## Transcribed Text

1. (1 pt) Find the Laplace transform of the following functions:
1. f(t) = −1
√
t −6t
F(s) = .
2. f(t) = 8t
3/2 −e
−4t
F(s) = .
3. f(t) = sin(5t) +cos(5t)
F(s) = .
2. (1 pt) Find the Laplace transform of the following functions:
1. f(t) = sin(8t) cos(8t)
F(s) = .
2. f(t) = cos2
(6t)
F(s) = .
3. f(t) = (8−4t)
2
F(s) = .
1. (1 pt) Find the inverse Laplace transform of
4s+2
s
2 +23
s > 0
y(t) = .
2. (1 pt) Find the inverse Laplace transform of
7s+8
s
2 −4
s > 2
y(t) = .
3. (1 pt) Consider the following initial value problem:
y
00 −6y
0 −7y = sin(2t) y(0) = 3, y
0
(0) = 2
Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solve for
Y(s) =
4. (1 pt) Consider the following initial value problem:
y
00 −4y
0 −32y = sin(9t) y(0) = 6, y
0
(0) = −2
Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solve for
Y(s) =
5. (1 pt) Use the Laplace transform to solve the following
initial value problem:
y
00 +7y
0 −18y = 0 y(0) = 4, y
0
(0) = −3
First, using Y for the Laplace transform of y(t), i.e., Y =
L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation
= 0
Now solve for Y(s) =
and write the above answer in its partial fraction decomposition,
Y(s) = A
s+a +
B
s+b where a < b
Y(s) = +
Now by inverting the transform, find y(t) = .
6. (1 pt) Use the Laplace transform to solve the following
initial value problem:
y
00 +2y
0 = 0 y(0) = 4, y
0
(0) = 8
First, using Y for the Laplace transform of y(t), i.e., Y =
L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation
= 0
Now solve for Y(s) =
and write the above answer in its partial fraction decomposition,
Y(s) = A
s+a +
B
s+b where a < b
Y(s) = +
Now by inverting the transform, find y(t) =
.
7. (1 pt) Use the Laplace transform to solve the following
initial value problem:
y
00 +16y = cos(7t) y(0) = 0, y
0
(0) = 0
First, using Y for the Laplace transform of y(t), i.e., Y =
L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solving for Y:
Y(s) =
Find the partial fraction decomposition of Y(s) and its inverse
Laplace transform to find the solution of the DE:
y(t) = .
8. (1 pt) Use the Laplace transform to solve the following
initial value problem:
x
0 = 9x+4y, y
0 = −5x+e
4t
x(0) = 0, y(0) = 0
Let X(s) = L{x(t)}, and Y(s) = L{y(t)}.
Find the expressions you obtain by taking the Laplace transform of both differential equations and solving for Y(s) and
X(s):
X(s) =
Y(s) =
Find the partial fraction decomposition of X(s) and Y(s) and
their inverse Laplace transforms to find the solution of the system of DEs:
x(t) =
y(t) = .
9. (1 pt) Find the Laplace transform of te6t
sin(9t).
L{te6t
sin(9t)} = .
10. (1 pt) Find the Laplace transform of t
2
sin(6t).
L{t
2
sin(6t)} = .
11. (1 pt) Given that
L{
cos(3
√
t)
√
πt
} =
e
−2.25/s
√
s
find the Laplace transform of q t
π
cos(3
√
t).
L{
q t
π
cos(3
√
t)} = .
1. (1 pt) Find the Laplace transform of
f(t) = −5u2(t) +2u3(t) +1u6(t)
F(s) = .
2. (1 pt) Find the inverse Laplace transform of
F(s) = −2e
−2s −1e
−4s −1e
−5s +5e
−8s
s
f(t) = . (Use step(t-c) for
uc(t) .)
3. (1 pt) Consider the function f(t) =
−6, 0 ≤ t < 1
3, 1 ≤ t < 5
4, t ≥ 5
;
1. Write the function in terms of unit step function
f(t) = . (Use step(t-c) for
uc(t) .)
2. Find the Laplace transform of f(t)
F(s) = .
4. (1 pt) Find the inverse Laplace transform of
F(s) = e
−9s
s
2 +4s−5
f(t) = . (Use step(t-c) for
uc(t) .)
