Chapter 09

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CHAPTER 9

Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. y = 9x2 ex = 3x2 y and y(0) = 3 by inspection. 3

2. y = x3 − 2 sin x, y(0) = 3 by inspection. dy dy = c; (1 + x) = (1 + x)c = y dx dx (b) second order; y = c1 cos t − c2 sin t, y + y = −c1 sin t − c2 cos t + (c1 sin t + c2 cos t) = 0

3. (a) first order;

c dy + y = 2 − e−x/2 + 1 + ce−x/2 + x − 3 = x − 1 dx 2 t (b) second order; y = c1 e − c2 e−t , y − y = c1 et + c2 e−t − c1 et + c2 e−t = 0

4. (a) first order; 2

5.

y3 1 dy dy dy dy = y 2 + 2xy , (1 − 2xy 2 ) = y 3 , = y dx dx dx dx 1 − 2xy 2

6. 2x + y 2 + 2xy

dy = 0, by inspection. dx

d 3x ye = 0, ye3x = C, y = Ce−3x dx dy separation of variables: = −3dx, ln |y| = −3x + C1 , y = ±e−3x eC1 = Ce−3x y including C = 0 by inspection

7. (a) IF: µ = e3

dx

= e3x ,

d [ye−2t ] = 0, ye−2t = C, y = Ce2t dt dy = 2dt, ln |y| = 2t + C1 , y = ±eC1 e2t = Ce2t separation of variables: y including C = 0 by inspection

(b) IF: µ = e−2

8. (a) IF: µ = e−4

dt

= e−2t ,

x dx

= e−2x , 2

2 d −2x2 ye = 0, y = Ce2x dx

2 2 dy = 4x dx, ln |y| = 2x2 + C1 , y = ±eC1 e2x = Ce2x y including C = 0 by inspection

separation of variables:

d t ye = 0, y = Ce−t dt dy = −dt, ln |y| = −t + C1 , y = ±eC1 e−t = Ce−t separation of variables: y including C = 0 by inspection

9. µ = e 4dx = e4x , e4x y = ex dx = ex + C, y = e−3x + Ce−4x

(b) IF: µ = e

10. µ = e2

x dx

dt

2

= ex ,

= et ,

2 2 2 d x2 1 2 1 ye = xex , yex = ex + C, y = + Ce−x dx 2 2

395

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396

Chapter 9

11. Âľ = e

12.

13.

dx

x

x

=e ,e y=

ex cos(ex )dx = sin(ex ) + C, y = e−x sin(ex ) + Ce−x

1 dy + 2y = , Âľ = e 2dx = e2x , e2x y = dx 2

1 2x 1 1 e dx = e2x + C, y = + Ce−2x 2 4 4

2 2 1 x dy + 2 y = 0, Âľ = e (x/(x +1))dx = e 2 ln(x +1) = x2 + 1, dx x + 1 d 2 C y x + 1 = 0, y x2 + 1 = C, y = √ dx x2 + 1

14.

1 dy +y = − , Âľ = e dx = ex , ex y = − x dx 1−e

15.

y

1 y 1

dy = dx, ln |y| = ln |x| + C1 , ln = C1 , = ÂąeC1 = C, y = Cx y x x x including C = 0 by inspection

16.

dy = 2x dx, tan−1 y = x2 + C, y = tan x2 + C 2 1+y

17.

√ √ dy x 2 2 dx, ln |1 + y| = − 1 + x2 + C1 , 1 + y = Âąe− 1+x eC1 = Ce− 1+x , = −√ 1+y 1 + x2 √ 1+x2

y = Ce− 18. y dy = 19.

20.

23.

24.

− 1, C = 0

x3 dx y 2 1 , = ln(1 + x4 ) + C1 , 2y 2 = ln(1 + x4 ) + C, y = Âą [ln(1 + x4 ) + C]/2 1 + x4 2 4

2(1 + y 2 ) y

dy = ex dx, 2 ln |y| + y 2 = ex + C; by inspection, y = 0 is also a solution

2 2 dy = −x dx, ln |y| = −x2 /2 + C1 , y = ÂąeC1 e−x /2 = Ce−x /2 , including C = 0 by inspection y

21. ey dy =

22.

ex dx = ln |1−ex |+C, y = e−x ln |1−ex |+Ce−x 1 − ex

sin x dx = sec x tan x dx, ey = sec x + C, y = ln(sec x + C) cos2 x

x2 dy −1 = (1 + x) dx, tan y = x + + C, y = tan(x + x2 /2 + C) 1 + y2 2

y − 1

1 dy dx 1

+ dy = csc x dx, ln = , −

y = ln | csc x − cot x| + C1 , y2 − y sin x y y−1 1 y−1 = ÂąeC1 (csc x − cot x) = C(csc x − cot x), y = , C = 0; y 1 − C(csc x − cot x) by inspection, y = 0 is also a solution, as is y = 1. 1 dy = cos x dx, ln |y| = sin x + C, y = C1 esin x y

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Exercise Set 9.1

25.

397

1 1 dy d + y = 1, Âľ = e (1/x)dx = eln x = x, [xy] = x, xy = x2 + C, y = x/2 + C/x dx x dx 2

3 1 + C, C = , y = x/2 + 3/(2x) 2 2

(a) 2 = y(1) =

(b) 2 = y(−1) = −1/2 − C, C = −5/2, y = x/2 − 5/(2x)

26.

2 2 dy x2 = x dx, ln |y| = + C1 , y = ÂąeC1 ex /2 = Cex /2 y 2 2

(a) 1 = y(0) = C so C = 1, y = ex

/2

1 2 1 = y(0) = C, so y = ex /2 2 2

(b)

−2

27. Âľ = e

x dx

−x2

=e

−x2

,e

y=

2xe−x dx = −e−x + C, 2

2

2

2

y = −1 + Cex , 3 = −1 + C, C = 4, y = −1 + 4ex

28. Âľ = e

dt

t

t

=e,ey=

2et dt = 2et + C, y = 2 + Ce−t , 1 = 2 + C, C = −1, y = 2 − e−t

29. (2y + cos y) dy = 3x2 dx, y 2 + sin y = x3 + C, π 2 + sin π = C, C = π 2 , y 2 + sin y = x3 + π 2 30.

dy 1 = (x + 2)ey , e−y dy = (x + 2)dx, −e−y = x2 + 2x + C, −1 = C, dx 2 1 1 1 2 2 −y 2 −y −e = x + 2x − 1, e = − x − 2x + 1, y = − ln 1 − 2x − x 2 2 2

31. 2(y − 1) dy = (2t + 1) dt, y 2 − 2y = t2 + t + C, 1 + 2 = C, C = 3, y 2 − 2y = t2 + t + 3 sinh x y = cosh x, Âľ = e (sinh x/ cosh x)dx = eln cosh x = cosh x, cosh x

1 1 1 (cosh 2x + 1)dx = sinh 2x + x + C = (cosh x)y = cosh2 x dx = 2 4 2 1 1 1 1 1 y = sinh x + x sech x + C sech x, = C, y = sinh x + x sech x + 2 2 4 2 2

32. y +

33. (a)

dx 1 dy = , ln |y| = ln |x| + C1 , y 2x 2

x = –0.5y2 x = –1.5y2

y 2 x = y2

|y| = C|x|1/2 , y 2 = Cx; by inspection y = 0 is also a solution.

x = – 3y2

x = 2y2

y=0 –2

2

1 1 sinh x cosh x + x + C, 2 2 1 sech x 4

x 2

2

(b) 1 = C(2) , C = 1/4, y = x/4 –2

x = 2.5y2

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398

Chapter 9

x2 y2 = − + C1 , y = Âą C 2 − x2 34. (a) y dy = −x dx, 2 2 √ (b) y = 25 − x2

y

y = √9 – x 2

3

y = √2.25 – x 2 y = √0.25 – x 2 x –3

3

y = – √1 – x 2 y = – √4 – x 2

–3

35.

dy x dx =− 2 , y x +4 1 ln |y| = − ln(x2 + 4) + C1 , 2 C y=√ x2 + 4

36. y + 2y = 3et , Âľ = e2 dt = e2t , d 2t ye = 3e3t , ye2t = e3t + C, dt y = et + Ce−2t 100 C=2 C=1 C=0

1.5 –2

C=2 C=1

C=0

–2

y = – √6.25 – x 2

2 C = –1 C = –2

2 C = –1 C = –2

–100

–1

37. (1 − y ) dy = x dx, 2

y−

2

38.

x3 y3 = + C1 , x3 + y 3 − 3y = C 3 3

1 +y y

yey

2

/2

dy = dx, ln |y| +

y2 = x + C1 , 2

= ÂąeC1 ex = Cex including C = 0 y

y

3

2

x –5

5

x –2

2

–3

–2

1 1 , all pass through the point 0, − and thus never through (0, 0). 2x2 − C C A solution of the initial value problem with y(0) = 0 is (by inspection) y = 0. The methods of Example 4 fail because the integrals there become divergent when the point x = 0 is included in the integral.

