00 − 2x0 + x = 0 which, by inspection, has the general solution x(t) = c1 et + c2 tet . Use the first equation to find y: y(t) = x0 (t) − x(t) = c1 et + c2 (tet + et ) − c1 et − c2 tet = c2 e t The general solution is (

x(t) = c1 et + c2 tet y(t) = c2 et

7. Substitute (6) to (4) ( d dt [c1 x1 (t) + c2 x2 (t)] = a1 (t)[c1 x1 (t) + c2 x2 (t)] + b1 (t)[c1 y1 (t) + c2 y2 (t)] d dt [c1 y1 (t) + c2 y2 (t)] = a2 (t)[c1 x1 (t) + c2 x2 (t)] + b2 (t)[c1 y1 (t) + c2 y2 (t)] Carry out the differentiation on the left sides and rearrange the right sides to obtain the following equivalent system. ( c1 x01 (t) + c2 x02 (t) = c1 [a1 (t)x1 (t) + b1 (t)y1 (t)] + c2 [a1 (t)x2 (t) + b1 (t)y2 (t)] c1 y10 (t) + c2 y20 (t) = c1 [a2 (t)x1 (t) + b2 (t)y1 (t)] + c2 [a2 (t)x2 (t) + b2 (t)y2 (t)] This system is satisfied for all t in [a, b] because of the assumption that the functions are solutions to the homogeneous equation. 8. Let x, y be any solution of (3). Consider x

= x(t) − xp (t)

= y(t) − yp (t) .

10.3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS109 We calculate that dx dx dxp = (t) − (t) = (a1 x + b1 y + f1 ) − (a1 x + b1 y) = f1 . dt dt dt Similarly dy = f2 . dt That proves the result.

10.3

Homogeneous Systems with Constant Coefficients

1. Find the general solution to each system. (a) The auxiliary equation is m− 1 = 0 with roots m1 = −1 and m2 = 1. With m1 = −1 the algebraic system is −2A + 4B = 0 −2A + 4B = 0 A = 2 and B = 1 is a nontrivial solution and ( x = 2e−t y = e−t is a nontrivial solution to the system of differential equations. With m2 = 1 a similar calculation yields ( x = et y = et as a second, independent solution. The general solution is ( x = 2c1 e−t + c2 et y = c1 e−t + c2 et . (b) The auxiliary equation is m2 − 6m + 18 = 0 with roots m1 = 3 + 3i and m2 = 3 − 3i. With m1 = 3 + 3i the algebraic system is (1 − 3i)A − 2B = 0 5A + (−1 − 3i)B = 0 A∗ = 2 and B ∗ = 1 − 3i has a nontrivial complex solution. Observe that A1 = 2, A2 = 0 and B1 = 1, B2 = −3. This is what is needed to

110

CHAPTER 10. SYSTEMS OF FIRST-ORDER EQUATIONS obtain two real solution. See Equations (16) and (17), Section 10.3. The solutions are ( x = e3t · 2 cos 3t y = e3t (cos 3t + 3 sin 3t) and ( x = e3t · 2 sin 3t y = e3t (sin 3t − 3 cos 3t)

The general solution is ( x = 2e3t [c1 cos 3t + c2 sin 3t] y = eet [c1 (cos 3t + 3 sin 3t) + c2 (sin 3t − 3 cos 3t)]. (c) The auxiliary equation is m2 − 6m + 9 = 0 with roots m1 = m2 = 3. With m1 = 3 the algebraic system is 2A + 4B = 0 −A − 2B = 0. A = 2 and B = −1 is a nontrivial solution and ( x = 2e3t y = −e3t is a nontrivial solution to the system of differential equations. We seek a second solution of the form ( x = (A1 + A2 t)e3t y = (B1 + B2 t)e3t . Substitute these into the system of differential equations to obtain (3A1 + A2 + 3A2 t)e3t = 5(A1 + A2 t)e3t + 4(B1 + B2 t)e3t (3B1 + B2 + 3B2 t)e3t = −(A1 + A2 t)e3t + (B1 + B2 t)e3t . The exponential terms can be canceled and, since these equations are identities in the variable t, it follows that 3A1 + A2 = 5A1 + 4B1 , 3A2 = 5A2 + 4B2 3B1 + B2 = −A1 + B1 , 3B2 = −A2 + B2 . The two equations on the right are actually the same: A2 = −2B2 , and can be solved with A2 = 2, B2 = −1. Substitute these values into

10.3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS111 the equations on the left and they can be solved with A1 = 1, B1 = 0. This yields a second solution of the form ( x = (1 + 2t)e3t y = −te3t . The general solution is ( x = [2c1 + c2 (1 + 2t)]e3t y = −(c1 + c2 t)e3t .

(d) The auxiliary equation is m2 + 2m = 0 with roots m1 = 0 and m2 = −2. With m1 = 0 the algebraic system is 4A − 3B = 0 8A − 6B = 0 A = 3 and B = 4 is a nontrivial solution and ( x=3 y=4 is a nontrivial (constant) solution to the system of differential equations. With m2 = −2 the algebraic system is 6A − 3B = 0 8A − 4B = 0 A = 1 and B = 2 is a nontrivial solution and ( x = e−2t y = 2e−2t is a second, independent solution. The general solution is ( x = 3c1 + c2 e−2t y = 4c1 + 2c2 e−2t . (e) This is an uncoupled system. By inspection, the general solution is ( x = c1 e2t y = c2 e3t . (f) The auxiliary equation is m2 +6m+9 = 0 with roots m1 = m2 = −3. With m1 = −3 the algebraic system is −A − B = 0 A + B = 0.

112

CHAPTER 10. SYSTEMS OF FIRST-ORDER EQUATIONS A = 1 and B = −1 is a nontrivial solution and ( x = e−3t y = −e−3t is a nontrivial solution to the system of differential equations. We seek a second solution of the form ( x = (A1 + A2 t)e−3t y = (B1 + B2 t)e−3t . Substitute these into the system of differential equations to obtain −(3A1 − A2 + 3A2 t)e−3t = −[4A1 + B1 + (4A2 + B2 )t]e−3t −(3B1 − B2 + 3B2 t)e−3t = [A1 − 2B1 + (A2 − 2B2 )t]e−3t . The exponential terms can be canceled and, since these equations are identities in the variable t, it follows that 3A1 − A2 = 4A1 + B1 , 3A2 = 4A2 + B2 −3B1 + B2 = A1 − 2B1 , −3B2 = A2 − 2B2 . The two equations on the right are actually the same: A2 = −B2 , and can be solved with A2 = 1, B2 = −1. Substitute these values into the equations on the left and they can be solved with A1 = −1, B1 = 0. This yields a second solution of the form ( x = (−1 + t)e−3t y = −te−3t . The general solution is ( x = [c1 + c2 (−1 + t)]e−3t y = −(c1 + c2 t)e−3t . (g) The auxiliary equation is m2 − 13m + 30 = 0 with roots m1 = 10 and m2 = 3. With m1 = 10 the algebraic system is −3A + 6B = 0 2A − 4B = 0 A = 2 and B = 1 is a nontrivial solution and ( x = 2e10t y = e10t

