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MA266 Practice Problems
- If y ′ +
t
y =
t
and y(1) = 0, then y(ln 2) =?
A. ln 2 − ln(ln 2) B. ln(ln 2) C. ln(ln 2) +
2 ln 2
D.
ln 2
e
2
E.
ln 2 − 1
- What is the largest open interval for which a unique solution of the initial value problem
ty
′
t + 1
y =
t − 2
t − 3
, y(1) = 0
is guaranteed?
A. 0 < t < 1 B. 0 < t < 2 C. 0 < t < 3 D. − 1 < t < 3 E. − 1 < t < 1
- An explicit solution of y′^ = y^2 − 1 is?
A. y =
Ce^2 t
1 − Ce^2 t^
B. y =
1 + Ce^2 t
1 − Ce^2 t^
C. y =
1 − Ce^2 t^
D. y =
1 + Ce^2 t
1 − e^2 t^
E.
y^3
3
− y = C
- If y′^ = y^3 and y(0) = 1, then y(−1) =?
A. 5
− (^14) B.
3 C. 1 D.
E. Does not exist
- Let y(x) be the solution to the initial value problem
xy ′ = 3y + 2x 4 , y(1) = 0.
Then, y(2) is
A. 4 B. 8 C. 16 D. 20 E. 32
- A tank initially contains 40 ounces of salt mixed in 100 gallons of water. A solution containing 4 oz of salt
per gallon is then pumped into the tank at the rate of 5 gal/min. The stirred mixture flows out of the tank at the same rate. How much salt is in the tank after 20 minutes?
A. 400 − 360 e − 1 B. 20 C. 80 D. 40 + 20e E. 400 + 360e 2
- Find the general solution of a homogeneous equation using substitution v =
y
x
dy
dx
5 x 2
2 xy
A. 3y 2
- 5x 2 = Cx 2 B. y 2
- 5x 2 = Cx 3 C. x 2
- 3y 2 = Cx D. 2y − 5 x 2 = Cx 4 E. y 2
- 3x 2 = Cx 3
- Suppose that
dy
dx
= (x + y) 2 − 1.
What is the implicit general solution to this differential equation? (Hint: use the substitution v(x) = x + y.)
A.
x + y
− x = C B.
x
y
x
y
− x = C D. x(x + y) + 1 = C E.
x + y
- An implicit solution of
y
2
dy
dx
is?
A. 2(xy 2
- y) = C B. xy 2
- y = C C. xy 2
- x + y = C D.
y^3
3
- y + x 2 y + x = C E. y = xy 2
- C
- Consider the autonomous differential equation
dy
dt
(y − 1)(y − 4) 2 .
Classify the stability of each equilibrium solution.
A. y = 1 and y = 4 both unstable B. y = 1 unstable; y = 4 stable C. y = 0 and y = 1 stable; y = 4 unstable D. y = 1 stable; y = 4 semistable E. y = 0 stable; y = 1 and y = 4 unstable
- Consider the following doomsday/extinction differential equation for a population P (t) with the initial
population P (0) = 4. dP
dt
= 3P (P − 2)
At what time t does “Doomsday” occur (which means the population explodes)?
A.
ln (2)
6
B.
ln (2)
3
C.
ln (4)
3
D.
ln (4)
6
E. ∞
- Use Euler’s method with step size h = 1 to find the approximate value of y(3), where y(x) solves the initial
value problem
y ′ = x +
y
2
, y(0) = − 8.
A. − 17 B. − 22. 5 C. − 23. 5 D. − 24. 5 E. − 27
- If the Wronskian W (f, g) = − 3 e 4 t and f (t) = 4e 2 t , then g(t) could be
A.
te 2 t B. 12e 2 t C. −
e 2 t D. −
te 4 t E. −
te 2 t
- The general solution of
y ′′ − 4 y ′
is?
A. y = C 1 e 2 t
- C 2 te 2 t B. y = C 1 e 2 t
- C 2 e 2 t C. y = C 1 e 2 t
- C 2 e − 2 t D. y = C 1 e − 2 t
- C 2 te − 2 t
E. y = C 1 t + C 2 t 2
- The general solution of
y ′′′
is?