39. Of the solutions y =

1 1 40. If y0 = 0 then, proceeding as before, we get C = 2x2 − , C = 2x20 − , and y y0 1 y= , which is deďŹ ned for all x provided 2x2 is never equal to 2x20 − 1/y0 ; this 2 2x − 2x20 + 1/y0 last condition will be satisďŹ ed if and only if 2x20 − 1/y0 < 0, or 0 < 2x20 y0 < 1. If y0 = 0 then y = 0 is, by inspection, also a solution for all real x.

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Exercise Set 9.1

399

41.

dy x2 = xe−y , ey dy = x dx, ey = + C, x = 2 when y = 0 so 1 = 2 + C, C = −1, ey = x2 /2 − 1 dx 2

42.

dy 3x2 = , 2y dy = 3x2 dx, y 2 = x3 + C, 1 = 1 + C, C = 0, dx 2y

2

y 2 = x3 , y = x3/2 passes through (1, 1).

0

1.6 0

43.

dy = rate in − rate out, where y is the amount of salt at time t, dt y dy 1 dy 1 = (4)(2) − (2) = 8 − y, so + y = 8 and y(0) = 25. dt 50 25 dt 25

µ = e (1/25)dt = et/25 , et/25 y = 8et/25 dt = 200et/25 + C, y = 200 + Ce−t/25 , 25 = 200 + C, C = −175, (a) y = 200 − 175e−t/25 oz

44.

(b) when t = 25, y = 200 − 175e−1 ≈ 136 oz

y 1 dy 1 dy = (5)(20) − (20) = 100 − y, so + y = 100 and y(0) = 0. dt 200 10 dt 10

1 µ = e 10 dt = et/10 , et/10 y = 100et/10 dt = 1000et/10 + C, y = 1000 + Ce−t/10 , 0 = 1000 + C, C = −1000; (a)

y = 1000 − 1000e−t/10 lb

(b)

when t = 30, y = 1000 − 1000e−3 ≈ 950 lb

45. The volume V of the (polluted) water is V (t) = 500 + (20 − 10)t = 500 + 10t; if y(t) is the number of pounds of particulate matter in the water, dt y 1 dy 1 dy = 0 − 10 = − y, + y = 0; µ = e 50+t = 50 + t; then y(0) = 50, and dt V 50 + t dt 50 + t d [(50 + t)y] = 0, (50 + t)y = C, 2500 = 50y(0) = C, y(t) = 2500/(50 + t). dt The tank reaches the point of overflowing when V = 500 + 10t = 1000, t = 50 min, so y = 2500/(50 + 50) = 25 lb. 46. The volume of the lake (in gallons) is V = 264πr2 h = 264π(15)2 3 = 178,200π gals. Let y(t) denote y dy y = 0 − 103 the number of pounds of mercury salts at time t, then = − lb/h and V 178.2π dt dt t dy =− , ln y = − + C1 , y = Ce−t/(178.2π) , and y0 = 10−5 V = 1.782π lb; y 178.2π 178.2π C = y(0) = y0 10−5 V = 1.782π, y = 1.782πe−t/(178.2π) lb of mercury salts. t 1 2 3 4 5 6 7 8 9 10 11 12 y(t) 5.588 5.578 5.568 5.558 5.549 5.539 5.529 5.519 5.509 5.499 5.489 5.480

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400

Chapter 9 c dv d ct/m gm ct/m + v = −g, µ = e(c/m) dt = ect/m , ve e = −gect/m , vect/m = − + C, dt m dt c gm gm gm gm −ct/m gm +Ce−ct/m , but v0 = v(0) = − +C, C = v0 + ,v = − + v0 + e v=− c c c c c mg (b) Replace with vτ and −ct/m with −gt/vτ in (23). c vτ (c) From Part (b), s(t) = C − vτ t − (v0 + vτ ) e−gt/vτ ; g vτ vτ vτ s0 = s(0) = C −(v0 +vτ ) , C = s0 +(v0 +vτ ) , s(t) = s0 −vτ t+ (v0 +vτ ) 1 − e−gt/vτ g g g

47. (a)

48. Given m = 240, g = 32, vτ = mg/c: with a closed parachute vτ = 120 so c = 64, and with an open parachute vτ = 24, c = 320. (a) Let t denote time elapsed in seconds after the moment of the drop. From Exercise 47(b), while the parachute is closed v(t) = e−gt/vτ (v0 + vτ ) − vτ = e−32t/120 (0 + 120) − 120 = 120 e−4t/15 − 1 and thus −20/3 v(25) = 120 e − 1 ≈ −119.85, so the parachutist is falling at a speed of 119.85 ft/s 120 when the parachute opens. From Exercise 47(c), s(t) = s0 − 120t + 120 1 − e−4t/15 , 32 s(25) = 10000 − 120 · 25 + 450 1 − e−20/3 ≈ 7449.43 ft. (b) If t denotes time elapsed after the parachute opens, then, by Exercise 47(c), 24 (−119.85 + 24) 1 − e−32t/24 = 0, with the solution (Newton’s s(t) = 7449.43 − 24t + 32 Method) t = 307.4 s, so the sky diver is in the air for about 25 + 307.4 = 332.4 s. 49.

R V (t) d V (t) Rt/L dI + I= , µ = e(R/L) dt = eRt/L , (eRt/L I) = e , dt L L dt L

t 1 t 1 IeRt/L = I(0) + V (u)eRu/L du, I(t) = I(0)e−Rt/L + e−Rt/L V (u)eRu/L du. L 0 L 0 t

t 1 20e2u du = 2e−2t e2u = 2 1 − e−2t A. (a) I(t) = e−2t 5 0 0

(b)

lim I(t) = 2 A

t→+∞

50. From Exercise 49 and Endpaper Table #42, t

t e2u 1 3e2u sin u du = 15e−2t + e−2t I(t) = 15e−2t + e−2t (2 sin u − cos u) 3 5 0 0 1 1 = 15e−2t + (2 sin t − cos t) + e−2t . 5 5 51. (a)

ck dv = − g, v = −c ln(m0 − kt) − gt + C; v = 0 when t = 0 so 0 = −c ln m0 + C, dt m0 − kt m0 C = c ln m0 , v = c ln m0 − c ln(m0 − kt) − gt = c ln − gt. m0 − kt

(b) m0 − kt = 0.2m0 when t = 100 so m0 − 9.8(100) = 2500 ln 5 − 980 ≈ 3044 m/s. v = 2500 ln 0.2m0

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Exercise Set 9.1

401

dv dx dv dv dv dv = = v so m = mv . dt dx dt dx dt dx mv m (b) dv = −dx, ln(kv 2 + mg) = −x + C; v = v0 when x = 0 so kv 2 + mg 2k

52. (a) By the chain rule,

C=

m m m m kv02 + mg ln(kv02 + mg), ln(kv 2 + mg) = −x + ln(kv02 + mg), x = ln . 2k 2k 2k 2k kv 2 + mg

(c) x = xmax when v = 0 so xmax =

m kv02 + mg 3.56 × 10−3 (7.3 × 10−6 )(988)2 + (3.56 × 10−3 )(9.8) ln = ln ≈ 1298 m mg 2(7.3 × 10−6 ) (3.56 × 10−3 )(9.8) 2k

√ √ dh π 53. (a) A(h) = π(1)2 = π, π = −0.025 h, √ dh = −0.025dt, 2π h = −0.025t + C; h = 4 when dt h √ √ 0.025 t = 0, so 4π = C, 2π h = −0.025t + 4π, h = 2 − t, h ≈ (2 − 0.003979 t)2 . 2π (b) h = 0 when t ≈ 2/0.003979 ≈ 502.6 s ≈ 8.4 min. 54. (a) A(h) = 6 2 4 − (h − 2)2 = 12 4h − h2 , √ √ dh = −0.025 h, 12 4 − h dh = −0.025dt, 12 4h − h2 dt 3/2 −8(4 − h) = −0.025t + C; h = 4 when t = 0 so C = 0, (4 − h)

3/2

= (0.025/8)t, 4 − h = (0.025/8)

2/3 2/3

t

2√4 − (h − 2)2

h−2

2 h

,

h ≈ 4 − 0.021375t2/3 ft 8 (b) h = 0 when t = (4 − 0)3/2 = 2560 s ≈ 42.7 min 0.025 55.

1 1 1 1 dv 1 1 = − v 2 , 2 dv = − dt, − = − t + C; v = 128 when t = 0 so − = C, dt 32 v 32 v 32 128 1 1 1 128 dx dx 128 − =− t− ,v = cm/s. But v = so = , x = 32 ln(4t + 1) + C1 ; v 32 128 4t + 1 dt dt 4t + 1 x = 0 when t = 0 so C1 = 0, x = 32 ln(4t + 1) cm.

56.