10.3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS113 is a nontrivial solution to the system of differential equations. With m2 = 3 a similar calculation yields ( x = 3e3t y = −2e3t as a second, independent solution. The general solution is ( x = 2c1 e10t + 3c2 e3t y = c1 e10t − 2c2 e3t . (h) The auxiliary equation is m2 − 6m + 13 = 0 with roots m1 = 3 + 2i and m2 = 3 − 2i. With m1 = 3 + 2i the algebraic system is (−2 − 2i)A − 2B = 0 4A + (2 − 2i)B = 0. A∗ = 1 − i and B ∗ = −2 is a nontrivial complex solution. Observe that A1 = 1, A2 = −1 and B1 = −2, B2 = 0. This is what is needed to obtain two real solutions. These solutions are ( x = e3t (cos 2t + sin 2t) y = −2e3t cos 2t and

(

x = e3t (sin 2t − cos 2t) y = −2e3t sin 2t.

The general solution is ( x = e3t [c1 (cos 2t + sin 3t) + c2 (sin 2t − cos 2t)] y = −2e3t (c1 cos 2t + c2 sin 2t). 2. We solve the system by differentiating the first equation: d2 x dx dy = a1 + b1 . dt2 dt dt Also b1 so

Thus

dy dx − b2 = (b1 a2 − b2 a1 )x dt dt

dy b2 dx b1 a2 − b2 a1 = + x. dt b1 dt b1 d2 x dx − (b2 + a1 ) − (b1 a2 − b2 a1 )x = 0 . 2 dt dt

114

CHAPTER 10. SYSTEMS OF FIRST-ORDER EQUATIONS Solving, we find that possible exponents are p b2 + a1 ± (b2 − a1 )2 + 4a2 b1 . 2 We see immediately that the condition a2 b1 > 0 guarantees two distince exponents, so the solutions will be linearly independent. A similar calculation obtains in solving for y.

3. In the Wronskian calculation all terms cancel except those with repeated sines and cosines. at e (A1 cos bt − A2 sin bt) eat (A1 sin bt + A2 cos bt) W (t) = det at e (B1 cos bt − B2 sin bt) eat (B1 sin bt + B2 cos bt) = e2at [(A1 cos bt − A2 sin bt)(B1 sin bt + B2 cos bt) (A1 sin bt + A2 cos bt)(B1 cos bt − B2 sin bt)] = e2at (A1 B2 cos2 bt − A2 B1 sin2 bt + A1 B2 sin2 bt − A2 B1 cos2 bt) = e2at (A1 B2 − A2 B1 )(cos2 bt + sin2 bt) = e2at (A1 B2 − A2 B1 ). If A1 B2 − A2 B1 = 0, then there is a constant k such that B1∗ = kA∗1 and the system has the solution ( x = A∗1 (e(a+ib)t y = kA∗1 e(a+ib)t . But then y = kx, which implies that the auxiliary equation has real roots. 5. (a) Substitute ( x = v1 (t)x1 (t) + v2 (t)x2 (t) y = v1 (t)y1 (t) + v2 (t)y2 (t) into the forced system. Because ( x = c1 x1 (t) + c2 x2 (t) y = c1 y1 (t) + c2 y2 (t) solves the homogeneous equation, most of the terms will cancel leaving the following system of equations ( v10 (t)x1 (t) + v20 (t)x2 (t) = f1 (t) v10 (t)y1 (t) + v20 (t)y2 (t) = f2 (t) The fact that the Wronskian of the homogeneous solutions is not zero guarantees that this system has a solution and, with luck, the solution functions can be integrated to yield a particular solution to the forced equation.

10.4. NONLINEAR SYSTEM

115

(b) Using the homogeneous solutions from Example 10.3.1, and the given forcing functions, the variation of parameters system has the following form. ( v10 e−3t + v20 e2t = −5t + 2 −4v10 e−3t + v20 e2t = −8t − 8 Subtract the second equation from the first to obtain the equation 5v10 e−3t = 3t + 10. Therefore, v10 = 51 (3t + 10)e3t , which integrates to v1 =

1 (t + 3)e3t . 5

The function v2 can be found by multiplying the first equation by 4 and adding to the second: 5v20 e2t = −28t. This integrates to v2 =

7 (2t + 1)e−2t . 5

The particular solution is ( xp = 15 (t + 3)e3t · e−3t + 75 (2t + 1)e−2t · e2t yp = 15 (t + 3)e3t · (−4e−3t ) + 57 (2t + 1)e−2t · e2t which simplifies to ( xp = 3t + 3 yp = 2t − 1.

10.4

Nonlinear System

1. Differentiate the first equation to obtain x00 = x0 (a − by) − bxy 0 . Use the second equation to eliminate y 0 : x00 = x0 (a − by) − bx[−y(c − gx)] = ax0 + (bcx − bx0 − bgx2 )y. Now use the first equation to eliminate the remaining y. x00 = ax0 + b(cx − x0 − gx2 ) ·

ax − x0 bx

This can also be expressed in the form xx00 = (x0 )2 + (gx2 − cx)x0 + acx2 − agx3 .

116

CHAPTER 10. SYSTEMS OF FIRST-ORDER EQUATIONS

2. Differentiating the second equation in (21) yields d2 y dy dx = − (c − gx) − y −g . dt2 dt dt Now plugging in the second equation yields dx d2 y = y(c − gx)2 + yg . 2 dt dt Since y, g > 0, we see that dx/dt > 0 implies d2 y/dt2 > 0.

Chapter 11

Partial Differential Equations and Boundary Value Problems 11.1

Introduction and Historical Remarks

11.2

Eigenvalues and the Vibrating String

1. Find the eigenvalues and eigenfunctions for y 00 + λy√= 0. In all√of the problems λ > 0 and the general solution is y = A sin λx + B cos λx. √ (a) y(0) = 0 implies that B√ = 0 so y = A sin λx. The condition y(2π) = 0 implies that λ · 2π = nπ for some integer n and the eigenvalues are λn = n2 /4, n = 1, 2, 3, · · · . The eigenfunctions are yn = sin(nx/2). √ (b) y(0) = 0 implies that B√ = 0 so y = A sin λx. The condition y(2π) = 0 implies that λ · 2π = nπ for some integer n and the eigenvalues are λn = n2 /4, n = 1, 2, 3, · · · . The eigenfunctions are yn = sin(nx/2). √ (c) y(0) = 0 implies that √ B = 0 so y = A sin λx. The condition y(1) = 0 implies that λ = nπ for some integer n and the eigenvalues are λn = n2 π 2 , n = 1, 2, 3, · · · . The eigenfunctions are yn = sin nπx. √ (d) y(0) = 0 implies that B √ = 0 so y = A sin λx. The condition y(L) = 0 implies that λ · L = nπ for some integer n and the eigenvalues are λn = n2 π 2 /L2 , n = 1, 2, 3, · · · . The eigenfunctions are yn = sin(nπx/L). (e) The condition y(−L) = 0 and y(L) = 0 imply that λ must solve the 117