5. (1 pt) Find the inverse Laplace transform of
F(s) = 4e
−6s
s
2 +49
f(t) = . (Use step(t-c) for
uc(t) .)
6. (1 pt) Find the Laplace transform of
f(t) = (
0, t < 2
(t −2)
2
, t ≥ 2
F(s) = .
7. (1 pt) Find the Laplace transform of
f(t) = (
0, t < 2
t
2 −4t +7, t ≥ 2
F(s) = .
1. (1 pt) Consider the following initial value problem:
y
00 +64y =
(
6, 0 ≤ t ≤ 9
0, t > 9
y(0) = 8, y
0
(0) = 0
Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solve for
Y(s) =
2. (1 pt) Take the Laplace transform of the following initial
value problem and solve for Y(s) = L{y(t)}:
y
00 −7y
0 −18y =
(
1, 0 ≤ t < 1
0, 1 ≤ t
y(0) = 0, y
0
(0) = 0
Y(s) = .
Now find the inverse transform to find y(t) =
. (Use step(t-c) for uc(t) .) Note:
1
s(s−9)(s+2)
=
−
1
18
s
+
1
22
s+2
+
1
99
s−9
3. (1 pt) Take the Laplace transform of the following initial
value and solve for Y(s) = L{y(t)}:
y
00 +1y =
(
sin(πt), 0 ≤ t < 1
0, 1 ≤ t
y(0) = 0, y
0
(0) = 0
Y(s) = . Hint: write the right hand
side in terms of the Heaviside function.
Now find the inverse transform to find y(t) =
. (Use step(t-c) for uc(t) .) Note:
π
(s
2 +π
2)(s
2 +1)
=
π
π
2 −1
1
s
2 +1
−
1
s
2 +π
2
4. (1 pt) Consider the following initial value problem:
y
00 +9y =
(
3t, 0 ≤ t < 2
0, t ≥ 2
y(0) = 0, y
0
(0) = 0
Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solve for
Y(s) =
5. (1 pt) Consider the following initial value problem:
y
00 +25y =
(
4t, 0 ≤ t ≤ 7
28, t > 7
y(0) = 0, y
0
(0) = 0
Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solve for
Y(s) =
6. (1 pt) Find the Laplace transform of
f(t) =
0, t < 4
5 sin(πt), 4 ≤ t < 5
0, t ≥ 5
F(s) = .
7. (1 pt) Take the Laplace transform of the following initial
value problem and solve for Y(s) = L{y(t)}:
y
00 −5y
0 −24y = S(t) y(0) = 0, y
0
(0) = 0
Where S(t) = (
1, 0 ≤ t < 1
0, 1 ≤ t < 2
, S(t +2) = S(t).
Y(s) = .
The graph of S(t) (a square wave function):
8. (1 pt) Take the Laplace transform of the following initial
value problem and solve for Y(s) = L{y(t)}:
y
00 +12y
0 +17y = T(t) y(0) = 0, y
0
(0) = 0
Where T(t) = (
t, 0 ≤ t < 1/2
1−t, 1/2 ≤ t < 1
, T(t +1) = T(t).
Y(s) = .
Graph of T(t) (a triangular wave function):
9. (1 pt) Take the Laplace transform of the following initial
value and solve for Y(s) = L{y(t)}:
y
00 +4y = R(t) y(0) = 0, y
0
(0) = 0
Where R(t) = sin(πt), R(t +1) = R(t).
Y(s) = .
Graph of R(t) (a rectified sine wave function):
1. (1 pt) Use the Laplace transform to solve the following
initial value problem:
y
00 −6y
0 −7y = δ(t −8) y(0) = 0, y
0
(0) = 0
Use step(t-c) for uc(t).
y(t) = .
y
00 +9y = 7δ(t −3) y(0) = 0, y
0
(0) = 0
Use step(t-c) for uc(t).
y(t) = .
3. (1 pt) Use the Laplace transform to solve the following
initial value problem:
y
00 +12y
0 +45y = δ(t −7) y(0) = 0, y
0
(0) = 0
Use step(t-c) for uc(t).
y(t) = .
2. (1 pt) Use the Laplace transform to solve the following
initial value problem:

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