MATH198 Solution Sheet 3 1. Check whether the solutions listed below satisfy the differential equations, showing your working. A, B and a are constants. dy = −4y ; dt An easy one to start with. (i)
y = Ae−4t
y = Ae−4t dy ⇒ = − 4Ae−4t dt dy LHS = = − 4Ae−4t , RHS = − 4y = − 4Ae−4t ⇒ LHS = RHS X dt The left-hand side and right-hand side are equal for all t, so the solution given was correct. d2 y dy + x −y =3; dx2 dx
2x2
(ii)
B y = Ax + √ − 3 x
Check: 1
y = Ax + Bx− 2 − 3 3 y 0 = A − 21 Bx− 2 5 3 Bx− 2 4
y 00 =
dy d2 y 2 3 − 12 − 25 − 32 1 +x − y = 2x 4 Bx + x A − 2 Bx − Ax + Bx − 3 2x dx2 dx 1 = 32 B − 12 B − B x− 2 + Ax − Ax + 3 = 3 X 2
⇒
y y 00 + (y 0 − y) y 0 = 0 ;
(iii)
y=
√ Aex + B
1
y
= (Aex + B) 2
y0
=
1 2
1
(Aex + B)− 2 Aex 3
(Chain rule, 1
y 00 = − 41 (Aex + B)− 2 A2 e2x + 21 (Aex + B)− 2 Aex
dy dy du = , with u = Aex + B.) dx du dx (Product rule and chain rule)
(When you take the second derivative the product rule means that there are two terms some people missed out the second term.) h i 1 3 1 y y 00 + (y 0 − y) y 0 = (Aex + B) 2 − 14 (Aex + B)− 2 A2 e2x + 12 (Aex + B)− 2 Aex h i 1 1 1 + 12 (Aex + B)− 2 Aex − (Aex + B) 2 21 (Aex + B)− 2 Aex = − 41 (Aex + B)−1 A2 e2x + 12 Aex + 14 (Aex + B)−1 A2 e2x − 21 Aex = 0 X
d2 y + a2 y = 0; dx2
(iv)
y = A cos ax + B sin ax
y = A cos ax + B sin ax y 0 = −Aa sin ax + Ba cos ax y 00 = −Aa2 cos ax − Ba2 sin ax ⇒ y 00 + a2 y = −Aa2 cos ax − Ba2 sin ax + a2 (A cos ax + B sin ax) = 0 X 2. Solve the following first-order differential equations. dy = y cos x dx This is separable: Z (i)
⇒ ⇒
Z dy = cos x dx y ln y = sin x + C y = esin x+C = eC esin x = Aesin x
dy = Aesin x cos x dx RHS = y cos x = Aesin x cos x = LHS
where A = eC
Check : LHS =
X
dy = (t2 + 1) y 4 dt Also separable: Z Z −4 y dy = (t2 + 1) dt (ii)
⇒
− 13 y −3 = 13 t3 + t + C
⇒
y 3 = − (t3 + 3t + 3C)−1 − 1 y = − t3 + 3t + A 3
⇒
Let A = 3C.
1
1
This solution can be written in other Rways, eg (−t3 − 3t − A)− 3 or (B − t3 − 3t)− 3 . Some people had trouble with the Rintegral y −4 dy. For any power, positive or negative, except 1 y p+1 + const. Logarithms only help when p = −1. for p = −1, the rule to use is y p dy = p+1 dy = dt
− 4 (3t2 + 3) = (t2 + 1) t3 + 3t + A 3 − 4 RHS = (t2 + 1) y 4 = (t2 + 1) t3 + 3t + A 3 = LHS X
Check : LHS =
1 3
t3 + 3t + A
− 34
x(1 + y 3 ) y = y2 0
(iii) Separable:
Z y 2 dy = x dx (Substitution y 3 = u simplifies the integral) 1 + y3 1 ln(1 + y 3 ) = 21 x2 + C Let A = e3C 3
Z ⇒ ⇒ ⇒
0
Check :
LHS = y =
RHS =
(iv)
3 2
1 + y 3 = Ae 2 x 3 2 13 y = Ae 2 x − 1
1 3
Ae
3 2 x 2
3 2 x(1 + y 3 ) = x Ae 2 x y2
− 23 3 2 − 23 3 2 3 2 x x 2 2 −1 A3x e = Ae −1 Ax e 2 x − 32 3 2 − 23 3 2 3 2 x x 2 2 1 + Ae −1 − 1 = Ae −1 Axe 2 x = LHS
dy x + 2y = dx x
This is not separable, but it is linear.