118

CHAPTER 11. PARTIAL DIFFERENTIAL EQUATIONS following two equations. −A sin A sin

√ √

√ λL + B cos

√

λL + B cos

λL = 0 λL = 0

It follows that either A or B must be 0 (verify). If B = 0, then √ λL = nπ and there are eigenvalues n2 π 2 /L2 with √ the eigenfunctions sin(nπx/L), n = 1, 2, 3, · · · . If A = 0, then λL = (2n − 1)π/2 and there are eigenvalues (2n − 1)2 π 2 /4L2 with the eigenfunctions cos((2n − 1)πx/2L), n = 1, 2, 3, · · · . (f ) Make the substitution t = x − a and, according to part (d), the eigenvalues are n2 π 2 /(b−a)2 with eigenfunctions yn (t) = sin nπt/(b− a). In terms of the variable x the eigenvalues are the same; the eigenfunctions are yn (x) = sin nπ(x − a)/(b − a). The solution technique used in (f) can also be applied to the problem in part (e) yielding eigenvalues λn = n2 π 2 /4L2 and eigenfunctions yn = sin nπ(x + L)/2L. Show that the solution formulas are equivalent to the ones found above. 2. (a) With u(x, t) = F (x + at) + G(x − at) we see that ux = F 0 (x + at) + G0 (x − at) ,

uxx = F 00 (x + at) + G00 (x − at)

and ut = aF 0 (x + at) − aG0 (x − at) ,

utt = a2 F 00 (x + at) + a2 G00 (x − at) .

Now it is clear that a2 uxx = utt . (b) With α = x + at and β = x − at we may define v(α, β) = u(x, t). Then ∂v ∂α ∂v ∂β ∂v ∂v ∂u = + = ·1+ ·1 ∂x ∂α ∂x ∂β ∂x ∂α ∂β and

∂2u ∂2v ∂2v ∂2v = + + 2 . ∂x2 ∂α2 ∂β 2 ∂α∂β

Likewise, ∂u ∂v ∂α ∂v ∂β ∂v ∂v = + = ·a+ · (−a) ∂t ∂α ∂t ∂β ∂t ∂α ∂β and

∂2u ∂2v ∂2v ∂2v = a2 2 + a2 2 − 2a2 . 2 ∂t ∂α ∂β ∂α∂β

Thus we see that a2 uxx − utt = 0 says precisely the same thing as vαβ = 0.

11.3. THE HEAT EQUATION

119

3. The solution has the form y(x, t) = F (x + at) + G(x − at). (a) The condition y(x, 0) = f (x) implies that F (x) + G(x) = f (x). The condition yt (x, 0) = 0 implies that aF 0 (x) − aG0 (x) = 0. Thus F 0 (x) = G0 (x) so there is a constant C such that F (x)−G(x) = C. It follows that 2F (x) = f (x) + C and 2G(x) = f (x) − C. This permits us to express the solution as y(x, t) = F (x + at) + G(x − at) 1 1 = [f (x + at) + C] + [f (x − at) − C] 2 2 1 = [f (x + at) + f (x − at)] 2 (b) If y(0, t) = 0 for all t, then f (at) = −f (−at) for all t and f is an odd function. If, in addition, y(π, t) = 0 for all t, then f (π + at) = −f (π−at) = f (at−π) for all t. Given x choose t such that x = at−π. Then f (x) = f (at − π) = f (at + π) = f (x + 2π). (c) Since f is odd, f (0) = f (−0) = −f (0). Thus 2f (0) = 0 implying that f (0) = 0. Because f has period 2π, f (−π) = f (π). It is also true that f (−π) = −f (π) (f is odd). Consequently, f (π) = 0. P∞ (d) Bernoulli’s solution is y(x, t) = the j=1 b+ j sin jx cos jt. Using P∞ given identity this can be expressed in the form y(x, t) = j=1 bj · 1 1 2 [sin j(x P+∞at) + sin j(x − at)] = 2 [f (x + at) + f (x − at)], where f (x) = j=1 bj sin jx. 4. (b) Setting t = 0, we see that the bj are the coefficients for the sine series of f (x) = (1/π)x(π − x). Thus Z Z Z π 2 2 2 π 1 x(π−x) sin jx dx = x sin jx dx− 2 bj = x2 sin jx dx π 0 π π π π 0 =

4 π2 j 3

[−1 + (−1)j ] .

5. If the initial shape is f (x) = c sin x, then the solution is y(x, t) = c sin x cos t. Thus for every t the shape of the solution is a single arch of a sine curve. Similarly, if the initial shape is f (x) = c sin nx, then the solution is y(x, t) = c sin nx cos nt. Thus for every t the shape of the solution is a sine curve with nodes at xk = π/k, k = 1, 2, · · · , n − 1.

11.3

The Heat Equation

1. The Fourier series solution to a2 wxx (x, t) = wt (x, t) satisfying the boundP∞ 2 2 ary conditions w(0, t) = w(π, t) = 0 is W (x, t) = j=1 bj e−j a t sin jx.

120

CHAPTER 11. PARTIAL DIFFERENTIAL EQUATIONS This can be seen by substituting w(x, t) = u(x)v(t) and separating variables. An easier way is to make the change of variables τ = a2 t in the heat equation to obtain wxx (x, τ ) = wτ (x, τ ) having the solution P∞ −j 2 τ W (x, τ ) = sin jx obtained in the text, and then express j=1 bj e the solution in terms of t. Now let w(x, t) = W (x, t) + g(x), where g(x) = w1 + π1 (w2 − w1 )x. By the superposition principal, w is also a solution to the heat equation and, since W (0, t) = W (π, t) = 0, w(x, t) satisfies the boundary conditions w(0, t) = g(0) = w1 , w(π, t) = g(π) = w2 . The initial temperature distribution, w(x, 0) = f (x),Pdetermines the val∞ ues of the coefficients bj as follows. Since f (x) = j=1 bj sin jx + g(x) P∞ the coefficients must be chosen so that j=1 bj sin jx = f (x) − g(x). ConRπ sequently, bj = π2 0 (f (x) − g(x)) sin jxdx, and the solution is w(x, t) = P∞ −j 2 a2 t sin jx + g(x). j=1 bj e

