MATH 308 Summer 2008 Practice Test II
1. The Existence and Uniqueness Theorem guarantees that the solution to uniquely exists on x
3 y ′′ + x sin x y ′ −
2 x − 5 y = 0 , y (2) = 6 , y ′ (2) = 7
(a) ( − π, π )
(b) (0 , π )
(c) (5 , ∞ )
(d) (0 , 5)
2. All of the following pairs of functions form a fundamental set of solutions to some second order differential equation on ( −∞ , ∞ ) EXCEPT
(a) 1 , e − t
(b) cos t, sin( t + 2 π )
(c) e − 2 t cos 2 t, e − 2 t sin 2 t
(d) e 5 t , e 5 t
− 1
3. Which of the following will be a particular solution to the equation
4 y ′′ + 4 y ′ + y = 24 x e x
2
?
(a) x 2 ( Ax + B )e x
2
(b) ( Ax + B )e x
2
(c) x ( Ax + B )e x
2
(d) ( Ax + B ) sin x
2
+ ( Cx + D ) cos x
2
4. All the following differential operators are linear EXCEPT
(a) L [ y ] = y ′′ − 3 y ′ + y 3
(b) L [ y ] = y ′′ + y ′ + 2 y
(c) L [ y ] = y ′′ + sin xy ′ + cos xy
(d) L [ y ] = y ′′ + xy ′ + ( x − 1) y

5. The motion of the mass-spring system with damping is governed by y ′′ + 2 y ′ + y =
0 , y (0) = 1 , y ′ (0) = − 3. This motion is
(a) undamped
(b) underdamped
(c) critically damped
(d) overdamped
6.
e 2+
3
π i
4
=
(a) π
√
2
(b) (1 + i ) e 2
(c)
(d)
√
2
(1 − i ) e
2
√
2
2
( − 1 + i ) e
2
7. A 2-kg mass is attached to a spring with stiffness k = 50 N/m. The mass is displaced
1/4 m to the left of the equilibrium point and given a velocity of 1 m/sec to the left. The damping force is negligible. The amplitude of this vibration is
√
41
(a)
20
(b) 1
√
20
(c)
41
1
(d)
4
8. The FSS to the equation y ′′ − 2 y ′ + 5 y = 0 is
(a) { cos x, sin x }
(b) { e x cos 2 x, e x sin 2 x }
(c) { e x , xe x }
(d) { e x , e − x }

9. Given that y a solution to
1
( y x
′′
) =
− y ′
−
1
2 x
2 +
1
2
− 2 y = e x
3 x
−
3
4 is a solution to y
. A solution to y ′′ − y
′′
′
−
− y
2
′ y
− 2 y
= 2 x
=
2 x
−
2 e and
3 x is y
2
( x ) =
1
4 e
3 x is
(a) − x 2 + x −
(b) x 2 − x − 3
2
3
2
− 1
4 e 3 x
− 1
4 e 3 x
(c) − x 2 + x +
(d) x − 3
2
3
4
− 1
4 e 3 x
− e 3 x
10. The Wronskian of two functions y
1
( x ) = x + 2 x 2 an y
2
( x ) = 2 x is
(a) 2 x (1 + 4 x − x (1 + 2 x ))
(b) − 2 x (1 + 4 x − x ln 2(1 + 2 x ))
(c) (1 + 4 x − x (1 + 2 x ))
(d) 2 x (1 + 2 x − x ln 2(1 + 4 x ))
11. Find a general solution to the equation e − 3 x y ′′ + 6 y ′ + 9 y =
1 + 2 x
12. Find a general solution to the equation
4 y ′′ + y ′ = 4 x
3 + 48 x
2 + 1
13. Given that y
1
( x ) = x is a solution to x 2 y ′′ + xy ′ − y = 0 , find the general solution to this equation on (0 , + ∞ ).