Math 252 Practice Questions for Test 2

1. Consider the differential equation
Math 252
Practice Questions for Test 2
x 2 y ′′ − 3x y ′ + k y = 0, (x > 0).
Find the general solution if (a) k = −5, (b) k = 4, (c) k = 13.
Answers:
(a) y = c 1 x 5 + c 2 x −1
(b) y = c 1 x 2 + c 2 x 2 ln x
(c) y = x 2 (c 1 cos(3 ln x) + c 2 sin(3 ln x))
2. Find the form of a particular solution y p in the following. (Leave your answers with undetermined
coefficients.)
(a) y ′′ + 4y = 5x + sin(2x)
(b) y ′′ − 2y ′ + y = 3x 2 + 5xe x
(c) y ′′ + 2y ′ + 5y = 3e −x + 4e −x sin(2x) − 2 cos(3x)
Answers:
(a) y p = Ax + B + Cx cos(2x) + Dx sin(2x)
(b) y p = Ax 2 + Bx + C + x 2 (Dx + E)e x
(c) y p = Ae −x + xe −x [B cos(2x) + C sin(2x)] + D cos(3x) + E sin(3x)
3. Find the solution of the initial-value problem
y ′′ − 4xy ′ + (4x 2 − 2)y = 0, y(0) = 1, y ′ (0) = 2
given that y 1 = e x2
is a solution of the equation.
Answer: y = e x2 + 2xe x2
4. Use the method of undetermined coefficients to find the general solution of
y ′′ + y ′ − 6y = 4e 3x + 5 cos(2x).
Answer: y = c 1 e −3x + c 2 e 2x + 2 3 e3x − 25
52 cos(2x) + 5
52 sin(2x)
5. Use the method of undetermined coefficients to find the general solution of
y ′′ − 5y ′ + 6y = x + 2 + 3e 2x .
Answer: y = c 1 e 3x + c 2 e 2x + x 6 + 17
36 − 3xe2x
6. Solve
y ′′ + 4y = sec(2x).
Answer: y = c 1 cos(2x) + c 2 sin(2x) + 1 2 x sin(2x) + 1 4
cos(2x) ln|cos(2x)|
7. Use the method of variation of parameters to find the general solution of
x 2 y ′′ − 2xy ′ + 2y = x 4 e −x .
Simplify your answer.
Answer: y = c 1 x 2 + c 2 x + xe −x (x + 2)

Math 252 Practice Questions for Test 2 Page 2 of 2
8. Find the solution of the initial-value problem
Answer: y = 2x + x ln x + x(ln x) 2
x 2 y ′′ − xy ′ + y = 2x, y(1) = 2, y ′ (1) = 3.
9. A mass of 2 kg is attached to a spring whose spring constant is 3 N/m. The medium offers a damping
force of magnitude 5 times the instantaneous velocity.
(a) Determine the steady-state position of the mass if it is driven by an external force (in Newton) of
magnitude F (t) = 4 cos(2t) + 3 sin(2t).
(b) Express your answer in (a) in the form x(t) = A sin(ωt + φ).
Answers:
(a) x p (t) = − 2 5 cos(2t) + 1 5 sin(2t)
(b) x p (t) = √ 5
5
sin(2t − 1.107)
10. A 1 kg mass is attached to a spring with spring constant 2 N/m and a damping device with β = 3
N/(m/s).
(a) Find the displacement x(t) of the mass if it is released 50 cm below equilibrium with no initial
velocity.
(b) A force (in Newton) of the form F (t) = F 0 cos t is applied to the system. Find the positive value of
F 0 so that the amplitude of the steady-state solution equals 1 m.
Answers:
(a) x(t) = e −t − 0.5e −2t
(b) F 0 = √ 10
11. A mass of 1 kg is attached to a spring whose spring constant is 5 N/m. The medium offers a damping
force of magnitude 4 times the instantaneous velocity.
(a) Find the displacement x(t) of the mass if it is released 50 cm below equilibrium with an initial
downward velocity of 1 m/s.
(b) Find all times t > 0 when the mass passes through equilibrium position.
Answers:
(a) x(t) = e −2t (2 sin t + 1 2
cos t)
(b) t = nπ − 0.245, for all n = 1, 2, 3 . . .
12. Find the first five nonzero terms of a power series solution of the initial-value problem
For which values of x does the series converge?
y ′′ − x 2 y ′ + 3y = 0, y(0) = 2, y ′ (0) = −1.
Answer: y = 2 − x − 3x 2 + 1 2 x3 + 2 3 x4 + · · · . The series converges for all x since there is no singular
point.
13. Solve the differential equation: y ′′ + x 2 y ′ + xy.
Answer: y = c 1 (1 − 1 6 x3 + 1
45 x6 − · · · ) + c 2 (x − 1 6 x4 + 5
252 x7 − · · · )
14. Solve the differential equation: y ′′ + e x y = 0.
Answer: y = c 1 (1 − x2
2 − x3
6 + · · · ) + c 2(x − x3
6 − x4
12 + · · · )