3. First find separated solutions to a2 wxx = wt + cw. The process is made √ easier by rescaling the variables. Let τ = ct and z = ac x to obtain the equation wzz = wτ + w. Substitute w(z, τ ) = α(z)β(τ ) to get α00 β = αβ 0 + αβ. Divide by αβ and 0 00 the variables are separated: αα = ββ + 1. Since the left side depends only on z and the right side depends only on τ , there is a constant K 0 00 such that αα = K = ββ + 1. Thus α00 = Kα and β 0 = (K − 1)β. This implies that √β(τ ) = Ce(K−1)τ . Regarding α, the √boundary conditions w(0, t) = w( ac π, t) = 0 require that α(0) = 0 and α( ac π) √ = 0. This forces K = −a2 n2 /c for some integer n and α(z) = A sin(anz/ c). In terms of z √ 2 2 and τ the separated solutions are w(z, r) = e−(a n /c+1)τ sin(anz/ c). In 2 2 terms of x and t, w(x, t) = e−ct e−n a t sin nx. Taking linear combinations P∞ 2 2 we have the formal series solution w(x, t) = e−ct j=1 bj e−j a t sin jx. P∞ The initial condition w(x, 0) = f (x) becomes j=1 bj sin jx = f (x) which R π is satisfied provided bj = π2 0 f (x) sin jxdx. 4. Now the initial boundary conditions are ∂u(0, t)/∂t = 0 and ∂u(1, t)/∂t = 0. The solution to such a system is that the rod will have temperature 100◦ at all points. 5. We seek separated solutions to the heat equation: a2 wxx = wt , satisfying the boundary conditions wx (0, t) = 0 = wx (π, t). Substitute w(x, t) = 0 00 α(x)β(t) to get a2 α00 β = αβ 0 or αα = αβ2 β . THus there is a constant K 00

such that αα = K = aβ2 β . That is, α00 = Kα and β 0 = Ka2 β, so β(t) = 2

CeKa t . Since the temperature is not expected to grow exponentially √ √ with time we assume K ≤ 0 so α(x) = sin −Kx or α(x) = cos −Kx or α(x) = C, a constant. The last possibility corresponds to K = 0.

11.4. THE DIRICHLET PROBLEM FOR A DISC

121

The boundary conditions require√α0 (0) = 0 = α0 (π). Consequently α(x) = C,√a constant, or α(x) = cos −Kx with K chosen so that α0 (0) = √ − K sin −Kπ = 0. Therefore, the eigenvalues are K = −n2 , n = 2 2 0, 1, 2, · · · . The separated solutions are w(x, t) = e−n a t cos nx. ThereP 2 2 ∞ fore, the series solution is w(x, t) = a20 + j=1 aj e−j a t cos jx where the P ∞ coefficients aj satisfy w(x, 0) = a20 + j=1 aj cos jx = f (x). That is, Rπ aj = π2 0 f (x) cos jxdx. 7. Let (x, y, z) be the point at the center of the box R. Its six faces are centered at x±∆x/2, y, z), (x, y±∆y/2, z), and (x, y, z±∆/2) respectively. The temperature at the center of the box at time t is w(x, y, z, t). Let ∆w denote the change in temperature at the center corresponding to a small change in time ∆t. The corresponding change in total heat energy in R is approximately proportional to ∆w · ∆V where ∆V = ∆x∆y∆z. By the principal of conservation of energy this change must equal the heat energy that has flowed into the box across its six faces. Taking into account the fact that the heat flow across a face is proportional to the area of the face, the temperature gradient across the face, and the length of the time interval, we have ∆w · ∆V ≈ a2 [∂x w(x + ∆x/2, y, z, t) − ∂x w(x − ∆x/2, y, z, t))∆y∆z + (∂y w(x, y + ∆y/2, z, t) − ∂y w(x, y − ∆y/2, t))∆x∆z + (∂z w(x, y, z + ∆z/2, t) − ∂z w(x, y, z − ∆z/2, t))∆x∆y]∆t Divide both sides by ∆V · ∆t to obtain ∂x w(x + ∆x/2, y, z, t) − ∂x w(x − ∆x/2, y, z, t) ∆w ≈ a2 [ ∆t ∆x ∂y w(x, y + ∆y/2, z, t) − ∂y w(x, y − ∆y/2, z, t) + ∆y ∂z w(x, y, z + ∆/2, t) − ∂z w(x, y, z − ∆z/2, t) ] + ∆z Now let ∆t → 0, then shrink the box to its center to get 2 ∂w ∂ w ∂2w ∂2w = a2 + + . ∂t ∂x2 ∂y 2 ∂z 2

11.4

The Dirichlet Problem for a Disc

1. Solve the Dirichlet problem for the unit disk for the given boundary function f (θ). Rπ j+1 (a) f (θ) = cos θ/2 is even; aj = π2 0 cos θ/2 · cos jθdθ = π4 · (−1) 4j 2 −1 , a0 = 4/π. Therefore, w(r, θ) =

∞ 2 4 X (−1)j j − r cos jθ. π π j=1 4j 2 − 1

122

CHAPTER 11. PARTIAL DIFFERENTIAL EQUATIONS (b) f (θ) = θ is odd; bj = π2

Rπ 0

θ sin jθdθ = 2 · (−1)j

w(r, θ) =

∞ X (−1)j+1 j=1

j+1

. Therefore,

rj sin jθ.

Rπ j (c) f (θ) is neither even nor odd; aj = π1 0 sin θ cos jθdθ = − π1 · 1+(−1) j 2 −1 , R π j 6= 1; a0 = 2/π, a1 = 0; bj = π1 0 sin θ sin jθdθ = 0, j 6= 1; b1 = 1/2. Therefore, w(r, θ) =

∞ 1 2X 1 1 − r2j cos 2jθ + r sin θ. π π j=1 4j 2 − 1 2

(d) The function f (θ) − 1/2 is odd; a0 /2 = 1/2; aj = 0, j > 1; bj = R j+1 +1 1 (−1) 1 π . Therefore, π 0 sin jθdθ = π · j w(r, θ) =

∞ 2X 1 1 + r2j−1 sin(2j − 1)θ. 2 π j=1 2j − 1

Rπ j 2 (e) f (θ) = θ2 /4 is even; aj = π2 0 θ2 /4 cos jθdθ = (−1) j 2 , a0 = π /6. Therefore, ∞ π 2 X (−1)j j + r cos jθ. w(r, θ) = 12 j=1 j 2 2. This is a simple change of variable in the radial direction. 3. Let the circle C be centered at (x0 , y0 ) (Cartesian coordinates) with radius R. The function u(x, y) = w(x0 + x, y0 + y) is harmonic on the disk centered at the origin of radius R. According to the Poisson integral formula for this disk (Exercise 2), u’s value at R πthe center of the disk: (0, θ), 1 (polar coordinates) is given by u(0, θ) = 2π u(R, φ)dφ. In terms of the −π original function w this formula can be expressed in the following form. Z 1 w(x0 , y0 ) = πw(x0 + R cos φ, y0 + R sin φ)Rdφ. 2πR −pi 4. This is an elementary application of the Chain Rule. 5. Neither Maple nor Mathematica can obtain useful formulas for the Poisson integrals.

