Math 175 Practice Questions for Test 3

Math 175
Practice Questions for Test 3
1. Find the general solution of
if (a) β = 1, (b) β = 2, (c) β = 3.
y ′′ (x) + βy ′ (x) + y(x) = 0
Answers:
(a) If β = 1, then y = e
(c −x/2 1 cos √ 3
2 x + c 2 sin √ )
3
2 x
(b) If β = 2, then y = (c 1 + c 2 x)e −x
(c) If β = 3, then y = c 1 e (−3+√ 5)x/2 + c 2 e (−3−√ 5)x/2
2. Solve the following differential equation.
(a) y ′′ + 3y ′ − 4y = e 2x , (use y p = Ae 2x )
(b) y ′′ + 4y ′ + 5y = 3x, (use y p = Ax + B)
(c) y ′′ + 6y ′ + 9y = sin 2x, (use y p = A cos 2x + B sin 2x)
(d) y ′′ + 9y = 5 cos x + sin x, (use y p = A cos x + B sin x)
(e) y ′′ − 5y ′ + 6y = x + 2 + 3e 2x , (use y p = Ax + B + Cxe 2x )
(f) y ′′ − 4y ′ + 4y = 3e 2x , (use y p = Ax 2 e 2x )
Answers:
(a) y(x) = c 1 e x + c 2 e −4x + 1 6 e2x .
(b) y(x) = c 1 e −2x sin x + c 2 e −2x cos x + 3 5 x − 12
25 .
(c) y(x) = c 1 e −3x + c 2 xe −3x − 12
5
169
cos 2x +
169
sin 2x
(d) y(x) = c 1 sin 3x + c 2 cos 3x + 1 8 sin x + 5 8 cos x
(e) y(x) = c 1 e 3x + c 2 e 2x + x 6 + 17
36 − 3xe2x
(f) y(x) = c 1 e 2x + c 2 xe 2x + 3 2 x2 e 2x
3. Evaluate the following.
(a) L { 2t 4 + 2 sin(3t) − cos(2t) } (b) L −1 {
{ } s − 2
(d) L −1 s 2 + 2s + 5
{
(g) L −1 s 2 }
+ s + 2
(s − 1)(s + 2)(s 2 + 4)
}
{ }
s + 3
s + 2
(c) L −1
(s − 1)(s + 2)
s 2 + 9
{ } e
(f) L −1 −2s
s + 3
⎧
⎪⎨ 1, 0 ≤ t < 2
(h) L {f(t)} for f(t) = 0, 2 ≤ t < 4
⎪⎩
−1, t ≥ 4
{ s
(e) L −1 2 }
+ 1
(s − 2) 3
Answers:
(a) L { 2t 4 + 2 sin(3t) − cos(2t) } = 48
s 5 + 6
s 2 + 9 − s
s 2 + 4
{
}
(b) L −1 s + 3
= 4
(s − 1)(s + 2)
3 et − 1 3 e−2t
{ } s + 2
(c) L −1 s 2 = cos(3t) + 2
+ 9
3 sin(3t)
(d) L −1 { s − 2
s 2 + 2s + 5
}
= e −t cos 2t − 3 2 e−t sin 2t
(e) L −1 { s 2 + 1
(s − 2) 3 }
= 5 2 t2 e 2t + 4te 2t + e 2t

Math 175 Practice Questions for Test 3 Page 2 of 2
{ } e
(f) L −1 −2s
= e −3(t−2) U(t − 2)
s + 3
{
(g) L −1 s 2 }
+ s + 2
(s − 1)(s + 2)(s 2 = 4
+ 4)
15 et − 1 6 e−2t − 1
10 cos 2t + 1 5
sin 2t
(h) L {f(t)} = 1 s − e−2s
s
− e−4s
s
4. Use Laplace transform to solve the initial value problems.
(a) y ′′ (t) − 2y ′ (t) + 2y(t) = e t , y(0) = 2, y ′ (0) = 0.
(b) y ′′ (t) − 4y ′ (t) = 3e −t , y(0) = 1, y ′ (0) = −1.
(c) y ′′ (t) + 2y ′ (t) + 5y(t) = 10, y(0) = 0, y ′ (0) = 0.
(d) y ′′ (t) − y ′ (t) − 2y(t) = cos(2t), y(0) = 1, y ′ (0) = −2.
(e) y ′′ (t) + y(t) = sin(2t), y(0) = 1, y ′ (0) = 0.
(f) y ′′ (t) + 9y(t) = cos(3t), y(0) = 1, y ′ (0) = 2.
(g) y ′ (t) + 2y(t) = f(t), y(0) = 3, where f(t) =
Answers:
(a) y(t) = e t + e t cos t − 2e t sin t
(b) y(t) = 1 2 + 3 5 e−t − 1
10 e4t
(c) y(t) = 2 − 2e −t cos(2t) − e −t sin(2t)
(d) y(t) = 7 5 e−t − 1 4 e2t − 3
20 cos(2t) − 1
20 sin(2t)
(e) y(t) = − 1 3 sin(2t) + cos t + 2 3 sin t
(f) y(t) = 2 3 sin(3t) + cos(3t) + 1 6 t sin(3t)
(g) y(t) = 1 2 + ( 5
2 e−2t − ) 1 2 1 − e
−2(t−1)
U(t − 1)
{
1, 0 ≤ t ≤ 1
0, t ≥ 1
5. The differential equation for the current i(t) in an LR series circuit is
L di + Ri(t) = V (t).
dt
If the inductance and resistance are respectively L = 1 H, R = 10 Ω and
{
1, 0 ≤ t < 2
V (t) =
0, t ≥ 2.
Find the current i(t) if i(0) = 0.
Answer: The current is
or
i(t) = 1 10 − 1
10 e−10t − 1 1
U(t − 2) + U(t − 2)e−10(t−2)
10 10
i(t) =
{
1
10 (1 − e−10t ), 0 ≤ t < 2
1
10 (e−10(t−2) − e −10t ), t ≥ 2
6. Find the current in an LR circuit if L = 2 H, R = 4 Ω and V (t) = 5e −t volts. Assume i(0) = 0.
Answer: i(t) = 5 2 e−t − 5 2 e−2t
7. Find the charge on the capacitor q(t) and the current i(t) in an LRC circuit if L = 5 3
H, R = 10 Ω,
C = 1
30
F and V = 300 V if q(0) = 0 and i(0) = 0.
Answer: The charge and current are
q(t) = 10 − 10e −3t (cos 3t + sin 3t) and i(t) = 60e −3t sin 3t.