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ACM 95b/100b Solutions for Problem Set 4
ACM 95b/100b
Solutions for Problem Set 4
Sean Mauch
sean@caltech.edu
February 8, 2002
1
Problem 1
Problem.
Use the shifting theorems to assist in solving the following initial value problems.
1.
4y
4y + 37y = 0,
y(0) = 3, y (0) = 3
2
2.
y + y = r(t),
y(0) = 0, y (0) = 0,
r(t) =
t,
0 < t < 1
0,
t > 1
Solution.
1. We consider the problem
4y
4y + 37y = 0,
y(0) = 3, y (0) = 3
2 .
We take the Laplace transform of the dierential equation and solve for
y.
4(s
2 y
sy(0) y (0)) 4(s
y
y(0)) + 37
y = 0
(4s
2
4s + 37)
y = 12s
6 y = 6
2s
1
4s
2
4s + 37 y = 3
s
1/2
s
2
s + 37/4 y = 3
s
1/2
(s
1/2)
2
+ 9
We take the inverse Laplace transform to obtain the solution.
y = 3 e
t/2
cos(3t)
Alternate Method. We could also solve this problem without the Laplace transform. We
substitute y = e
t
into the dierential equation.
4
2
4 + 37 = 0
= 1
2 ± 3
1 The general solution of the dierential equation is
y = c
1
e
t/2
cos(3t) + c
2
e
t/2
sin(3t).
We substitute this solution into the initial conditions to determine the constants.
y(0) = 3
c
1
= 3
y (0) = 3
2
3
2 + 3c
2
= 3
2
c
2
= 0
y = 3 e
t/2
cos(3t)
2. We consider the problem
y + y = r(t),
y(0) = 0, y (0) = 0,
r(t) =
t,
0 < t < 1
0,
t > 1
We take the Laplace transform of the dierential equation.
s
2 y
sy(0) y (0) +
y =
1
0
t e
st
dt
s
2
+ 1
y = 1
s
2
1s + 1s
2
e
s y =
1
s
2
(s
2
+ 1)
1
s(s
2
+ 1) +
1
s
2
(s
2
+ 1)
e
s y = 1
s
2 1
s
2
+ 1 +
s
s
2
+ 1
1
s +
1
s
2
+ 1
1
s
2
e
s
We take the inverse Laplace transform to obtain the solution.
y = t
sin t + (cos(t 1) 1 + sin(t 1) (t 1)) H(t 1)
y = t
sin t + (cos(t 1) + sin(t 1) t) H(t 1)
Alternate Method. We could also solve this problem without the Laplace transform. First
we solve the problem on the interval [0 . . . 1]. The homogeneous solution is
y
h
= c
1
cos t + c
2
sin t.
We determine a particular solution y
p
= t by inspection. The general solution is
y = c
1
cos t + c
2
sin t + t.
We substitute this solution into the initial conditions to determine the constants.
y(0) = 0
c
1
= 0
y (0) = 0
c
2
+ 1 = 0
y = t
sin t,
for t
[0 . . . 1]
2 Now we solve problem on the interval [1 . . .
). The general solution is
y = c
1
cos t + c
2
sin t.
We determine initial conditions at t = 1 from the solution on the interval [0 . . . 1].
y(1) = 1
sin(1),
y (1) = 1
cos(1)
We substitute the solution into the initial conditions to determine the constants.
y(1) = 1
sin(1),
y (1) = 1
cos(1)
c
1
cos(1) + c
2
sin(1) = 1
sin(1),
c
1
sin(1) + c
2
cos(1) = 1
cos(1)
c
1
= cos(1)
sin(1),
c
2
= cos(1) + sin(1)
1
y = (cos(1)
sin(1)) cos t + (cos(1) + sin(1) 1) sin t
Now we have the solution of the problem.
y =
t
sin t,
0
t < 1
(cos(1)
sin(1)) cos t + (cos(1) + sin(1) 1) sin t, 1 t
2
Problem 2
Problem.
Laplace transforms of integrals.
1. Suppose f (t) has Laplace transform
f (s). Verify the following relationship:
L
t
0
f ( )d
= 1
s f (s)
2. The current i(t) in an electric circuit is governed by the following integral equation
Ri(t) + 1
C
t
0
i( )d = v(t)
where R is the resistance and C is the capacitance. Use the Laplace transform to solve for i(t)
when the imposed voltage drop over the circuit is
v(t) =
0,
0 < t < 2
e
t
,
t > 2.
Solution.
1.