Chapter 12

The Nonlinear Theory 12.1

Some Motivating Examples

12.2

Specializing Down

1. Derive equation (2). Let s be the distance from the bob to its equilibrium position, measured along the path of motion (the circle). The force on the bob in the direction of motion is Fg = −mg sin x. The damping force is Fd = −cs0 so, according to Newton’s Second Law, ms00 = −mg sin x − cs0 . Using the fact that s = ax this becomes max00 = −mg sin x − cax0 . Therefore, x00 +

c 0 g x + sin x = 0. m a

2. This is a standard application of the Picard existence and uniqueness theorem. See Chapter 3. 3. The same curves, traversed in positive directions. 5. Let x0 = y. Equation (1) is equivalent to ( x0 = y y 0 = − ag sin x. The critical points are (nπ, 0), n an integer. Equation (2) is equivalent to ( x0 = y c y 0 = − ag sin x − m y. It has the same critical points as Equation (1), (nπ, 0), n an integer. 123

124

CHAPTER 12. THE NONLINEAR THEORY Equation (3) is equivalent to ( x0 = y y 0 = −x − µ(x2 − 1)y. It has one critical point, (0, 0).

6. (a) We have y = 0 and −y + x3 + x2 − 2x = 0. Thus x3 + x2 − 2x = 0 so x = 0, −2, 1. The critical points are thus (0, 0), (−2, 0), (1, 0). (b) We have x = y and x2 − 5x + 6 = 0. Hence x = 2, 3 and the critical points are (2, 2) and (3, 3). 7. Clearly x = x0 et . Substitute this into the second equation, y 0 = x0 et + et , to see that y = x0 et + et + A. The constant A can be expressed in terms of y(0) = y0 : A = y0 − x0 − 1, so the solution starting at (x0 , y0 ) at t = 0 is ( x(t) = x0 et y(t) = (x0 + 1)et + y0 − x0 − 1. The trajectories are straight lines. Some of them are sketched below. The vertical trajectory starts at (0, −3) and the horizontal trajectories all start at a point of the form (−1, k). There are no equilibrium points.

12.3

Types of Critical Points: Stability

1. (a) All points (x, 0) are critical points. None are isolated. Eliminating t, 2xy 2 dy dy 2xdx 2 dx = y(x2 +1) so y = x2 +1 . Consequently, ln y = ln(x + 1) + C and 2 y = C(x + 1). The paths are parabolas. If y > 0 x is increasing and if y < 0, x is decreasing. (b) The origin (0, 0) is the only critical point. Eliminating t, dy/dx = −x/y so ydy = −xdx. Consequently, y / 2 = −x2 /2 + C and y 2 + x2 = C. The paths are circles. If y > 0, then x is increasing so the circles are traversed in the clockwise direction. (c) There are no critical points. Eliminating t, dy/dx = cos x so y = sin x + C. The paths are sine waves. Since x0 > 0 the tractories are traversed from left to right. (d) All points (0, y) are critical points. None are isolated. Eliminating t, dy/dx = −2xy 2 so −dy/y 2 = 2xdx and 1/y = x2 + C. The paths lie on the curves y = 1(x2 +C). Since x0 > 0 when x < 0 the trajectories in the left half-plane are traversed from left to right. The tractories in the right half-plane are traversed from right to left. All points move toward the y-axis.

12.3. TYPES OF CRITICAL POINTS: STABILITY

125

2. (a) We calculate the determinant of 1−m 0 0 −1 − m The resulting roots are m = ±1. For m = 1 we find that A = 1 and B = 0. So the corresponding solution is x = et , y = 0. For m = −1 we find that A = 0, B = 1. The corresponding solution is x = 0, y = e−t . The general solution is then x = c1 et , y = c2 e−t . The differential equation for the paths is −y dy = . dx x We solve this equation by separation of variables to obtain y=

C . x

This critical point (0, 0) is clearly a saddle point. (b) We calculate the determinant of −1 − m 0

0 −2 − m

The resulting roots are m = −1, −2. For m = −1 we find that A = 1, B = 0. The corresponding solution is x = e−t , y = 0. For m = −2 we find that A = 0, B = 1. The corresponding solution is x = 0, y = e−2t . The general solution is then x = c1 et , y = c2 e−2t . The differential equation for the paths is 2y dy = . dx x We solve this equation by separation of variables to obtain y = Cx2 . This critical point (0, 0) is clearly a node. . 3. This second order equation is equivalent to the following system. (

x0 = y y 0 = 2x3 .

126

CHAPTER 12. THE NONLINEAR THEORY Therefore, the origin is an isolated critical point. Eliminating t, dy/dx = 2x3 /y, so ydy = 2x3 dx and y 2 /2 = x4 /2 + C. The solution trajectories lie on the curves y 2 − x4 = C. If y > 0, then x is increasing; if y < 0, then x is decreasing. Several solution curves are sketched below. Note that the four trajectories that lie on the parabolas y = ±x2 not only form a natural boundary between two distinct classes of solution curves, but also seem to be attracting all trajectories to them as t → ∞ and as t → −∞.

12.4

Critical Points and Stability for Linear Systems

1. (a) The auxiliary equation is (m − 2)(m − 3) = 0 so m1 = 2 and m2 = 3. The origin is an unstable node. (b) THe auxiliary equation is m2 +6m+13 = 0 with roots m1,2 = −1±2i. The origin is an asymptotically stable spiral. (c) The auxiliary equation is m2 −1 = 0 with roots m1 = 1 and m2 = −1. The origin is an unstable saddle points. (d) The auxiliary equation is m2 + 9 = 0 with roots ±3i. The origin is a stable center. (e) The auxiliary equation is m2 + 6m + 9 = 0 with roots −3, −3. The origin is an asymptotically stable borderline node. (f) The auxiliary equation is m2 + 2m = 0 with roots m1 = 0 and m2 = −2. The origin is not an isolated critical point. It can be classified as stable. (g) The auxiliary equation is m2 − 6m + 18 = 8 with roots m1,2 = 3 ± 3i. The origin is an unstable spiral. 2. In case a1 b2 − a2 b1 = 0 then the two equations coincide and the simultaneous solutions form a line. So there are certainly infinitely many (indeed uncountably many) solutions, and none of them is isolated. 3. (a) The condition a1 b1 − a2 b1 6= 0 guarantees that there is only one point (x0 , y0 ) where x0 = 0 and y 0 = 0 simultaneously. (b) Make the substitution to obtain ( dx̄ dt = a1 (x̄ + x0 ) + b1 (ȳ + y0 ) + c1 dȳ dt = a2 (x̄ + x0 ) + b2 (ȳ + y0 ) + c2 . By the choice of x0 and y0 in part (a), this simplifies to ( dx̄ dt = a1 x̄ + b1 ȳ dȳ dt = a2 x̄ + b2 ȳ.

12.4. CRITICAL POINTS AND STABILITY FOR LINEAR SYSTEMS 127 (c) The critical point is (−3, 2)–find the common zeros of the two linear terms on the right √ side. The eigenvalues for the system are negative: m1,2 = −3 ± 3, so the critical point is an asymptotically stable node. 4. (a) The auxiliary equation is given by the determinant of 0−m 1 −a2 −2b − m That is, m2 + 2bm + a2 = 0 . The roots of this equation are m = −b ±

b2 − a2 .