√ √ 1 dv = −0.02 v, √ dv = −0.02dt, 2 v = −0.02t + C; v = 9 when t = 0 so 6 = C, dt v √ dx dx 2 v = −0.02t + 6, v = (3 − 0.01t)2 cm/s. But v = so = (3 − 0.01t)2 , dt dt 100 100 x=− (3 − 0.01t)3 + C1 ; x = 0 when t = 0 so C1 = 900, x = 900 − (3 − 0.01t)3 cm. 3 3

57. Differentiate to get

2 dy = − sin x + e−x , y(0) = 1. dx

1 dP [H(x) + C] where µ = eP (x) , = p(x), µ dx constant. Then dy µ 1 + p(x)y = H (x) − 2 [H(x) + C] + p(x)y = q − dx µ µ

58. (a) Let y =

d H(x) = µq, and C is an arbitrary dx p [H(x) + C] + p(x)y = q µ

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402

Chapter 9

1 [H(x)+C] is a solution µ of the initial value problem with y(x0 ) = y0 . This shows that the initial value problem has a solution.

(b) Given the initial value problem, let C = µ(x0 )y0 −H(x0 ). Then y =

To show uniqueness, suppose u(x) also satisfies (5) together with u(x0 ) = y0 . Following the 1 arguments in the text we arrive at u(x) = [H(x) + C] for some constant C. The initial µ condition requires C = µ(x0 )y0 − H(x0 ), and thus u(x) is identical with y(x). 59. Suppose that H(y) = G(x) + C. Then

dH dy dH dG = G (x). But = h(y) and = g(x), hence dy dx dy dx

y(x) is a solution of (10). 60. (a) y = x and y = −x are both solutions of the given initial value problem.

(b) y dy = − x dx, y 2 = −x2 + C; but y(0) = 0, so C = 0. Thus y 2 = −x2 , which is impossible.

EXERCISE SET 9.2 1. y 1 = xy/4 −2 ≤ x ≤ +2 −2 ≤ y ≤ +2

y

2. 4 3

y 2

2

1 x

x –2

1

2

2

3

4

–2

3.

y 2

y(0) = 2

y(0) = 1

x 5 –1

4.

y(0) = –1

dy + y = 1, µ = e dx = ex , dx d [yex ] = ex , dx yex = ex + C, y = 1 + Ce−x

(a) −1 = 1 + C, C = −2, y = 1 − 2e−x (b) 1 = 1 + C, C = 0, y = 1 (c) 2 = 1 + C, C = 1, y = 1 + e−x

y

5.

10 y(1) = 1

y(–1) = 0

x -2

2

y(0) = –1

–10

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Exercise Set 9.2

6.

403

dy d −2x − 2y = −x, µ = e−2 dx = e−2x , ye = −xe−2x , dx dx 1 1 ye−2x = (2x + 1)e−2x + C, y = (2x + 1) + Ce2x 4 4

(a) 1 = 3/4 + Ce2 , C = 1/(4e2 ), y = (b) −1 = 1/4 + C, C = −5/4, y =

1 1 (2x + 1) + e2x−2 4 4

5 1 (2x + 1) − e2x 4 4

(c) 0 = −1/4 + Ce−2 , C = e2 /4, y =

1 1 (2x + 1) + e2x+2 4 4

7.

lim y = 1

8.

x→+∞

lim y =

x→+∞

+∞

if y0 ≥ 1/4

−∞,

if y0 < 1/4

9. (a) IV, since the slope is positive for x > 0 and negative for x < 0. (b) VI, since the slope is positive for y > 0 and negative for y < 0. (c) V, since the slope is always positive. (d) II, since the slope changes sign when crossing the lines y = ±1. (e) I, since the slope can be positive or negative in each quadrant but is not periodic. (f )

III, since the slope is periodic in both x and y.

11. (a) y0 = 1, yn+1 = yn + (xn + yn )(0.2) = (xn + 6yn )/5

n xn yn

d −x ye = xe−x , dx ye−x = −(x + 1)e−x + C, 1 = −1 + C, C = 2, y = −(x + 1) + 2ex

(b) y − y = x, µ = e−x ,

0 0 1

xn y(xn) abs. error perc. error

1 0.2 1.20

0 1 0 0

2

3

4

5

0.4 1.48

0.6 1.86

0.8 2.35

1.0 2.98

0.2 1.24 0.04

0.4 1.58 0.10

0.6 2.04 0.19

0.8 2.65 0.30

1.0 3.44 0.46

3

6

9

11

13

y

(c) 3

x 0.2

0.4

0.6

0.8

1

12. h = 0.1, yn+1 = (xn + 11yn )/10 n xn yn

0 0 1.00

1 0.1 1.10

2

3

4

5

6

7

8

0.2 1.22

0.3 1.36

0.4 1.53

0.5 1.72

0.6 1.94

0.7 2.20

0.8 2.49

9 0.9

10 1.0

2.82

3.19

In Exercise 11, y(1) ≈ 2.98; in Exercise 12, y(1) ≈ 3.19; the true solution is y(1) ≈ 3.44; so the absolute errors are approximately 0.46 and 0.25 respectively.

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404

Chapter 9 1/3

13. y0 = 1, yn+1 = yn + 12 yn n xn yn

0 0 1

1 0.5 1.50

2 1 2.11

3 1.5 2.84

4 2

5 2.5 4.64

3.68

6 3 5.72

7 3.5 6.91

y

8 4

9

8.23

x 1

2

3

4

1.5

2

14. y0 = 1, yn+1 = yn + (xn − yn2 )/4 n xn yn

0 0 1

1 0.25 0.75

2 0.50 0.67

3 0.75 0.68

4 1.00 0.75

5 1.25 0.86

6 1.50 0.99

7 1.75 1.12

y

8 2

2.00 1.24

1.5 1 0.5 x 0.5

15. y0 = 1, yn+1 = yn + n tn yn

0 0 1

1 0.5 1.27

2 1 1.42

1

1 cos yn 2 3 1.5 1.49

y

4 2

3

1.53

t 3

16. y0 = 0, yn+1 = yn + e−yn /10 n tn yn

0 0 0

1 0.1 0.10

2

3

4

5

6

7

8

9

10

0.2 0.19

0.3 0.27

0.4 0.35

0.5 0.42

0.6 0.49

0.7 0.55

0.8 0.60

0.9 0.66

1.0 0.71

y 1

t 1

17. h = 1/5, y0 = 1, yn+1 = yn + n tn yn

0 0 1

1 0.2 1

2 0.4 1.12

1 sin(πn/5) 5

3

4

5

0.6 1.31

0.8 1.50

1.0 1.62

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Exercise Set 9.2

18. (a) By inspection,

405 2 dy = e−x and y(0) = 0. dx

(b) yn+1 = yn + e−xn /20 = yn + e−(n/20) /20 and y20 = 0.7625. From a CAS, y(1) = 0.7468. 2

2

19. (b) y dy = −x dx, y 2 /2 = −x2 /2 + C1 , x2 + y 2 = C; if y(0) = 1 then C = 1 so y(1/2) =

√

3/2.

√ 20. (a) y0 = 1, yn+1 = yn + ( yn /2)∆x √ ∆x = 0.2 : yn+1 = yn + yn /10; y5 ≈ 1.5489 √ ∆x = 0.1 : yn+1 = yn + yn /20; y10 ≈ 1.5556 √ ∆x = 0.05 : yn+1 = yn + yn /40; y20 ≈ 1.5590 (c)

dy 1 √ √ = dx, 2 y = x/2 + C, 2 = C, y 2 √ y = x/4 + 1, y = (x/4 + 1)2 , y(1) = 25/16 = 1.5625

21. (a) The slope field does not vary with t, hence along a given parallel line all values are equal since they only depend on the height y. (b) As in part (a), the slope field does not vary with t; it is independent of t. d 1 dy d dy (c) From G(y) − x = C we obtain (G(y) − x) = −1= C = 0, i.e. = f (y). dx f (y) dx dx dx dy √ 22. (a) Separate variables: √ = dx, 2 y = x + C, y = (x/2 + C1 )2 is a parabola that opens up, y and is therefore concave up. √ (b) A curve is concave up if its derivative is increasing, and y = y is increasing. dy y 3 − 2xy dy dy − 2xy − x2 = 0, = 2 . dx dx dx x − 3xy 2 (b) If y(x) is an integral curve of the slope field in part (a), then d {x[y(x)]3 − x2 y(x)} = [y(x)]3 + 3xy(x)2 y (x) − 2xy(x) − x2 y (x) = 0, so the integral curve dx must be of the form x[y(x)]3 − x2 y(x) = C.

23. (a) By implicit differentiation, y 3 + 3xy 2

(c) x[y(x)]3 − x2 y(x) = 2 24. (a) By implicit differentiation, ey + xey

dy dy + ex + yex = 0, dx dx

dy ey + yex =− y dx xe + ex

(b) If y(x) is an integral curve of the slope field in part (a), then d {xey(x) + y(x)ex } = ey(x) + xy (x)ey(x) + y (x)ex + y(x)ex = 0 from part (a). Thus dx xey(x) + y(x)ex = C. (c) Any integral curve y(x) of the slope field above satisfies xey(x) + y(x)ex = C; if it passes through (1, 1) then e + e = C, so xey(x) + y(x)ex = 2e defines the curve implicitly. 25. Euler’s Method is repeated application of local linear approximation, each step dependent on the previous step. 26. (a) For any n, yn is the value of the discrete approximation at the right endpoint, that, is an approximation of y(1). By increasing the number of subdivisions of the interval [0, 1] one might expect more accuracy, and hence in the limit y(1).