L
t
0
f ( ) d
= 0
e
st
t
0
f ( ) d dt
=
e
st
s
t
0
f ( ) d 0
0
e
st
s
d
dt
t
0
f ( ) d
dt
= 1
s 0
e
st
f (t) dt
= 1
s f (s)
3 2. We consider the problem
Ri(t) + 1
C
t
0
i( )d = v(t),
v(t) =
0,
0 < t < 2
e
t
,
t > 2.
We take the Laplace transform of the integral equation.
Ri + 1
C
i
s = 0
e
st
v(t) dt
R + 1
Cs
i = 2
e
st
e
t
dt
RCs + 1
Cs
i = e
2(s+1)
s + 1
i =
Cs e
2(s+1)
(s + 1)(RCs + 1)
i =
C e
2
e
2s
(CR
1)(s + 1)
e
2
e
2s
R(CR
1)(s + 1/(CR))
We take the inverse Laplace transform to determine the current.
i(t) = C e
2
CR
1 e
(t2)
H(t
2)
e
2
R(CR
1) e
(t2)/(CR)
H(t
2)
i(t) =
1
e
2
R(CR
1) CR e
2
t
e
(2
t)/(CR)
H(t
2)
3
Problem 3
Problem.
Impulsive forcing.
1. Solve the following initial value problem and describe the behavior of y, y , and y
at t = 1
and t = 2
y + 5y + 6y = H(t
1) + (t 2),
y(0) = 0, y (0) = 1
2. Solve the following initial value problem and plot your solution
y + y = g(t),
y(0) = y (0) = 0,
g(t) =
10
k=1
(
1)
k+1
(t
k)
Solution.
1. We consider the problem
y + 5y + 6y = H(t
1) + (t 2),
y(0) = 0, y (0) = 1.
We take the Laplace transform of the dierential equation.
s
2 y
sy(0) y (0) + 5(s
y
y(0)) + 6
y = e
s
s
+ e
2s
s
2
+ 5s + 6
y = 1 + e
s
s
+ e
2s y = 1 +
e
s
s
+ e
2s
s
2
+ 5s + 6
4 We take the inverse Laplace transform to determine the solution. y =
1
s
2
+ 5s + 6 +
e
s
s(s
2
+ 5s + 6) +
e
2s
s
2
+ 5s + 6 y =
1 + e
2s
s
2
+ 5s + 6 +
e
s
s(s
2
+ 5s + 6) y = 1 + e
2s
1
s + 2
1
s + 3
+ e
s
1
6s
1
2(s + 2) +
1
3(s + 3)
y = e
2t
e
3t
+
1
6
1
2 e
2(t1)
+ 1
3 e
3(t1)
H(t
1) + e
2(t2)
e
3(t2)
H(t
2)
At t = 1, y is smooth, y has a corner and y
has a jump discontinuity. At t = 2, y has a
corner, y has a jump discontinuity and y has a Dirac delta type discontinuity.
2. We consider the problem,
y + y = g(t),
y(0) = y (0) = 0,
g(t) =
10
k=1
(
1)
k+1
(t
k).
We take the Laplace transform of the dierential equation.
s
2 y
sy(0) y (0) +
y =
10
k=1
(
1)
k+1
e
ks y =
1
s
2
+ 1
10
k=1
(
1)
k+1
e
ks
We take the inverse Laplace transform to determine the solution.
y =
10
k=1
(
1)
k+1
sin(t
k)H(t k)
The solution is plotted in Figure 1.
5
10
15
20
25
30
35
-10
-7.5
-5
-2.5
2.5
5
7.5
Figure 1: Harmonic oscillator with impulse forcing.
5 4
Problem 4
Problem.
Dierence-dierential equations are important in control theory. Suppose that y(t) =
y
0
(t) for
1 t 0 and then subsequently y(t) satises
dy(t)
dt
+ 2y(t)
2y(t 1) = 0,
t
0.
Note how the rate of change of y depends on its value at a previous time.
1. Show that the Laplace transform of y(t) satises y(s) = y
0
(0) + 2 e
s
0
1
e
st
y
0
(t) dt
2 + s
2 e
s
2. Consider the case y
0
(t) = 1 and solve for y(t).
Solution.
1. We consider the problem
dy(t)
dt
+ 2y(t)
2y(t 1) = 0,
t
0.
First we compute the Laplace transform of y(t
1).