(b) (i) Here the critical point is (0, 0). And m = ±ia. The motion is oscillatory. (iii) Here the critical point is (0, 0). And m = −b. Now the oscillation is dying out. 5. The hypotheses of Case E imply that b2 = −a1 and a21 + a2 b1 = −c2 for some positive number c. Begin by replacing b2 with −a1 and making the substitution y/x = u. This implies that dy/dx = xdu/dx + u and we have x

a2 − a1 u du +u= . dx a1 + b1 u

A little algebra yields u + ab11 u2 + 2 ab11 u − ab12

du = −

dx . x

Complete the square in the denominator, u + ab11 u2 + 2 ab11 u − ab12

du = −

dx , x

and replace a21 + a2 b1 with −c2 u + ab11 2 (u + ab11 )2 + cb2 1

du = −

dx . x

This integrates to 1 ln 2

a1 u+ b1

c2 + 2 b1

# = − ln x + A,

128

CHAPTER 12. THE NONLINEAR THEORY which, in terms of x and y, can be written in the form

y a1 + x b1

2 +

c2 B = 2, b21 x

where B > 0. Multiply both sides by b21 x2 to get (b1 y+a1 x)2 +c2 x2 = b21 B. Now expand the squared term and remove c in favor of −a21 −a2 b1 to obtain b21 y 2 + 2a1 b1 xy − a2 b1 x2 = b21 B 2 . This determines a one parameter family of ellipses because 4a21 b21 + 4b21 a2 b1 = 4b21 (a21 + a2 b1 ) < 0.

12.5

Stability by Liapunov’s Direct Method

1. Each one can be classified using Theorem 11.5.4. (a) Neither since a > 0 and b2 − 4ac > 0. (b) Of positive type since a > 0 and b2 − 4ac < 0. (c) Neither since a < 0 and b2 − 4ac > 0. (d) Of negative type because a < 0 and b2 − 4ac < 0. 2. The presence of the cubic terms means that F is of one sign when x is large and positive and the opposite sign when x is large and negative. Similarly for y. So the function is neither of positive nor of negative type. 3. (a) As in Example 11.5.3, we seek a Liapunov function of the form E(x, y) − ax2m + by 2n . OBserve that for such a function ∂E ∂E F+ G = 2max2m−1 (−3x3 − y) + 2nby 2n−1 (x5 − 2y 3 ) ∂x ∂y = (2nby 2n−1 x5 − 2max2m−1 y) − (6max2m+2 + 4nby 2n+2 ). We seek m and n (positive integers) and positive values for a and b that make the first term zero. Clearly n = 1, m = 3, a = 1, and b = 3 will do the job. Thus the function E(x, y) = x6 + 3y 2 is of positive type and (∂E/∂x)F + (∂E/∂y)G = −18x8 − 12y 4 is of negative type. According to Theorem 11.3, the origin is asymptotically stable. (b) As above, consider E(x, y) = ax2m + by 2n . Observe that for such a function ∂E ∂E F+ G = 2max2m−1 (−2x + xy 3 ) + 2nby 2n−1 (x2 y 2 − y 3 ) ∂x ∂y = (2nbx2 y 2n+1 + 2max2m y 3 ) − (4max2m + 2nby 2n+2 ).

12.6. SIMPLE CRITICAL POINTS OF NONLINEAR SYSTEMS

129

Let n = 1, m = 1, a = 1, and b = 1 and the function E(x, y) = x2 +y 2 is of positive type while (∂E/∂x)F + (∂E/∂y)G = 4x2 y 3 − 4x2 − 2y 4 = −4(1 − y 3 )x2 − 2y 4 is of negative type when |y| < 1. The origin is asymptotically stable. 4. The positivity of the derivative in (d) implies that any path beginning near the origin (at a point where E is positive) will continue out to infinity, and that E will blow up along that path. So the critical point is unstable. 5. Consider E(x, y) = ax2m + by 2n . Observe that for such a function ∂E ∂E F+ G = 2max2m−1 (2xy + x3 ) + 2nby 2n−1 (−x2 + y 5 ) ∂x ∂y = (4max2m y − 2nbx2 y 2n−1 ) + (2max2m+2 + 2nby 2n+4 ). Let n = 1, m = 1, a = 1, and b = 2 and the function E(x, y) = x2 + 2y 2 is of positive type and so is (∂E/∂x)F + (∂E/∂y)G = 2x4 + 4y 6 . According to Exercise 4, the origin is unstable.

12.6

Simple Critical Points of Nonlinear Systems

1. E(x, y) = x2 +y 2 is a Liapunov function for the first system with (∂E/∂x)F + (∂E/∂y)G = −2x4 − 2y 4 . Thus (0, 0) is an asymptotically stable critical point. When applied to the second system, E(x, y) = x2 + y 2 has the property that (∂E/∂x)F + (∂E/∂y)G = 2x4 + 2y 4 implying that (0, 0) is unstable. See Exercise 4 in Section 11.5. 2. The requested figure is above. Now we write r = a sin 2θ = 2a sin θ cos θ hence r3 = 2a(r sin θ)(r cos θ) so (x2 + y 2 )3/2 = 2axy . Differentiating both sides gives 1/2 3 2 dy x + y2 · (2x + 2ydy/dx) = 2xy + 2ax 2 dx

130

CHAPTER 12. THE NONLINEAR THEORY

The function r = a sin 2θ. or

p dy 2ay − 3x x2 + y 2 p = . dx −2ax + 3y x2 + y 2

That is the differential equation that we seek. 3. (a) Observe that (0, 0) is an isolated singular point of the linearized sysp 2 + y2 ≤ tem. Moreover, when polar coordinates are used, |2xy|/ x p 2r and 3y 2 / x2 + y 2 ≤ 3r. Both approach 0 as r → 0 so (0, 0) is a simple critical points. The auxiliary equation √ of the linearized system is m2 − 2m + 3 = 0 with roots m1,2 = 1 ± 2i. Thus (0, 0) is an unstable spiral for the linearized system and for the original system as well. p (b) Using polar p coordinates, |3x2 y|/ x2 + y 2 ≤ 3r2 and, since | sin x| ≤ |x|, |y sin x|/ x2 + y 2 ≤ r. Both approach 0 as r → 0. Moreover, the linearized system has an isolated critical point at (0, 0) so the origin is a simple critical point. The auxiliary equation for the linearized √ −5± 17 2 system is m + 5m + 2 = 0 with roots m1,2 = . Both roots 2 are negative so the origin is an asymptotically stable node for the linearized system and for the original system as well. 4. We examine the associated linear system dx dt dy dt

−x + µy .