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406

Chapter 9

1 n+1 (b) For a ďŹ xed value of n we have, for k = 1, 2, . . . , n, yk = yk−1 + yk−1 = yk−1 In particn n 2 n n n+1 n+1 n+1 n+1 yn−1 = yn−2 = . . . = y0 = . Consequently, ular yn = n n n n

n

n+1 lim yn = lim = e, which is the (correct) value y = ex

. n→+∞ n→+∞ n x=1

EXERCISE SET 9.3 1. (a)

dy = ky 2 , y(0) = y0 , k > 0 dt

(b)

dy = −ky 2 , y(0) = y0 , k > 0 dt

3. (a)

1 ds = s dt 2

(b)

d2 s ds =2 dt2 dt

4. (a)

dv = −2v 2 dt

(b)

d2 s = −2 dt2

dy = 0.02y, y0 = 10,000 dt 1 (c) T = ln 2 ≈ 34.657 h 0.02

5. (a)

1 1 ln 2 = ln 2 T 20 dy (a) = ((ln 2)/20)y, y(0) = 1 dt

ds dt

2

(b) y = 10,000e2t/100 (d) 45,000 = 10,000e2t/100 , 45,000 ≈ 75.20 h t = 50 ln 10,000

6. k =

(c) y(120) = 26 = 64

(b) y(t) = et(ln 2)/20 = 2t/20 (d) 1,000,000 = 2t/20 , t = 20

7. (a)

ln 106 ≈ 398.63 min ln 2

dy ln 2 1 = −ky, y(0) = 5.0 Ă— 107 ; 3.83 = T = ln 2, so k = ≈ 0.1810 dt k 3.83

(b) y = 5.0 Ă— 107 e−0.181t (c) y(30) = 5.0 Ă— 107 e−0.1810(30) ≈ 219,000 (d) y(t) = (0.1)y0 = y0 e−kt , −kt = ln 0.1, t = − 8. (a) k =

ln 0.1 = 12.72 days 0.1810

1 dy 1 ln 2 = ln 2 ≈ 0.0050, so = −0.0050y, y0 = 10. T 140 dt

(b) y = 10e−0.0050t (c) 10 weeks = 70 days so y = 10e−0.35 ≈ 7 mg. (d) 0.3y0 = y0 e−kt , t = −

ln 0.3 ≈ 240.8 days 0.0050

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Exercise Set 9.3

407

9. 100e0.02t = 10,000, e0.02t = 100, t =

1 ln 100 ≈ 230 days 0.02

1 10. y = 10,000ekt , but y = 12,000 when t = 5 so 10,000e5k = 12,000, k = ln 1.2. y = 20,000 when 5 ln 2 ln 2 2 = ekt , t = =5 ≈ 19, in the year 2017. k ln 1.2 3.5 1 1 11. y(t) = y0 e−kt = 10.0e−kt , 3.5 = 10.0e−k(5) , k = − ln ≈ 0.2100, T = ln 2 ≈ 3.30 days 5 10.0 k 1 12. y = y0 e−kt , 0.7y0 = y0 e−5k , k = − ln 0.7 ≈ 0.07 5 ln 2 (a) T = ≈ 9.90 yr k y ≈ e−0.07t , so e−0.07t × 100 percent will remain. (b) y(t) ≈ y0 e−0.07t , y0 ln 2 ≈ 0.1155; y ≈ 3e0.1155t 6

13. (a) k =

(b) y(t) = 4e0.02t 1 ln 200 ≈ 0.5887, 9 ≈ 0.5550, y ≈ 0.5550e0.5887t .

(c) y = y0 ekt , 1 = y0 ek , 200 = y0 e10k . Divide: 200 = e9k , k = y ≈ y0 e0.5887t ; also y(1) = 1, so y0 = e−0.5887

ln 2 ≈ 0.1155, 2 = y(1) ≈ y0 e0.1155 , y0 ≈ 2e−0.1155 ≈ 1.7818, y ≈ 1.7818e0.1155t T

(d) k =

ln 2 ≈ 0.1386, y ≈ 10e−0.1386t T

14. (a) k =

(b) y = 10e−0.015t

(c) 100 = y0 e−k , 1 = y0 e−10k . Divide: e9k = 100, k = y0 = e10k ≈ e5.117 ≈ 166.83, y = 166.83e−0.5117t .

1 ln 100 ≈ 0.5117; 9

ln 2 ≈ 0.1386, 10 = y(1) ≈ y0 e−0.1386 , y0 ≈ 10e0.1386 ≈ 11.4866, y ≈ 11.4866e−0.1386t T

(d) k =

16. (a) None; the half-life is independent of the initial amount. (b) kT = ln 2, so T is inversely proportional to k. ln 2 ; and ln 2 ≈ 0.6931. If k is measured in percent, k = 100k, k 69.31 ln 2 70 ≈ then T = ≈ . k k k

17. (a) T =

(b) 70 yr

(c) 20 yr

(d) 7%

18. Let y = y0 ekt with y = y1 when t = t1 and y = 3y1 when t = t1 + T ; then y0 ekt1 = y1 (i) and 1 y0 ek(t1 +T ) = 3y1 (ii). Divide (ii) by (i) to get ekT = 3, T = ln 3. k 19. From (11), y(t) = y0 e−0.000121t . If 0.27 = if 0.30 =

y(t) ln 0.27 = e−0.000121t then t = − ≈ 10,820 yr, and y0 0.000121

ln 0.30 y(t) then t = − ≈ 9950, or roughly between 9000 B.C. and 8000 B.C. y0 0.000121

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408

Chapter 9

20. (a)

(b) t = 1988 yields

1

y/y0 = e−0.000121(1988) ≈ 79%.

0

50000 0

ln 2 ≈ 0.00012182; T2 = 5730 + 40 = 5770, k2 ≈ 0.00012013. T1 1 y 1 = 684.5, 595.7; t2 = − ln(y/y0 ) = 694.1, 604.1; in With y/y0 = 0.92, 0.93, t1 = − ln k1 y0 k2 1988 the shroud was at most 695 years old, which places its creation in or after the year 1293.

21. (a) Let T1 = 5730 − 40 = 5690, k1 =

(b) Suppose T is the true half-life of carbon-14 and T1 = T (1 + r/100) is the false half-life. Then ln 2 ln 2 we have the formulae y(t) = y0 e−kt , y1 (t) = y0 e−k1 t . At a certain with k = , k1 = T T1 point in time a reading of the carbon-14 is taken resulting in a certain value y, which in the case of the true formula is given by y = y(t) for some t, and in the case of the false formula is given by y = y1 (t1 ) for some t1 . 1 y If the true formula is used then the time t since the beginning is given by t = − ln . If k y0 1 y the false formula is used we get a false value t1 = − ln ; note that in both cases the k1 y0 value y/y0 is the same. Thus t1 /t = k/k1 = T1 /T = 1 + r/100, so the percentage error in the time to be measured is the same as the percentage error in the half-life. 22. (a)

dp = −kp, p(0) = p0 dh

(b) p0 = 1, so p = e−kh , but p = 0.83 when h = 5000 thus e−5000k = 0.83, ln 0.83 ≈ 0.0000373, p ≈ e−0.0000373h atm. k=− 5000 23. (a) y = y0 bt = y0 et ln b = y0 ekt with k = ln b > 0 since b > 1. (b) y = y0 bt = y0 et ln b = y0 e−kt with k = − ln b > 0 since 0 < b < 1. (c) y = 4(2t ) = 4et ln 2

(d) y = 4(0.5t ) = 4et ln 0.5 = 4e−t ln 2

24. If y = y0 ekt and y = y1 = y0 ekt1 then y1 /y0 = ekt1 , k = y = y1 = y0 e−kt1 then y1 /y0 = e−kt1 , k = −

ln(y1 /y0 ) ; if y = y0 e−kt and t1

ln(y1 /y0 ) . t1

25. (a) If y = y0 ekt , then y1 = y0 ekt1 , y2 = y0 ekt2 , divide: y2 /y1 = ek(t2 −t1 ) , k = (t2 − t1 ) ln 2 ln 2 = . If y = y0 e−kt , then y1 = y0 e−kt1 , y2 = y0 e−kt2 , k ln(y2 /y1 ) (t2 − t1 ) ln 2 1 ln 2 =− ln(y2 /y1 ), T = y2 /y1 = e−k(t2 −t1 ) , k = − . t2 − t1 k ln(y2 /y1 )

(t2 − t1 ) ln 2

. In either case, T is positive, so T =

ln(y2 /y1 )

1 ln(y2 /y1 ), t2 − t1

T =

(b) In Part (a) assume t2 = t1 + 1 and y2 = 1.25y1 . Then T =

ln 2 ≈ 3.1 h. ln 1.25

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Exercise Set 9.3

409

26. (a) In t years the interest will be compounded nt times at an interest rate of r/n each time. The value at the end of 1 interval is P + (r/n)P = P (1 + r/n), at the end of 2 intervals it is P (1 + r/n) + (r/n)P (1 + r/n) = P (1 + r/n)2 , and continuing in this fashion the value at the end of nt intervals is P (1 + r/n)nt . (b) Let x = r/n, then n = r/x and lim P (1 + r/n)nt = lim P (1 + x)rt/x = lim P [(1 + x)1/x ]rt = P ert . + + n→+∞

x→0

x→0

rt

(c) The rate of increase is dA/dt = rP e

= rA.