⇒
⇒ ⇒ ⇒ ⇒
x + 2y dy = We must put it into standard form: dx x dy 2 − y=1 dx x Find the integrating factor: Z 2 µ = exp − dx = exp[−2 ln x] = x−2 x Multiply whole equation by µ dy x−2 − 2x−3 y = x−2 dx d −2 (x y) = x−2 dx Integrate both sides (don’t forget the consant) x−2 y = −x−1 + A y = Ax2 − x dy = 2Ax − 1 dx x + 2y x + 2(Ax2 − x) RHS = = = 2Ax − 1 = LHS x x
Check : LHS =
X
X
3. Check whether your answers to question 2 satisfy the given differential equations, showing your working. Which ones did you get right? If you checked properly, there should have been no surprises when you looked at this solution sheet - you should know already whether your answers were correct or not.
4. [Short problems] The following are examples of expressions found when solving separable equations. Solve them to find y. The split is about 50/50 between people who can do these and people who can’t. If you got any of these wrong, please look carefully at the solutions below. 1 = x + x2 + C y+3
(i)
The LHS and RHS of this equation have the same numerical value. If we perform exactly the same operations on both sides, then both sides will still have the same value, and we will have a new true equation. First operation: Invert both sides. y+3=
1 x + x2 + C
Inverting the RHS is unambiguous, the only correct answer is x+x12 +C . The answer is NOT 1 + C1 or x1 + x12 + C1 . x+x2 Second operation: Subtract 3 from both sides to get the final answer. y=
(ii)
1 − 3. x + x2 + C
ln(y +1) = 4 ln(t+3)−2 ln(t)+A
Again we must be careful to perform the exact same operations on both sides. To get rid of the ln on the LHS we need to exponentiate, we have to do the same thing to the RHS, so we must exponentiate the whole RHS. exp{ln(y + 1)} = exp{ 4 ln(t + 3) − 2 ln(t) + A } = exp{4 ln(t + 3)} × exp{−2 ln(t)} × exp{A} ⇒
y + 1 = (t + 3)4 t−2 eA ⇒
y = eA
(t + 3)4 (t + 3)4 − 1 = B −1 t2 t2
where B ≡ eA
This was the worst one: if algebra were a democracy the answer would be y + 1 = (t + 3)4 t−2 + eA or y + 1 = (t + 3)4 + t−2 + eA . (iii)
2
ey = x2 + C
First take the logarithm of both sides n 2o ln ey = ln x2 + C ⇒ y 2 = ln x2 + C Again: We apply the log to the whole LHS and the whole RHS. Expressions such as ln x2 +ln C are wrong. Finally, take the square root of both sides. p y = ¹ ln {x2 + C}. 1 1 = +B (iv) y t This is like 3(i). We invert the whole LHS, so we must invert the whole RHS as one object y=
1 t
1 t = 1 + Bt +B
(The popular answer t + B −1 is wrong.) (v)
log(y+1)−log(y) = 3 log(x+1)−3 log(x−1)+log C
Easiest is to use the standard rules about logs to group each side into a single term log(y + 1) − log(y) = 3 log(x + 1) − 3 log(x − 1) + log C (x + 1)3 y+1 = log C ⇒ log y (x − 1)3 ⇒
y+1 (x + 1)3 = C y (x − 1)3
⇒
1+ ⇒
1 (x + 1)3 = C y (x − 1)3 1 (x + 1)3 −1 = C y (x − 1)3
Again, when we invert the RHS we must invert the whole expression: 1 (x − 1)3 y= = (x + 1)3 C(x + 1)3 − (x − 1)3 C − 1 (x − 1)3
Checking A good way of checking algebraic problems like these is by “putting in numbers”. To check 2(iv) put any values for t and B into the answer, and see what y value it gives. For example let’s try t = 0.24 and B = 1.62. This gives y=
t 0.24 = = 0.172811 1 + Bt 1 + 1.62 × 0.24
Now see what happens when we put these same numbers into the starting equation 1 1 = = 5.78667 y 0.172811 1 1 +B = + 1.62 = 5.78667 X t 0.24 If we have made no mistakes the numbers must agree. If they don’t, something is wrong. (Try what happens with the wrong answer t + B −1 .)