The determinant of the matrix 0−m −1

1 µ−m

12.7. NONLINEAR MECHANICS: CONSERVATIVE SYSTEMS

131

is m2 − µm + 1 and the roots are m=

µ±

µ2 − 4 . 2

If µ ≥ 2 then it is clear that the critical point (0, 0) is unstable. If 0 < µ < 2 then the critical point is unstable. If µ < 0 then the critical point is stable. p a1 b1 5. Because det 6= 0, T (x, y) = (a1 x + b1 y)2 + (a2 x + b2 y)2 ata)2 b2 tains a positive minimum value m on the circle C of radius 1 centered at the origin. Suppose that there are nonzero critical points p for the system (xn , yn ) such that (xn , yn ) → (0, 0) as n → ∞. Let rn = x2n + yn2 and we have the following contradiction

0 < m ≤ T (xn /rn , yn /rn ) p (a1 xn + b1 yn )2 + (a2 xn + b2 yn )2 = rn p 2 f (xn , yn ) + g(xn , yn )2 = rn |f (xn , yn )| |g(xn , yn )| ≤ + → 0 as n → ∞. rn rn

12.7

Nonlinear Mechanics: Conservative Systems

1. The nonlinear spring equation is equivalent to the system ( dx dt = y dy 3 dt = −kx − αx . The solution curves satisfy the equation dy/dx = −(kx + αx3 )/y so ydy = (−kx − αx3 )dx and y 2 = 2(−kx2 /2 − αx4 /4) + E. p The solution trajectories lie on the curves y = ± E − (kx2 + αx4 /2). The following pictures display the integral curves for k = 1, α = 1 and for k = 1, α = −1. ** 2. We can write this as dx dt dy dt

x − 2x3 .

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CHAPTER 12. THE NONLINEAR THEORY

Sketch of the paths. This leads to the differential equation dy x − 2x3 = . dx y Solving by separation of variables yields p y = ± x2 − x4 + C . The figure shows these curves for various values of C. It is clear that the critical point at the origin is stable. 3. For the hard spring in Exercise 1, z = V (x) = x2 + x4 /2 and V 0 (x) = 0 only when x = 0; this is a minimum yielding a stable center at (0, 0) as shown in the picture on the left. In the picture on the right in Exercise 1, the spring is soft. z = V (x) = x2 − x4 /2 and V 0 (x) = 2x(1 − x2 ) which is 0 at x = 0 and at x = ±1. The former is a local minimum for V –(0, 0) is a stable center–and the latter two are absolute maxima yielding unstable saddle points at (1, 0) and (−1, 0) in the (x, y)-phase plane. The phase plane trajectories and the energy curve for an inflection point are shown below. The critical point (1, 0) is an unstable cusp.

12.8

The Poincaré-Bendixson Theorem

1. Introduce polar coordinates as was done for system (3) in this section. This yields ( dr r2 dt = r(3 − e ) dθ dt = 1.

12.8. THE POINCARÉ-BENDIXSON THEOREM

133

√ Clearly there are periodic solutions: r(t) = ln 3, θ(t) = t + t0 . In terms of x and y, ( √ x(t) = ln 3 cos(t + t0 ) √ y(t) = ln 3 sin(t + t0 ). 2. (a) We apply Theorem 11.8.5. Observe that both f and g are continuously differentiable, that g is odd, and that f is even. √ Furthermore, F (x) = x5 − 3x3 = x3 (x2 − 3) is negative on (0, 3), positive on (a, ∞) and increases to ∞ on the latter interval. We may conclude that the differential equation has a unique closed path surrounding the origin. So there is a periodic solution. (b) We apply Theorem 11.8.2. Rewrite the equation as dx dt dy dt We see that

= y =

(x2 + 1)y − x5

∂F ∂G + = x2 + 1 , ∂x ∂y

which is always positive. So the differential equation has no periodic solutions. (e) We apply Theorem 11.8.5. Observe that both f and g are continuously differentiable, that g is odd, and that f is even. Furthermore, F (x) = x7 /7 is negative on (−∞, 0), positive on (0, ∞) and increases to ∞ on the latter interval. We may conclude that the differential equation has a unique closed path surrounding the origin. So there is a periodic solution. 3. Rescale the variable t with the substitution τ = αt, α > 0, to obtain d2 x dx + bα(x2 − 1) + cx = 0. 2 dτ dτ p Divide this equation by aα2 and let α = c/a to get aα2

d2 x dx + µ(x2 − 1) + x = 0, 2 dτ dτ √ where µ = b/ ac.

134

CHAPTER 12. THE NONLINEAR THEORY

Chapter 13

Qualitative Properties and Theoretical Aspects 13.1

A Bit of Theory

1. The associated polynomial is r2 − functions solve the equation. 1x so both e e−x Their Wronskian is W (x) = det x = −2. Therefore, they are e −e−x independent solutions. 2. On any interval not containing 0, the equation can be rewritten as y 00 − 2y 0 2y 2 x + x2 = 0. Both x and x are solutions (verify). The Wronskian is 2 W (x) = x so they are linearly independent and y = c1 x + c2 x2 is the general solution. Initial conditions determine that c1 = 12 and c2 = 25 . the particular solution is y = 21 x + 25 x2 . 3. Both ex and e2x are solutions (verify). The Wronskian is W (x) = e3x so they are linearly independent and y = c1 ex + c2 e2x is the general solution. The initial conditions yield the system c1 + c2 = −1, c1 + 2c2 = 1 with solutions c1 = −3, c2 = 2. The particular solution is y = −3ex + 2e2x . 4. Both e2x and xe2x are solutions (verify). The Wronskian is W (x) = e4x so they are linearly independent and y = c1 e2x + c2 xe2x is the general solution. 5. By inspection one solution if y1 = x2 . The formula developed in R Section 2.4 can be used to obtain a second solution: p(x) = 0 so y2 = x2 x14 dx = 1 3x . These solutions are linearly independent on [1, 2] (verify) so the general solution is y = c1 x2 +c2 x−1 . The initial conditions yield the particular solution y = 3x2 − 2x−1 . 6. Verify y1 and y2 are linearly independent solutions on the interval [0, 2]. Find the particular solution satisfying the stated initial conditions. 135