27. (a) A = 1000e(0.08)(5) = 1000e0.4 ≈ $1, 491.82 (b) P e(0.08)(10) = 10, 000, P e0.8 = 10, 000, P = 10, 000e−0.8 ≈ $4, 493.29 (c) From (11), with k = r = 0.08, T = (ln 2)/0.08 ≈ 8.7 years. 28. Let r be the annual interest rate when compounded continuously and r1 the eective annual interest rate. Then an amount P invested at the beginning of the year is worth P er = P (1 + r1 ) at the end of the year, and r1 = er − 1. 29. (a)

dT dT = −k(T − 21), T (0) = 95, = −k dt, ln(T − 21) = −kt + C1 , dt T − 21 T = 21 + eC1 e−kt = 21 + Ce−kt , 95 = T (0) = 21 + C, C = 74, T = 21 + 74e−kt

64 32 (b) 85 = T (1) = 21 + 74e−k , k = − ln = − ln , T = 21 + 74et ln(32/37) = 21 + 74 74 37 t 32 ln(30/74) 30 = ≈ 6.22 min , t= T = 51 when 74 37 ln(32/37) 30.

32 37

t ,

dT = k(70 − T ), T (0) = 40; − ln(70 − T ) = kt + C, 70 − T = e−kt e−C , T = 40 when t = 0, so dt 5 70 − 52 = ln ≈ 0.5, 30 = e−C , T = 70 − 30e−kt ; 52 = T (1) = 70 − 30e−k , k = − ln 30 3 T ≈ 70 − 30e−0.5t

31. Let T denote the body temperature of McHam’s body at time t, the number of hours elapsed after dT dT = −k(T − 72), 10:06 P.M.; then = −kdt, ln(T − 72) = −kt + C, T = 72 + eC e−kt , T − 72 dt 3.6 ≈ 0.4940, 77.9 = 72 + eC , eC = 5.9, T = 72 + 5.9e−kt , 75.6 = 72 + 5.9e−k , k = − ln 5.9 ln(26.6/5.9) T = 72+5.9e−0.4940t . McHam’s body temperature was last 98.6â—Ś when t = − ≈ −3.05, 0.4940 so around 3 hours and 3 minutes before 10:06; the death took place at approximately 7:03 P.M., while Moore was on stage. dT dT = k(Ta − T ) where k > 0. If T0 > Ta then = −k(T − Ta ) where k > 0; dt dt both cases yield T (t) = Ta + (T0 − Ta )e−kt with k > 0.

32. If T0 < Ta then

dy y y as both 33. (a) Both y(t) = 0 and y(t) = L are solutions of the logistic equation = k 1− dt L sides of the equation are then zero. dy (b) If y is very small relative to L then y/L ≈ 0, and the logistic equation becomes ≈ ky, dt which is a form of the equation for exponential growth.

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Chapter 9

(c) All the terms on the right-hand-side of the logistic equation are positive, except perhaps y 1 − , which is positive if y < L and negative if y > L. L (d) The rate of change of y is a function of only one variable, y itself. The right-hand-side of the dierential equation is a quadratic equation in y, which can be thought of as a parabola in y which opens down and crosses the y-axis at y = 0 and y = L. The parabola thus takes its maximum midway between the two y-intercepts, namely at y = L/2. dy y 1 1 1 k 34. (a) Given = k 1− y, separation of variables yields + dy = dt so that dt L L y L−y L ln y − ln(L − y) = kt + C. The initial condition yields y0 y y0 ln y0 − ln(L − y0 ) = C, C = ln and thus ln , = kt + ln L − y0 L−y L − y0 y y0 y0 L = ekt + , with solution y(t) = . L−y L − y0 y0 + (L − y0 )e−kt (b) From part (a), lim y(t) = t→+∞

y0 L =L y0 + (L − y0 ) lim e−kt t→+∞

35. (a) k = L = 1, y0 = 2

(b) k = L = y0 = 1 2

2

0

2

0

2 0

0

(c) k = y0 = 1, L = 2

(d) k = y0 = 1, L = 4 4

2

0

5

0

12 0

0

36. y0 = 400, L ≈ 1000, and y(300) ≈ 700; solve to obtain k ≈ 0.042 37. y0 ≈ 2, L ≈ 8; since the curve y = 6e−2k = 2, k =

1 ln 3 ≈ 0.5493. 2

2¡8 16 passes through the point (2, 4), 4 = , −kt 2 + 6e 2 + 6e−2k

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Exercise Set 9.4

411

400,000 passes through the point (200, 600), 400 + 600e−kt 800 1 = , k= ln 2.25 ≈ 0.00405. 3 200

38. y0 ≈ 400, L ≈ 1000; since the curve y = 600 =

400,000 , 600e−200k 400 + 600e−200k

39. (a) y0 = 5

(b) L = 12

(d) L/2 = 6 = (e)

60 , 5 + 7e−t = 10, t = − ln(5/7) ≈ 0.3365 5 + 7e−t

1 dy = y(12 − y), y(0) = 5 dt 12

40. (a) y0 = 1 (d) 750 = (e)

(c) k = 1

(b) L = 1000

(c) k = 0.9

1 1000 ln(3 · 999) ≈ 8.8949 , 3(1 + 999e−0.9t ) = 4, t = −0.9t 0.9 1 + 999e

dy 0.9 = y(1000 − y), y(0) = 1 dt 1000

41. Assume y(t) students have had the flu t days after semester break. Then y(0) = 20, y(5) = 35. dy = ky(L − y) = ky(1000 − y), y0 = 20 dt 20000 1000 (b) Part (a) has solution y = = ; −kt 20 + 980e 1 + 49e−kt 1000 1000 35 = , k = 0.115, y ≈ . −5k 1 + 49e 1 + 49e−0.115t

(a)

(c)

t 0 1 2 3 y(t) 20 22 25 28

4 5 6 7 8 9 10 11 12 13 14 31 35 39 44 49 54 61 67 75 83 93

y

(d) 100 75 50 25

t 3

6

9

12

EXERCISE SET 9.4 1. (a) y = e−2x , y = −2e−2x , y = 4e−2x ; y + y − 2y = 0 y = ex , y = ex , y = ex ; y + y − 2y = 0. (b) y = c1 e−2x + c2 ex , y = −2c1 e−2x + c2 ex , y = 4c1 e−2x + c2 ex ; y + y − 2y = 0 2. (a) y = e−2x , y = −2e−2x , y = 4e−2x ; y + 4y + 4y = 0 y = xe−2x , y = (1 − 2x)e−2x , y = (4x − 4)e−2x ; y + 4y + 4y = 0. (b) y = c1 e−2x + c2 xe−2x , y = −2c1 e−2x + c2 (1 − 2x)e−2x , y = 4c1 e−2x + c2 (4x − 4)e−2x ; y + 4y + 4y = 0.

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Chapter 9

3. m2 + 3m − 4 = 0, (m − 1)(m + 4) = 0; m = 1, −4 so y = c1 ex + c2 e−4x . 4. m2 + 5m + 6 = 0, (m + 2)(m + 3) = 0; m = −2, −3 so y = c1 e−2x + c2 e−3x . 5. m2 − 2m + 1 = 0, (m − 1)2 = 0; m = 1, so y = c1 ex + c2 xex . 6. m2 − 6m + 9 = 0, (m − 3)2 = 0; m = 3 so y = c1 e3x + c2 xe3x . 7. m2 + 1 = 0, m = ±i so y = c1 cos x + c2 sin x. √ √ √ 8. m2 + 5 = 0, m = ± 5 i so y = c1 cos 5 x + c2 sin 5 x. 9. m2 − m = 0, m(m − 1) = 0; m = 0, 1 so y = c1 + c2 ex . 10. m2 + 3m = 0, m(m + 3) = 0; m = 0, −3 so y = c1 + c2 e−3x . 11. m2 − 4m + 4 = 0, (m − 2)2 = 0; m = 2 so y = c1 e2t + c2 te2t . 12. m2 − 10m + 25 = 0, (m − 5)2 = 0; m = 5 so y = c1 e5t + c2 te5t . 13. m2 + 4m + 13 = 0, m = −2 ± 3i so y = e−2x (c1 cos 3x + c2 sin 3x). 14. m2 − 6m + 25 = 0, m = 3 ± 4i so y = e3x (c1 cos 4x + c2 sin 4x). 15. 8m2 − 2m − 1 = 0, (4m + 1)(2m − 1) = 0; m = −1/4, 1/2 so y = c1 e−x/4 + c2 ex/2 . 16. 9m2 − 6m + 1 = 0, (3m − 1)2 = 0; m = 1/3 so y = c1 ex/3 + c2 xex/3 . 17. m2 + 2m − 3 = 0, (m + 3)(m − 1) = 0; m = −3, 1 so y = c1 e−3x + c2 ex and y = −3c1 e−3x + c2 ex . Solve the system c1 + c2 = 1, −3c1 + c2 = 9 to get c1 = −2, c2 = 3 so y = −2e−3x + 3ex . 18. m2 − 6m − 7 = 0, (m + 1)(m − 7) = 0; m = −1, 7 so y = c1 e−x + c2 e7x , y = −c1 e−x + 7c2 e7x . Solve the system c1 + c2 = 5, −c1 + 7c2 = 3 to get c1 = 4, c2 = 1 so y = 4e−x + e7x . 19. m2 + 6m + 9 = 0, (m + 3)2 = 0; m = −3 so y = (c1 + c2 x)e−3x and y = (−3c1 + c2 − 3c2 x)e−3x . Solve the system c1 = 2, −3c1 + c2 = −5 to get c1 = 2, c2 = 1 so y = (2 + x)e−3x . √ √ √ 20. m2 + 4m + 1 = 0, m = −2 ± 3 so y = c1 e(−2+ 3)x + c2 e(−2− 3)x , √ √ √ √ y = (−2 + 3)c1 e(−2+ 3)x + (−2 − 3)c2 e(−2− 3)x . Solve the system c1 + c2 = 5, √ √ √ √ (−2 + 3)c1 + (−2 − 3)c2 = 4 to get c1 = 52 + 73 3, c2 = 52 − 73 3 so √ √ √ √ y = 52 + 73 3 e(−2+ 3)x + 52 − 73 3 e(−2− 3)x .