L[y(t 1)] = 0
e
st
y(t
1) dt
= 1
e
s(t+1)
y(t) dt
= e
s
0
1
e
st
y(t) dt + e
s 0
e
st
y(t) dt
= e
s
0
1
e
st
y
0
(t) dt + e
s y(s)
We take the Laplace transform of the dierence-dierential equation.
s
y(s)
y(0) + 2
y(s)
2 e
s
0
1
e
st
y
0
(t) dt + 2 e
s y(s) = 0
(2 + s
2 e
s
)
y(s) = y
0
(0) + 2 e
s
0
1
e
st
y
0
(t) dt y(s) = y
0
(0) + 2 e
s
0
1
e
st
y
0
(t) dt
2 + s
2 e
s
2. Consider the case y
0
(t) = 1. y(s) = 1 + 2 e
s
0
1
e
st
dt
2 + s
2 e
s y(s) = 1 + 2 e
s
( e
s
1)/s
2 + s
2 e
s y(s) = 1
s
u(t) = 1
Clearly this solution satises the dierence-dierential equation.
6 5
Problem 5
Problem.
Inverting Laplace transforms in the complex plane.
1. Suppose that f (t) has Laplace transform
f (s) for
(s) > a. Show that
L
1
[
f (s)] = 0 for t < 0
by inverting the Laplace transform using the Mellin inversion formula. Close the Bromwich
contour from s = c
R to s = c + R with a semicircle to the right and assume that
|
f (s)
| K
R
0 as R .
2. Use the Mellin inversion formula and the Bromwich contour to invert the following Laplace
transforms:
(a)
1
(s
2
a
2
)
2
(b)
1
(s + b)
2
+ a
2
3. Consider the Laplace transform of
f (s) = log
s+1
s
. Select the branch that is analytic except
on the branch cut joining s = 0 and s =
1 and is real valued on the positive real axis. By
appropriately closing the Bromwich contour to the left, verify that
f (t) =
L
1
[
f (s)] = 1 e
t
t
,
t > 0.
Solution.
1. We suppose that f (t) has Laplace transform
f (s) for
(s) > a. We evaluate
L
1
[
f (s)] for
t < 0 by inverting the Laplace transform using the Mellin inversion formula.
L
1
[
f (s)] =
c+ c e
st f (s) ds
Here c > a. We close the Bromwich contour from s = c
R to s = c + R with a semicircle to
the right C
R
. C is the resulting closed contour. We assume that
|
f (s)
| K
R
0 as R .
First show that the integral along C
R
vanishes as R
. We parameterize the semicircle
with s = c + R e ,
(/2 . . . /2).
C
R
e
st f (s) ds /2
/2
e
t(c+R e )
K
R
R d
= K
R
R e
tc
/2
/2
e
tR cos
d
= K
R
R e
tc 0
e
tR sin
d
We use Jordans Lemma to bound the integral.
= K
R
R e
tc
1
tR
= K
R
e
tc
t
0 as R
7 Then we use Cauchys Theorem to evaluate the inverse Laplace transform.
L
1
[
f (s)] =
c+ c e
st f (s) ds
= lim
R
C
e
st f (s) ds
= 0 for t < 0
2. If
f (s) is analytic except for poles at s
1
, s
2
, . . . , s
N
and
f (s)
0 as |s| then the inverse
Laplace transform of
f (s) is
f (t) =
L
1
[
f (s)] =
N
n=1
Res ( e
st f (s), s
n
),
for t > 0.
(a)
L
1
1
(s
2
a
2
)
2
= Res
e
st
(s
2
a
2
)
2
, s = a
+ Res
e
st
(s
2
a
2
)
2
, s =
a
=
d
ds
e
st
(s + a)
2
s=a
+
d
ds
e
st
(s
a)
2
s=
a
=
t e
st
(s + a)
2 2 e
st
(s + a)
3
s=a
+
t e
st
(s
a)
2 2 e
st
(s
a)
3
s=
a
= (at 1) e
at
4a
3
+ (at + 1) e
at
4a
3
= at cosh(at) sinh(at)
2a
3
(b)
L
1
1
(s + b)
2
+ a
2
= Res
e
st
(s + b)
2
+ a
2
, s =
b a + Res
e
st
(s + b)
2
+ a
2
, s =
b + a
=
e
st
s + b
a
s=
ba
+
e
st
s + b + a
s=
b+a
= e
(
ba)t
2a
+ e
(
b+a)t
2a
= e
bt
sin(at)
a
3. We consider the Laplace transform of
f (s) = log
s+1
s
. We select the branch that is analytic
except on the branch cut joining s = 0 and s =
1 and is real valued on the positive real axis.
Let be any positive number. The in