MATH 267 Section P
Fall 2015
PRACTICE FINAL EXAM
1. Find the general solution of
y 0 = 3y 4 − 8xy 4
Solution:
y(x) = √
3
12x2
1
− 9x + C
2. Solve the initial value problem
y 00 + 4y 0 + 13y = 0
y(0) = 2 y 0 (0) = 1
Solution:
y(x) = e−2x (2 cos 3x +
5
sin 3x)
3
3. Use the substitution z = ln y to find the general solution of the equation
y 0 + ex y = y ln y
Solution:
y(x) = e(C−x)e
x
4. Find the general solution of
y 00 + 2y 0 + y = t2 − cos 4t
Solution:
y(t) = C1 e−t + C2 te−t + t2 − 4t + 6 +
15
8
cos 4t −
sin 4t
289
289
5. Use the variation of parameters method to find a solution of
y 00 + 4y = sec 2t
Solution:
yp (t) =
1
1
(cos 2t) ln (cos 2t) + t sin 2t
4
2
MATH 267 Section P
Fall 2015
6. Let

1
A = 6
0
PRACTICE FINAL EXAM

0
1
−2
0
1
4
(a) Find the eigenvalues and eigenvectors of A.
(b) Find the general solution of x0 = Ax.
Solution: a) The eigenvalues are λ = 1, 2, −3 with corresponding eigenvectors
 
 
 
−2
0
0
9
1
1
12
1
−4
b)
 
 
 
−2
0
0
x(t) = C1 et  9  + C2 e2t 1 + C3 e−3t  1 
12
1
−4
7. Let
A=
1
5
−1
−3
(a) Find a fundamental matrix for x0 = Ax.
(b) Find the solution of x0 = Ax satisfying the initial condition
1
x(0) =
−1
Solution: a) One fundamental matrix is
cos t
Φ(t) = e−t
2 cos t + sin t
sin t
− cos t + 2 sin t
b)
x(t) = e
−t
cos t + 3 sin t
7 sin t − cos t
8. The system
x0
y
0
= y
= y−x
has (0, 0) as its only equilibrium point. Classify this equilibrium point by stability and type, and sketch
the phase portrait near (0, 0).
Solution: Unstable spiral point
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MATH 267 Section P
Fall 2015
9. Let
(
f (t) =
2
0
PRACTICE FINAL EXAM
0<t<3
t>3
(a) Find the Laplace transform of f .
(b) Use the Laplace transform method to solve
y 00 − 9y = f (t)
Solution:
L{f (t)} =
y(t) =
y(0) = 0
y 0 (0) = 1
2
(1 − e−3s )
s
e3t + e−3t − 2 e3(t−3) + e−3(t−3) − 2
1 3t
(e − e−3t ) +
+
U(t − 3)
6
9
9
10. We seek a solution of
y 00 + x2 y 0 − 3y = 0
in the power series form y(x) =
P∞
n=0
an xn .
(a) What is the recurrence relation for the coefficients?
(b) If y(0) = 2 and y 0 (0) = 0 find an for n = 0, 1, 2, 3.
Solution:
a2 =
3a0
2
an+2 =
−(n − 1)an−1 + 3an
(n + 1)(n + 2)
n = 1, 2, . . .
a0 = 2 a1 = 0 a2 = 3 a3 = 0
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