21. m2 + 4m + 5 = 0, m = −2 ± i so y = e−2x (c1 cos x + c2 sin x), y = e−2x [(c2 − 2c1 ) cos x − (c1 + 2c2 ) sin x]. Solve the system c1 = −3, c2 − 2c1 = 0 to get c1 = −3, c2 = −6 so y = −e−2x (3 cos x + 6 sin x). 22. m2 − 6m + 13 = 0, m = 3 ± 2i so y = e3x (c1 cos 2x + c2 sin 2x), y = e3x [(3c1 + 2c2 ) cos 2x + (−2c1 + 3c2 ) sin 2x]. Solve the system c1 = −2, 3c1 + 2c2 = 0 to get c1 = −2, c2 = 3 so y = e3x (−2 cos 2x + 3 sin 2x). 23. (a) m = 5, −2 so (m − 5)(m + 2) = 0, m2 − 3m − 10 = 0; y − 3y − 10y = 0. (b) m = 4, 4 so (m − 4)2 = 0, m2 − 8m + 16 = 0; y − 8y + 16y = 0. (c) m = −1 ± 4i so (m + 1 − 4i)(m + 1 + 4i) = 0, m2 + 2m + 17 = 0; y + 2y + 17y = 0.

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Exercise Set 9.4

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24. c1 ex + c2 e−x is the general solution, but cosh x = 12 ex + 12 e−x and sinh x = 12 ex − 12 e−x so cosh x and sinh x are also solutions. √ 25. m2 + km + k = 0, m = −k Âą k 2 − 4k /2 (a) k 2 − 4k > 0, k(k − 4) > 0; k < 0 or k > 4 (b) k 2 − 4k = 0; k = 0, 4

(c) k 2 − 4k < 0, k(k − 4) < 0; 0 < k < 4

dy dz 1 dy dy = = and dx dz dx x dz d 1 dy 1 d2 y dz 1 dy 1 d2 y d2 y d dy 1 dy = = − 2 = 2 , = − 2 2 2 2 dx dx dx dx x dz x dz dx x dz x dz x dz

26. z = ln x;

substitute into the original equation to get

d2 y dy + (p − 1) + qy = 0. 2 dz dz

dy d2 y + 2y = 0, m2 + 2m + 2 = 0; m = −1 Âą i so +2 dz 2 dz 1 y = e−z (c1 cos z + c2 sin z) = [c1 cos(ln x) + c2 sin(ln x)]. x √ dy d2 y −2 (b) − 2y = 0, m2 − 2m − 2 = 0; m = 1 Âą 3 so 2 dz dz

27. (a)

y = c1 e(1+

√ 3)z

√ 3)z

+ c2 e(1−

= c1 x1+

√ 3

√ 3

+ c2 x1−

28. m2 + pm + q = 0, m = 12 (−p Âą p2 − 4q ). If 0 < q < p2 /4 then y = c1 em1 x + c2 em2 x where −px/2 m1 < 0 and m2 < 0, if q = p2 /4 then y = c1 e + c2 xe−px/2 , if q > p2 /4 then y = e−px/2 (c1 cos kx + c2 sin kx) where k = 12 4q − p2 . In all cases lim y(x) = 0. x→+∞

29. (a) Neither is a constant multiple of the other, since, e.g. if y1 = ky2 then em1 x = kem2 x , e(m1 −m2 )x = k. But the right hand side is constant, and the left hand side is constant only if m1 = m2 , which is false. (b) If y1 = ky2 then emx = kxemx , kx = 1 which is impossible. If y2 = y1 then xemx = kemx , x = k which is impossible. 30. y1 = eax cos bx, y1 = eax (a cos bx − b sin bx), y1 = eax [(a2 − b2 ) cos bx − 2ab sin bx] so

y1 + py1 + qy1 = eax [(a2 − b2 + ap + q) cos bx − (2ab + bp) sin bx]. But a = − 12 p and b = 12 4q − p2 so a2 − b2 + ap + q = 0 and 2ab + bp = 0 thus y1 + py1 + qy1 = 0. Similarly, y2 = eax sin bx is also a solution. Since y1 /y2 = cot bx and y2 /y1 = tan bx it is clear that the two solutions are linearly independent.

31. (a) The general solution is c1 eÂľx + c2 emx ; let c1 = 1/(Âľ − m), c2 = −1/(Âľ − m). (b)

eÂľx − emx = lim xeÂľx = xemx . ¾→m ¾→m Âľâˆ’m lim

32. (a) If Îť = 0, then y = 0, y = c1 + c2 x. Use y(0) = 0 and y(Ď€) = 0 to get c1 = c2 = 0. If Îť < 0, then let Îť = −a2 where a > 0 so y − a2 y = 0, y = c1 eax + c2 e−ax . Use y(0) = 0 and y(Ď€) = 0 to get c1 = c2 = 0. √ √ √ (b) If Îť > 0, then m2 + Îť = 0, m2 = âˆ’Îť = Îťi2 , m√= Âą Îťi, y √ = c1 cos Îť x + c2√sin Îť x. If y(0) = √ 0 and y(Ď€) = 0, then c1 = 0 and c1 cos Ď€ √ Îť + c2 sin √ Ď€ Îť = 0 so c2 sin Ď€ Îť = 0. But c2 sin Ď€ Îť = 0 for arbitrary values of c2 if sin Ď€ Îť = 0, Ď€ Îť = nĎ€, Îť = n2 for n = 1, 2, 3, . . ., otherwise c2 = 0.

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414

Chapter 9

33. k/M = 0.5/2 = 0.25 (b) T = 2π · 2 = 4π s, f = 1/T = 1/(4π) Hz (d) y = 0 at the equilibrium position, so t/2 = π/2, t = π s. (e) t/2 = 2π at the maximum position below the equlibrium position, so t = 4π s.

(a) From (20), y = 0.4 cos(t/2) y (c) 0.3

x o

4c

–0.3

34. 64 = w = −M g, M = 2, k/M = 0.25/2 = 1/8, √ (a) From (20), y = cos(t/(2 2)) √ √ M (b) T = 2π = 2π(2 2) = 4π 2 s, k √ f = 1/T = 1/(4π 2) Hz y (c)

√ k/M = 1/(2 2)

1

t 2p

6p

10p

–1

√ √ (d) y = 0 at the equilibrium position, so t/(2 2) = π/2, t = π 2 s √ √ (e) t/(2 2) = π, t = 2π 2 s 35. l = 0.05, k/M = g/l = 9.8/0.05 = 196 s−2 (a) From (20), y = −0.12 cos 14t. y

(c)

(b) T = 2π M/k = 2π/14 = π/7 s, f = 7/π Hz (d) 14t = π/2, t = π/28 s

0.15

(e)

14t = π, t = π/14 s

t π 7

2π 7

–0.15

36. l = 2, k/M = g/l = 32/2 = 16, (a) From (20), y = −2 cos 4t.

k/M = 4 (b) T = 2π M/k = 2π/4 = π/2 s; f = 1/T = 2/π Hz

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Exercise Set 9.4

(c)

415

y

(d) 4t = π/2, t = π/8 s

2

(e)

4t = π, t = π/4 s

t 3

6

–2

37. (a) From the graph it appears that the maximum velocity occurs at the equilibrium position, t = π/8. dy dv dv = 0. But v = = 8 sin 4t, = To show this mathematically, vmax occurs when dt dt dt dv 32 cos 4t, = 0 when 4t = π/2, t = π/8. dt (b) From the graph it appears that the minimum velocity occurs at the equilibrium position as the dv = 32 cos 4t = 0 when 4t = π/2, 3π/2, . . .. block is falling, i.e. t = 3π/8. Mathematically, dt When 4t = 3π/2 the block is falling, so t = 3π/8. 38. (a) T = 2π

4Ď€ 2 4Ď€ 2 w 4Ď€ 2 w + 4 M 4Ď€ 2 w , k = 2 M = 2 , so k = = , 25w = 9(w + 4), k T T g g 9 g 25

25w = 9w + 36, w =

(b) From Part (a), w =

9 4π 2 w 4π 2 1 π2 ,k = = = 4 g 9 32 4 32 9 4

39. By Hooke’s Law, F (t) = −kx(t), since the only force is the restoring force of the spring. Newton’s Second Law gives F (t) = M x (t), so M x (t) + kx(t) = 0, x(0) = x0 , x (0) = 0.

