math problem
MiracleFY
Math275Test2ReviewKeySpring2020.pdf
MATH 275 TEST 2 REVIEW KEY
INSTRUCTIONS : REDUCE / SIMPLIFY ALL ANSWERS AND KEEP ALL
EXPONENTS POSITIVE WHERE APPLICABLE.
1. Find a general solution to the differential equation.
y′′(x) − 3y′(x) + 2y(x) = ex sin(x)
Answer : y(x) = (cos(x) − sin(x))ex
2 + c1e
x + c2e 2x
2. Find a general solution to the differential equation.
y′′(θ) + 2y′(θ) + 2y(θ) = e−θ cos(θ)
Answer : y(θ) = θ
2eθ sin(θ) +
c1 cos(θ) + c2 sin(θ)
eθ
3. Find the solution to the initial value problem.
y′′(θ) −y(θ) = sin(θ) −e2θ, y(0) = 1, y′(0) = −1
Answer : y(θ) = − 1
2 sin(θ) −
1
3 e2θ +
3
4 eθ +
7
12eθ
4. Find a general solution to the differential equation using the method of variation of
parameters.
y′′ + 2y′ + y = e−t
Answer : y(t) = t2
2et + c1 + c2t
et
5. Find a general solution to the differential equation using the method of variation of
parameters.
y′′ + 9y = sec2(3t)
Answer : y(t) = sin(3t) ln |sec(3t) + tan(3t)|− 1
9 + c1 cos(3t) + c2 sin(3t)
6. Find a general solution to the differential equation using the method of variation of
parameters.
y′′ + 4y′ + 4y = e−2t ln(t)
Answer : y(t) = 2t2 ln(t) − 3t2
4e2t + c1 + c2t
e2t
7. Find a general solution to the differential equation to the given Cauchy-Euler equation
for t > 0. d2w
dt2 +
6
t
dw
dt +
4
t2 w = 0
Answer : w(t) = c1 t
+ c2 t4
8. Find a general solution to the differential equation to the given Cauchy-Euler equation
for t > 0.
9t2y′′(t) + 15ty′(t) + y(t) = 0
Answer : y(t) = c1 + c2 ln(t)
3 √ t
9. Find a general solution to the differential equation to the given Cauchy-Euler equation
for t < 0.
y′′(t) − 1
t y′(t) +
5
t2 y(t) = 0
Answer : y(t) = c1t cos[2 ln(−t)] + c2t sin[2 ln(−t)], t < 0 3 c1 = −C1 and c2 = −C2 are arbitrary constants.
10. Find a general solution to the differential equation to the given Cauchy-Euler equation
for t < 0.
t2y′′(t) + 9ty′(t) + 17y(t) = 0
Answer : y(t) = c1 cos[ln(−t)] + c2 sin[ln(−t)]
t4 , t < 0
11. Solve the given initial value problem for the Cauchy-Euler equation.
t2y′′(t) − 4ty′(t) + 4y(t) = 0; y(1) = −2 , y′(1) = −11
Answer : y(t) = t− 3t4
12. Find a general solution to the differential equation with x as the independent variable.
y′′′ + 3y′′ + 28y′ + 26y = 0
Answer : y(x) = c1 + c2 cos(5x) + c3 sin(5x)
ex
13. Find a general solution to the differential equation with x as the independent variable.
y(4) + 4y′′′ + 6y′′ + 4y′ + y = 0
Answer : y(x) = c1 + c2x + c3x
2 + c4x 3
ex
14. Find a general solution to the given homogeneous equation.
(D + 4)(D − 3)(D + 2)3(D2 + 4D + 5)2D5[y] = 0
Answer : y(x) = c1 e4x
+c2e 3x+
c3 + c4x + c5x 2 + c6 cos(x) + c7x cos(x) + c8 sin(x) + c9x sin(x)
e2x +
+ c10 + c11x + c12x 2 + c13x
3 + c14x 4
15. Solve the given initial value problem.
y′′′(x) −y′′(x) − 4y′(x) + 4y(x) = 0; y(0) = −4 , y′(0) = −1 , y′′(0) = −19
Answer : y(x) = ex − 2
e2x − 3e2x
16. Solve the given initial value problem.
y′′′(x) − 4y′′(x) + 7y′(x) − 6y(x) = 0; y(0) = 1 , y′(0) = 0 , y′′(0) = 0
Answer : y(x) = e2x −ex √
2 sin(x √
2)
17. Use the annihilator method to determine the form of a particular solution for the given
equation.
y′′ + 2y′ + 2y = e−x cos(x) + x2
Answer : yp(x) = c1 cos(x) + c2 sin(x) + c3x cos(x) + c4x sin(x)
ex + c5x
2 + c6x + c7
18. Use the annihilator method to determine the form of a particular solution for the given
equation.
z′′′ − 2z′′ + z′ = x−ex
Answer : zp(x) = c2x + c3x 2 + c6x
2ex
19. Use the method of variation of parameters to determine a particular solution to the
given equation.
z′′′ + 3z′′ − 4z = e2x
Answer : zp(x) = e2x
16
20. Use the method of variation of parameters to determine a particular solution to the
given equation.
y′′′ + y′′ = tan(x), 0 < x < π 2
Answer : yp(x) = ln(sec(x)) − sin(x) ln(sec(x) + tan(x))
