EXAM
Exam 2
Math 3350, Summer 2011
June 20, 2011
• Write all of your answers on the blank paper
provided. Do not write on the exam questions
handout. When finished, write your name and the
section number on the first page of your answers.
Staple the exam questions and your sheet of notes to
the back of your answers. You may leave when
finished.
• You must show enough work to justify your answers.
Unless otherwise instructed,
give exact answers, not
√
approximations (e.g., 2, not 1.414).
• This exam has 6 problems. There are 320 points
total.
Good luck!
60 pts.
Problem 1. Consider the Euler-Cauchy equation
d2 y
1 dy
1
−
+ y = 0,
dx2
x dx x2
x > 0.
One solution is y1 = x.
A. Use the method of Reduction of Order to find a second solution. [No
credit for using another method!]
B. Show these two solutions are independent, using the Wronskian.
80 pts.
Problem 2. In each part, find the general solution of the equation. If initial
conditions are given, find the solution of the initial value problem.
A.
B.
C.
40 pts.
d2 y
dy
−
− 2y = 0,
2
dx
dx
y(0) = 5,
y 0 (0) = 1
d2 y
dy
+6
+ 9y = 0,
dx2
dx
d2 y
dy
+4
+ 13y = 0,
dx2
dx
Problem 3. In each part, find the general solution of the given Euler-Cauchy
equation. (As usual, we assume x > 0 in Euler-Cauchy equations.)
A.
x2
dy
d2 y
+ 2x
− 2y = 0
dx2
dx
x2
d2 y
dy
− 3x
+ 4y = 0
dx2
dx
B.
1
40 pts.
Problem 4. In each part, find the general solution.
A.
D2 (D − 2)(D − 3)3 y = 0
B.
(D + 1)(D2 − 4D + 5)3 y = 0
60 pts.
Problem 5. Use the method of undetermined coefficients (either version) to
find the general solution.
A.
B.
C.
40 pts.
d2 y
dy
+
− 2y = −x2 + 1.
2
dx
dx
dy
d2 y
+
− 2y = sin(x).
dx2
dx
d2 y
dy
− 2y = ex .
+
dx2
dx
Problem 6. Use Shifting Rule version of the method of undetermined coefficients to find the general solution of the given equation. No
credit for using any other method (including the book’s version of undetermined
coefficients).
(D2 − 3D + 2)y = 3x2 e2x
2