40. 0 = v(0) = y (0) = c2 41. y = y0 cos

k , so c2 = 0; y0 = y(0) = c1 , so y = y0 cos M

k t, T = 2Ď€ M

k t. M

M 2Ď€t , y = y0 cos k T

2π 2πt y0 sin has maximum magnitude 2π|y0 |/T and occurs when T T 2πt/T = nπ + π/2, y = y0 cos(nπ + π/2) = 0.

(a) v = y (t) = −

4π 2 2πt y0 cos has maximum magnitude 4π 2 |y0 |/T 2 and occurs when T2 T 2πt/T = jπ, y = y0 cos jπ = ¹y0 .

(b) a = y (t) = −

42.

d 1 1 k 2 2 k[y(t)] + M (y (t)) = ky(t)y (t)+M y (t)y (t) = M y (t)[ y(t)+y (t)] = 0, as required. dt 2 2 M

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416

Chapter 9

43. (a) m2 + 2.4m + 1.44 = 0, (m + 1.2)2 = 0, m = −1.2, y = C1 e−6t/5 + C2 te−6t/5 , 6 16 16 C1 = 1, 2 = y (0) = − C1 + C2 , C2 = , y = e−6t/5 + te−6t/5 5 5 5 y 2

1 x 1

2

3

4

5

(b) y (t) = 0 when t = t1 = 25/48 ≈ 0.520833, y(t1 ) = 1.427364 cm 16 −6t/5 (t + 5/16) = 0 only if t = −5/16, so y = 0 for t ≥ 0. e 5 √ 44. (a) m2 + 5m + 2 = (m + 5/2)2 − 17/4 = 0, m = −5/2 ± 17/2, (c) y =

y = C1 e(−5+

√ 17)t/2

√ 17)t/2

+ C2 e(−5−

, √

√ −5 + 17 −5 − 17 C1 + C2 C1 + C2 = 1/2, −4 = y (0) = 2 2 √ √ 17 − 11 17 17 + 11 17 , C2 = C1 = 68 68 √ √ √ 17 − 11 17 (−5+ 17)t/2 17 + 11 17 −(5+√17)t/2 e e y= + 68 68

y 0.5 0.4 0.3 0.2 0.1

x 1

2

3

4

5

–0.1 –0.2 –0.3 –0.4 –0.5

(b) y (t) = 0 when t = t1 = 0.759194, y(t1 ) = −0.270183 cm so the maximum distance below the equilibrium position is 0.270183 cm. (c) y(t) = 0 when t = t2 = 0.191132, y (t2 ) = −1.581022 cm/sec so the speed is |y (t2 )| = 1.581022 cm/s.

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Exercise Set 9.4

417

√ √ √ 45. (a) m2 + m + 5 = 0, m = −1/2 ± ( 19/2)i, y = e−t/2 C1 cos( 19t/2) + C2 sin( 19t/2) , √ √ 1 = y(0) = C1 , −3.5 = y (0) = −(1/2)C1 + ( 19/2)C2 , C2 = −6/ 19, √ √ √ y = e−t/2 cos( 19 t/2) − (6/ 19 )e−t/2 sin( 19 t/2) y 1 x 1

2

3

4

5

–1

(b) y (t) = 0 for the first time when t = t1 = 0.905533, y(t1 ) = −1.054466 cm so the maximum distance below the equilibrium position is 1.054466 cm. (c) y(t) = 0 for the first time when t = t2 = 0.288274, y (t2 ) = −3.210357 cm/s. (d) The acceleration is y (t) so from the differential equation y = −y − 5y. But y = 0 when the object passes through the equilibrium position, thus y = −y = 3.210357 cm/s2 . √ 11 1 46. (a) m2 + m + 3 = 0, m = − ± i, 2 2 √ √ y = e−t/2 (c1 cos( 11t/2) + c2 sin( 11t/2)) √ √ Solve c1 = −2, (−c1 /2) + c2 11/2 = v0 to obtain c1 = −2, c2 = 2(v0 − 1)/ 11 and √ √ √ y = e−t/2 (−2 cos(3 11t/2) + [(2v0 − 1)/ 11] sin(3 11t/2)) (b) v0 ≈ 2.436

y

(c) 2 1

x 2

4

6

8

–1 –2

47. (a) m2 + 3.5m + 3 = (m + 1.5)(m + 2), y = C1 e−3t/2 + C2 e−2t , 1 = y(0) = C1 + C2 , v0 = y (0) = −(3/2)C1 − 2C2 , C1 = 4 + 2v0 , C2 = −3 − 2v0 , y(t) = (4 + 2v0 )e−3t/2 − (3 + 2v0 )e−2t (b) v0 = 2, y(t) = 8e−3t/2 − 7e−2t , v0 = −1, y(t) = 2e−3t/2 − e−2t , v0 = −4, y(t) = −4e−3t/2 + 5e−2t y 2 1

v0 = 2 v0 = –1 1

–1

2 v0 = –4

3

x 4

5

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418

Chapter 9

48. (a) y0 = y(0) = c1 , v0 = y (0) = c2

k , c2 = M

M v0 , y = y0 cos k

(b) l = 0.5, k/M = g/l = 9.8/0.5 = 19.6, √ √ 1 sin( 19.6 t) y = − cos( 19.6 t) + 0.25 √ 19.6

k t + v0 M

M sin k

k t M

1.1

0

3.1

–1.1

√ sin( 19.6 t), so

√ 1 (c) y = − cos( 19.6 t) + 0.25 √ 19.6 2 0.25 ≈ 1.10016 m is the maximum displacement. |ymax | = (−1)2 + √ 19.6 49.

dy1 dy dy1 + p(x)y = c + p(x)(cy1 ) = c + p(x)y1 = c ¡ 0 = 0 dt dt dt

50. The case p(x) = 0 has solutions y = C1 y1 + C2 y2 = C1 x + C2 . So assume now that p(x) = 0. dy d2 y dy = 0. Let Y = The dierential equation becomes + p(x) so that the equation becomes dx dx2 dx dY + p(x)Y = 0, which is a ďŹ rst order separable equation in the unknown Y . We get dx

dY = −p(x) dx, ln |Y | = − p(x) dx, Y = Âąe− p(x)dx . Y Let P (x) be a speciďŹ c antiderivative of p(x); then any solution Y is given by Y = Âąe−P (x)+C1 for some C1 . Thus all solutions are given by Y (t) = C2 e−P (x) including C2 = 0. Consequently

dy −P (x) −P (x) = C2 e , y = C2 e dx + C3 . If we let y1 (x) = e−P (x) dx and y2 (x) = 1 then dx y1 and y2 are both solutions, and they are linearly independent (recall P (x) = 0) and hence y(x) = c1 y1 (x) + c2 y2 (x).

REVIEW EXERCISES, CHAPTER 9 3. (a) linear

(b) both

(c) separable

(d) neither

4. (a) yes

(b) yes

(c) no

(d) yes

5.

dy − 4xy = x dx

(a) IF: e

(−4x)dx

= e−2x , 2

2 2 d 2 [ye−2x ] = xe−2x , ye−2x = dx

2 2 1 xe−2x dx = − e−2x + C, 4

2 1 y = − + Ce2x 4 dy dy 1 1 (b) = 4xy + x, = x dx, ln(4y + 1) = x2 + C, ln(4y + 1) = 2x2 + C1 , dx 4y + 1 4 2 2

2

2

4y + 1 = C2 e2x , y = 14 (C2 e2x − 1) = C3 e2x − 14 , same as in part (a)

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Review Exercises, Chapter 9

6. Âľ = e

419

3dx

3x

ex dx = ex + C, y = e−2x + Ce−3x

3x

=e ,e y=

7.

dy 1 = x2 dx, tan−1 y = x3 + C, y = tan 2 1+y 3

8.

dy 1 , Âľ = e dx = ex , ex y = +y = x dx 1+e

9.

10.

1 +y y

1 3 x +C 3

ex dx = ln(1 + ex ) + C, y = e−x ln(1 + ex ) + Ce−x 1 + ex

dy = ex dx, ln |y| + y 2 /2 = ex + C; by inspection, y = 0 is also a solution

1 3 cos y dy = dx, dy = 3 cos x dx, ln | sin y| = 3 sin x + C1 , tan y sec x sin y sin y = Âąe3 sin x+C1 = ÂąeC1 e3 sin x = Ce3 sin x , C = 0, y = sin−1 Ce3 sin x , as is y = 0 by inspection

11. Âľ = e−

x dx

= e−x

2

2

y = −1 + Cex 12.

