Final Exam Study Guide Solution on Ordinary Differential Equations-M302

M302 Study Guide (Final) (S. Zhang) . By the DE y 0 = ±∞ only when
1. Solve 6 − 2y = 0 ⇒ y=3
0 2
2xy − 3y = 9x , y(1) = 3 When y = 3, by above solution
• ans: Warning, we must make the coefficient of y 0 1: 9 = −t2 + 9
3 9 t2 = 0
y0 − y= x
2x 2 t=0
The integrating factor is
R Since the answer is interval containing t0 = 1, (0, ∞).
µ=e −3/(2x)
= e−(3/2) ln x = x−3/2
3. Show the differential equation is exact, and solve it as an
The general solution exact equation.
9 y
Z Z
y = µ−1 µg = x3/2 x−3/2 x (x3 + )dx + (y 2 + ln x)dy = 0
2 x
= x3/2 (9x1/2 + c)
• ans: Exact equation: we write it in the form F 0 = 0:

c = −6 M dx + N dy = 0

where F = F (x, y), and x and y are independent variable.
y = 9x2 − 6x3/2 Then we must have
dsolve({2*x*diff(y(x),x)-3*y(x)=9*x^2, Fx = M, Fy = N, My = Nx = Fxy
y(1)=3},y(x));
Here we check
y 1 1
2. Determine an interval for which the initial value problem My = (x3 + )y = , Nx = (y 2 + ln x)x =
is certain to have a unique solution. x x x
How to find F ?
(a)
Note, the above part can be omitted if you can do the sec-
0 sin t t+2 ond part below correctly. Because the second part, found
y = y+ , y(−2) = 3.
t+1 (t + 6)(t − 4) a solution, implies the equation is exact!
(b)
y 1
Z Z
−2t F = M dx = x3 +
dx = x4 + y ln x + f (y)
y0 = , y(1) = 2. x 4
6 − 2y
• ans: Fy = ln x + f 0 (y) = y 2 + ln x, f 0 (y) = y 2

(a) So f = 13 y 3 . Solution of dF = 0 is
1 F =C
p=− t 6= −1
t+1
t+2
q= t 6= −6, 4
(t + 6)(t − 4)
1 4 1
x + y ln x + y 3 = C
The answer is: the interval containing t0 = −2: 4 3
(−6, −1).
(b) We solve the nonlinear equation first, then locate 4. Find THE set of fundamental solution at t0 and use it to
point t at which y 0 = ±∞. solve the IVP

(6 − 2y)dy = (−2t)dt y 00 − 5y 0 + 4y = 0, t0 = 1
Z Z
(6 − 2y)dy = (−2t)dt
y 00 − 5y 0 + 4y = 0, y(1) = 2, y 0 (1) = 3
6y − y 2 = −t2 + C
• ans:
y(1) = 2 ⇒ 12 − 4 = −1 + C; C = 9
6y − y 2 = −t2 + 9 y 00 − 5y + 4y = 0, y(1) = 1, y 0 (1) = 0


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, r2 − 5r + 4 = 0, r = 1, 4 The solution is
1
y = yc + yp = C1 ex + C2 e−x + ex x
y = Aet + Be4t 2

For yp by VP,
4 t 1
y1 = e − 4 e4t
3e 3e yp = u1 ex + u2 e−x
u01 ex + u02 e−x = 0
00 0
y − 5y + 4y = 0, y(1) = 0, y (1) = 1
u01 ex − u02 e−x = ex

1 t 1 Add the two equations, 2u01 = 1 and we get u01 = x2 .
y2 = − e + 4 e4t Subtracting, we get 2u02 = −e2x and u2 = − 41 e2x
3e 3e
1 x 1
5 1 yp = u1 ex + u2 e−x = e x − ex
y = 2y1 + 3y2 = et + 4 e4t 2 4
3e 3e
Finally, we get the solution again
5. Find the general solution of the homogeneous equation yc
and write down the form for undetermined coefficients for 1 1
y = yc + yp = C1 ex + C2 e−x + ex x − ex
yp . (Do not solve for yp .) 2 4
0 x −x 1 x
= C1 e + C2 e + e x
(D2 − 4)2 (D − 2)2 (D2 − 2D + 5)2 y 2
=ex sin 3x − e2x x + cos 2x + ex x2 cos 2x.
7. Find the general solution (resolve all constants):
• ans: First, we find yH = yc . From the characteristic
y 000 + 4y 0 = 12, y(0) = 1, y 0 (0) = 0, y 00 (0) = 0.
equation (replace D by r):

(D2 − 4)2 (D − 2)2 (D2 − 2D + 5)2 = 0 • ans: For yH = yc (coplementary solution, or solu-
tion for the homogeneous equation), we find the roots of
we get roots: chracteristic eq
r = 2, 2, 2, 2, −2, −2, 1 ± 2i, 1 ± 2i r3 + 4r = 0, r = 0, ±2i

yH = yc =e2x (A + Bx + Cx2 + Dx3 ) + e−2x (E + F x) yH = c1 + c2 cos 2x + c3 sin 2x
+ ex ((G + Hx) cos 2x + (I + Jx) sin 2x).

Then we can write down a yp form, (note 1 ± 2i is a double Since r = 0 is a root, we need an extra x in yp form,
root, and 2 is a 4-fold root)
yp = x(A + Bx)
x
yp =e (A cos 3x + B sin 3x)
Plug it into the equaion,
+ (C + Dx)x4 e2x + (E cos 2x + F sin 2x)
ex ((G + Hx + Ix2 ) cos 2x + (J + Kx + Lx2 ) sin 2x)x2 4A = 12
8B = 0
6. Find the general solution by both the undetermined coef-
ficients method and the variation of parameters method.
yp = 3x
y 00 − y = ex .
y = y H + yp =
• ans: For yc , r2 − 1 = 0,
= c1 + c2 cos 2x + c3 sin 2x + 3x
yc = C1 ex + C2 e−x .
3
By initial condition y = 1 − 2 cos 2x + 3x.
For yp by UC,
8. Find the first 4 nonzero terms in the series solutions (with-
yp = Aex x ⇒ Aex x + 2Aex − Aex x = ex out finding the recurrent relation)
1
⇒ y p = ex x y 00 + xy 0 + y = 0, y(0) = 1, y 0 (0) = 0
2

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