A. y = C 1 e − 2 t cos t + C 2 e − 2 t sin t B. y = C 1 + C 2 e − 2 t cos t + C 3 e − 2 t sin t C. y = C 1 + C 2 e t cos 2t + C 3 e t sin 2t D. y = C 1 + C 2 cos t + C 3 sin t E. y = C 1 + C 2 e 2 t cos t + C 3 e 2 t sin t
- Let y(x) be the solution to the reducible second-order differential equation
y ′′
- (y ′ ) 2 = 0, y(0) = 0, y ′ (0) = 1.
Find y(2). (Use the substitution p = y′^ > 0.)
A. ln 3 B. e−^2 C. ln 5 D. e^4 E. 4
- The solution of
x ′ =
x, x(0) =
is?
A. 2e 3 t
B. 2e 3 t
C. e 3 t
D. 3e 3 t
− e −t
E. 3e 3 t
- Solve
x ′ =
x, x(0) =
A. x(t) = 2e t
sin t cos t
− e t
cos t sin t
B. x(t) = 2e t
sin t cos t
cos t sin t
C. x(t) = 2e t
sin t cos t
− e t
cos t − sin t
D. x(t) = e t
sin t cos t
− e t
cos t sin t
E. x(t) = e t
− sin t cos t
− e t
cos t sin t
- Solve the initial value problem
x ′ = Ax, x(0) =
, where A =
A. e t
− 2 te t
. B. e t
. C. e t
. D. e t
. E. e t
− 2 te t
- What values of the parameter α in the system below make the origin a saddle point in the phase plane:
x ′ =
α 2
x
A. α > 2 B. α > −
C. α < −
D. 2 > α > −
E. α < − 2
- Find a particular solution of x 1 x 2
x 1 x 2
A. xp =
B. xp =
C. xp =
D. xp =
E. xp =
- Find the general solution of x 1 x 2
x 1 x 2
6 e−t 1
A. c 1
e t
e 2 t
e −t
B. c 1
e t
e 2 t
C. c 1
e t
e 2 t −
6 e−t 1
D. c 1
e t
e 2 t
e −t
E. c 1
e t
e 2 t
e −t
- L{e t (1 + cos 2t)} =?
A.
s − 1
(s − 1)^2 + 4
B.
s − 1
s
s − 1
(s − 1)^2 + 4
C.
s − 1
s − 1
s^2 − 2 s + 5
D.
s
s
(s − 1)^2 + 4
E.
s − 1
s − 1
s^2 − 2 s + 5
- Find the Laplace transform of
f (t) =
t, 0 ≤ t < 1 0 , 1 ≤ t < ∞
A. e−s
s
s − 2
B.
s^2
−e−s^
s^2
C.
s^2
−e−s
s
s^2
D.
s^2
+2e−s
s
s^2
E. e−s
s
s^2
- Solve
y ′′
y(0) = 0, y ′ (0) = 1.
A. u 1 (t)
2 − 4 e−(t−1)^ + 2e−2(t−1)
B. u 1 (t)
2 − 4 e −(t−1)
C. u 0 (t)
2 − 4 e −(t−1)
D.
2 − 4 e−(t−1)^ + 2e−2(t−1)
E. e−t^ − e−^2 t
- Find the solution of the initial value problem
y ′′
y(0) = 0, y ′ (0) = 1.
A. y = sin t + u 0 (t) sin(t − π) B. y = sin t + uπ (t) sin(πt) C. y = uπ (t)(sin t + sin(t − π)) D. y = uπ (t) sin t E. y = sin t + uπ (t) sin(t − π)
- The inverse Laplace transform of
F (s) =
se −s
s^2 + 2s + 5
is?
A. u 1 (t)
e t− 1 cos 2(t − 1) − 1 2 e
t− 1 sin 2(t − 1)
B. u 1 (t) (e −t cos 2t) − 1 2 e
−t sin 2t
C. u 1 (t)
e −t+ cos 2(t − 1) − 1 2 e
−t+ sin 2(t − 1)
D. u 1 (t)
e −t cos 2(t − 1) − 1 2 e
−t sin 2(t − 1)
E. e −t+ cos 2(t − 1) − 1 2 e
−t+ sin 2(t − 1)
35. L
Z (^) t
0
sin 2(t − τ ) cos(3τ )dτ
A.
s^2 + 4
s
s^2 + 9
B.
2 s
(s^2 + 4)(s^2 + 9)
C.
s^2 + 4
s
s^2 + 9
D.
(s^2 + 4)(s^2 + 9)
E.
s
(s^2 + 4)(s^2 + 9)