/2

/2

, e−x

2

/2

y=

xe−x

2

/2

dx = −e−x

2

2

, 3 = −1 + C, C = 4, y = −1 + 4ex

/2

+ C,

/2

dy = dx, tan−1 y = x + C, Ď€/4 = C; y = tan(x + Ď€/4) y2 + 1 d 1 1 1 (y cosh x) = cosh2 x = (1 + cosh 2x), y cosh x = x + sinh 2x + C. When x = 0, y = 2 so dx 2 2 4 1 1 2 = C, and y = x sechx + sinh 2x sechx + 2 sechx. 2 4

13. IF:

14.

2 d 2 dy + y = 4x, Âľ = e (2/x)dx = x2 , yx = 4x3 , yx2 = x4 + C, y = x2 + Cx−2 , dx x dx

2 = y(1) = 1 + C, C = 1, y = x2 + 1/x2 15.

16.

1 1 + y5 y

dy =

dx 1 1 , − y −4 + ln |y| = ln |x| + C; − = C, y −4 + 4 ln(x/y) = 1 x 4 4

Ď€ Ď€ 1 dy 2 + C = 2 tan + C = 2 + C, C = −3, = 4 sec 2x dx, − = 2 tan 2x + C, −1 = 2 tan 2 2 y y 8 4 1 y= 3 − 2 tan 2x

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420

Chapter 9

d −x ye = xe−x sin 3x, dx

3 2 3 1 −x −x −x e cos 3x + − x + e−x sin 3x + C; ye = xe sin 3x dx = − x − 10 50 10 25 53 3 3 1 2 53 3 , y = − x− cos 3x + − x + sin 3x + ex 1 = y(0) = − + C, C = 50 50 10 50 10 25 50

17. (a) Âľ = e−

dx

= e−x ,

y

(c) 4

x –10

–2 –2

18. (a) 2ydy = dx, y 2 = x + C; if y(0) = 1 then C = 1, y 2 = x + 1, y = √ C = 1, y 2 = x + 1, y = − x + 1. y 1

x 1

–1

(b)

x + 1; if y(0) = −1 then

y

1

–1

√

x –1

1

–1 y

dy 1 = −2x dx, − = −x2 + C, −1 = C, y = 1/(x2 + 1) 2 y y

1

x –1

1

19. Assume the tank contains y(t) oz of salt at time t. Then y0 = 0 and for 0 < t < 15, dy y = 5 ¡ 10 − 10 = (50 − y/100) oz/min, with solution y = 5000 + Ce−t/100 . But y(0) = 0 so dt 1000 C = −5000, y = 5000(1 − e−t/100 ) for 0 ≤ t ≤ 15, and y(15) = 5000(1 − e−0.15 ). For 15 < t < 30, dy y = 0− 5, y = C1 e−t/200 , C1 e−0.075 = y(15) = 5000(1−e−0.15 ), C1 = 5000(e0.075 −e−0.075 ), dt 1000 y = 5000(e0.075 − e−0.075 )e−t/100 , y(30) = 5000(e0.075 − e−0.075 )e−0.3 ≈ 556.13 oz.

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Review Exercises, Chapter 9

421

20. (a) Assume the air contains y(t) ft3 of carbon monoxide at time t. Then y0 = 0 and for y d t/12000 1 t/12000 dy = 0.04(0.1) − (0.1) = 1/250 − y/12000, ye e = , t > 0, dt 1200 dt 250 yet/12000 = 48et/12000 + C, y(0) = 0, C = −48; y = 48(1 − e−t/12000 ). Thus the percentage y of carbon monoxide is P = 100 = 4(1 − e−t/12000 ) percent. 1200 (b) 0.012 = 4(1 − e−t/12000 ), t = 36.05 min y

21. 4 3 2 1

x 1

22.

2

3

4

2 dy 1 1 2 = x dx, ln |y| = x + C, y = C1 ex /16 y 8 16

23. y0 = 1, yn+1 = yn + n xn yn

0 0 1

1 0.5 1.50

2 1 2.11

√

yn /2 3

1.5 2.84

4 2 3.68

5 2.5 4.64

6 3 5.72

7 3.5 6.91

8 4

y 9

8.23

x 1

24. y0 = 1, yn+1 = yn + n tn yn

0 0 1

1 0.5 1.42

2 1 1.92

1 sin yn 2 3 1.5 2.39

2

3

4

y 3

4 2 2.73

t 3

25. h = 1/5, y0 = 1, yn+1 = yn + n tn yn

0 0 1.00

1 0.2 1.06

2 0.4 0.90

1 cos(2πn/5) 5

3

4

5

0.6 0.74

0.8 0.80

1.0 1.00

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422

Chapter 9

26. (a) yn+1 = yn + 0.1(1 + 5tn − yn ), y0 = 5 n tn yn

0 1 5.00

1 1.1 5.10

2

3

4

5

6

7

8

9

1.2 5.24

1.3 5.42

1.4 5.62

1.5 5.86

1.6 6.13

1.7 6.41

1.8 6.72

1.9 7.05

10 2 7.39

(b) The true solution is y(t) = 5t − 4 + 4e1−t , so the percentage errors are given by tn yn y(tn) abs. error rel. error (%)

27. (a) k =

1 5.00 5.00 0.00 0.00

1.1 5.10 5.12 0.02 0.38

1.2 5.24 5.27 0.03 0.66

1.3 5.42 5.46 0.05 0.87

1.4 5.62 5.68 0.06 1.00

1.5 5.86 5.93 0.06 1.08

1.6 6.13 6.20 0.07 1.12

ln 2 ≈ 0.1386; y ≈ 2e0.1386t 5

1.7 6.41 6.49 0.07 1.13

1.8 6.72 6.80 0.08 1.11

1.9 7.05 7.13 0.08 1.07

2 7.39 7.47 0.08 1.03

(b) y(t) = 5e0.015t 1 ln 100 ≈ 0.5117, 9 ≈ 0.5995, y ≈ 0.5995e0.5117t .

(c) y = y0 ekt , 1 = y0 ek , 100 = y0 e10k . Divide: 100 = e9k , k = y ≈ y0 e0.5117t ; also y(1) = 1, so y0 = e−0.5117 (d) k = 28. (a)

ln 2 ≈ 0.1386, 1 = y(1) ≈ y0 e0.1386 , y0 ≈ e−0.1386 ≈ 0.8706, y ≈ 0.8706e0.1386t T

d y(t) = 0.01y, y(0) = 5000 dt

(b) y(t) = 5000e0.01t (c) 2 = e0.01t , t = 100 ln 2 ≈ 69.31 h (d) 30,000 = 5000e0.01t , t = 100 ln 6 ≈ 179.18 h 29. From section 9.3 formula (11), y(t) = y0 e−0.000121t , so 0.785y0 = y0 e−0.000121t , t = − ln 0.785/0.000121 ≈ 2000.6 yr 30. A = /w,

d/ dw dw dA = w+/ = (w + /) = dt dt dt dt

31. (a) y = C1 ex + C2 e2x √ (c)

y = e−x/2 C1 cos

√ 7 7 x + C2 sin x 2 2

1 dw 2 dt

Ă— perimeter, since

dw d/ = . dt dt

(b) y = C1 ex/2 + C2 xex/2

32. (a) (m + 3)(m − 1), m = 1, −3, y = c1 et + c2 e−3t ; 1 = c1 + c2 , 5 = c1 − 3c2 , c1 = 2, c2 = −1, y = 2et − e−3t (b) (m − 3)2 = 0, m = 3, 3, y = (c1 + c2 t)e3t , 2 = c1 , 1 = 3c1 + c2 , c1 = 2, c2 = −5, y = (2 − 5t)et (c) (m − 2)2 + 9 = 0, m = 2 Âą 3i, y = e2t (c1 cos 3t + c2 sin 3t), 1 = c1 , 5 = 2c1 + 3c2 , c1 = 1, c2 = 2, y = e2t (cos 3t + 2 sin 3t)

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Review Exercises, Chapter 9

423

33. k/M = 0.25/1 = 0.25 (a) From (20) in Section 9.4, y = 0.3 cos(t/2) (c)

y

(b) T = 2π · 2 = 4π s, f = 1/T = 1/(4π) Hz (d) y = 0 at the equilibrium position, so t/2 = π/2, t = π s.

0.3

(e)

x o

4c

t/2 = π at the maximum position below the equlibrium position, so t = 2π s.

–0.3

34. l = 0.5, k/M = g/l = 32/0.5 = 64,

k/M = 8

(a) From (20) in Section 9.4, y = −1.5 cos 8t. y

(c)

(d) 8t = π/2, t = π/16 s

1

(e) t 2p

–1

(b) T = 2π M/k = 2π/8 = π/4 s; f = 1/T = 4/π Hz

6p

10p

8t = π, t = π/8 s

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