136CHAPTER 13. QUALITATIVE PROPERTIES AND THEORETICAL ASPECTS (a) Both ex and e−2x are solutions (verify). The Wronskian is W (x) = −3e−x so they are linearly independent and y = c1 ex + c2 e−2x is a general solution. Initial conditions determine that c1 = 6, and c2 = 2. The particular solution is y = 6ex + 2e−2x . (b) Both ex and e−2x are solutions (verify). The Wronskian is W (x) = −3e−x so are linearly independent and y = c1 ex + c2 e−2x is a general solution. Initial conditions determine that c1 = 0, and c2 = 0. The particular solution is a constant y = 0. (c) Both e−2x and e−3x are solutions (verify). The Wronskian is W (x) = −e−5x so they are linearly independent and y = c1 e−2x + c2 e−5x is a general solution. Initial conditions determine that c1 = e−2 and B = −1. The particular solution is y = e−2 − e−x . (d) Both 1 and e−x are solutions (verify). The Wronskian is W (x) = −e−x so they are linearly independent and y = c1 + c2 e−x . Initial conditions determine that c1 = e−2 and c2 = −1. The particular solution is y = e−2 − e−x . 7. Solutions to y 00 + (y 0 )2 = 0. dp (a) Let p = y 0 so p0 − −p2 , a separable equation. Integrate −p 2 = dx 1 to obtain p1 = x + C or y 0 = x+C . Integrate once more to find that y = ln |x+C|+D. Every solution is either of this form or is a solution to y 0 = 0 (these solutions were lost when we divided by p to separate variables). Thus there are also constant solutions of the form y = A. (b) Clearly y1 = 1 and y2 = ln x are solutions on an interval I in the positive real half-line. They are also linearly independent on I: c1 · 1 + c2 · ln x = 0 for all x in I implies that c2 · x1 = 0 on I so c2 = 0 (and c1 = 0 also). Unfortunately, y = c1 + c2 ln x is not the general solution on I. This is because there are solutions that are not of this form, y = ln(x + 1) for example.

8. If two solutions y1 , y2 have a common zero x0 in the interval, then W (x0 ) = y1 (x0 )y20 (x0 ) − y2 (x0 )y10 (x0 ) = 0. If they have maxima or minima at the same point x0 in the interval, then y10 (x0 ) = y20 (x0 ) = 0 and W (x0 ) = 0. Since W is either nowhere vanishing or vanishes identically, W = 0 and y1 , y2 are linearly dependent. 3 x ,x ≥ 0 9. Observe that g(x) = −x3 , x < 0 3x5 − 3x5 ,x ≥ 0 (a) W (f, g) = f (x)g 0 (x) − f 0 (x)g(x) = −3x5 + 3x5 , x < 0 (b) Sketch the graphs of f and g on [−1, 1] to see that they are linearly independent in this interval. (One is not a scalar multiple of the other.)

13.2. PICARD’S EXISTENCE AND UNIQUENESS THEOREM

137

(c) his does not contradict Proposition 3.1.1 because f and g do not satisfy the hypotheses of that statement: They are not solutions to a second-order, linear, homogeneous differential equation on the interval in question. 00

u 10. (a) Substituting y = uv and simplify to get v 00 + v 0 ( 2u u + P (x)) + v( u + u0 2u0 P (x) u + Q(x)) = 0. Solve the separable equation u + P (x) = 0 R

P (x)

to get u(x) = e − 2 dx . Then ( uu )0 = uu − ( uu )2 implies that u00 u

u u 2 = − P 2(x) + [ P (x) 2 ] . Substituting u and u to the equation 0

yields v 00 + v[− P 2(x) − P 4(x) + Q(x)] = 0. R

P (x)

(b) PR(x) = 2x, P 0 (x) = 2 and Q(x) = 1+x2 . By (a), u(x) = e − 2 dx = 1 e −xdx = e− x2 . Substituting y = uv yields v 00 = 0. The general 1 1 solution is v = c1 + c2 x and y = c1 e− x2 + Bxe− x2 .

13.2

Picard’s Existence and Uniqueness Theorem

13.3

Oscillations and the Sturm Separation Theorem

1. To verify (1), let let f (x) = y1 (x + b) and g(x) = y1 (x)y2 (b) + y2 (x)y1 (b). Then both f and g are solutions to the equation y 00 + y = 0. Moreover, they both satisfy the initial conditions y(0) = y1 (b) and y 0 (0) = y10 (b) (the function g satisfies the second condition because y2 = y10 ). The uniqueness theorem implies that f = g. A similar argument will verify property (2). Properties (3) and (4) follow from (1) and (2), as do (5) and (6). 2. The Wronskian of y1 = a sin x + b cos x and y2 = c sin x + d cos x is W = −(ad − bc) (verify). If ad − bc 6= 0, then w 6= 0 and y1 , y2 are linearly independent. By Theorem 3.3.1, zeros of y1 , y2 are distinct and occur alternately. 2

3. Writing Bessel’s equation as y 00 + x1 y 0 + (1 − xp2 )y = 0 we have p(x) = x1 2

and q(x)R = 1 − xp2 . Therefore, solutions are of the form y = uv with 1 1 v = e− 2 x dx = √1x , and u a solution to the equation u00 + qu = 0 where 2

1−4p q = 1 − xp2 − 14 · x12 − 12 · −1 x2 = 1 + x2 . Given any Rp value, there is a ∞ positive x0 such that q(x) is positive for x ≥ x0 and x0 q(x)dx = +∞. The argument used to verify Theorem 3.3.3 can be used to show that if y is a non-trivial solution, then u (and y) have infinitely any zeros greater than x0 .

138CHAPTER 13. QUALITATIVE PROPERTIES AND THEORETICAL ASPECTS 4. Substituting y = ez yields ÿ− ẏ+k = 0 (see Problem 5 in Section 2.1). The associated polynomial is r2 − r + k. When k ≤ 41 , the polynomial has two √ √ 1 1 real roots 12 ± 12 1 − 4k and the general solution is y = Ae( 2 + 2 1−4k)z + 1

√

√ 1+ 1−4k

√ 1− 1−4k

Be( 2 − 2 1−4k)z = Ax 2 + Bx 2 , which has finite √many pos1 itive zeros. When k > 4 , the polynomial has two roots 1± 24k−1 i and √ √ the general solution is y = Aez/2 cos( 4k−1 z) + Bez/2 sin( 4k−1 z) = 2 2 √ √ √ √ 4k−1 4k−1 A x cos( 2 ln x) + B x sin( 2 ln x), which has infinitely many positive zeros.

13.4

The Sturm Comparison Theorem

1. Let x1 and x2 be successive positive zeros of a nontrivial solution yp of x2 y 00 + xy 0 + (x2 − p2 )y = 0. The solution has the form yp = x−1/2 up 2 where up is a solution to the normal equation u00 + (1 + 1−4p 4x2 )u = 0. (a) Let 0 ≤ p < 1/2. Theorem 3.4.3 implies that x2 − x1 is less than π. 2 < k 2 . Therefore, Let k > 1. For all x sufficiently large, 1 + 1−4p 4x2 if x1 is sufficiently large, then the next zero of up : x2 , cannot lie between x1 and x1 +π/k (we are comparing up to a nontrivial solution ofy 00 + k 2 y = 0 that also vanishes at x1 ). In other words, x2 − x1 ≥ π/k. (b) The argument is similar to the one used in part (a). 2. By Problem 4 in Section 3.3, y 00 + (k/x2 )y = 0 has infinite number of positive zeros if k > 1/4 and finitely many positive zeros if k ≤ 1/4. Apply Theorem 3.4.2 to y 00 + q(x)y = 0 and y 00 + (k/x2 )y = 0 yields the results. 3. If f (x) ≥ 0 for all x and k > 0, then every nontrivial solution of the differential equation y 00 +(f (x)+k)y = 0 has an infinite number of positive zeros.