MAT 275 Written Homework #7 SOLUTIONS
6.1, 6.2 Due: April 11
Solve the following problems, showing any necessary work.
1. [1 point] Use the definition of the Laplace Transform to find L(t).
Solution:
L(t) = Z∞
0
te−st dt =t·e−st
−s
∞
0
−Z∞
0
1·e−st
−sdt
(by integration by parts: u=t,v0=e−st, so u0= 1 and v=e−st
−s)
=t·e−st
−s
∞
0
−e−st
s2
∞
0
= lim
t→∞ t·e−st
−s−e−st
s2−0·e−s·0
−s−e−s·0
s2
= 0 −0−0−1
s2=1
s2.
(The first limit is zero because lim
t→∞
t·e−st
−s= lim
t→∞
t
−sest
L0H
=== lim
t→∞
1
−s2est = 0.)
2. [1 point] Use a table of Laplace Transforms to find the Laplace Transform of 5e7t−8 cos(3t) + 7.
Solution:
L(5e7t−8 cos(3t) + 7) = 5L(e7t)−8L(cos(3t)) + 7L(1) = 5·1
s−7−8·s
s2+ 9 + 7 ·1
s.
3. [1 point] Use Laplace Transforms to find the solution to the initial value problem below.
d2y
dt2+ 3 dy
dt −4y= 50tet
y(0) = 1
y0(0) = 9
Solution: (Compare this problem to the on Written Homework #5.) First, take the Laplace Trans-
form of both sides of the differential equation:
Ld2y
dt2+ 3 dy
dt −4y=L50tet
L[y00]+3L[y0]−4L[y] = 50 Ltet
[s2Y−s·y(0) −y00(0)] + 3[sY −y(0)] −4Y=−50 d
ds L[et] = −50 d
ds 1
s−1
s2Y−s−9+3sY −3−4Y=50
(s−1)2
=⇒
1
Now solve for Y:
(s2+ 3s−4)Y=50
(s−1)2+s+ 12
Y=
50
(s−1)2+s+ 12
s2+ 3s−4=
50
(s−1)2+s+ 12
(s+ 4)(s−1)
This expression for Ycan be put into “polynomial over polynomial” form by multiplying the nu-
merator and denominator by (s−1)2. You will then want to keep the denominator factored, and
multiply out the numerator:
Y=50 + (s+ 12)(s−1)2
(s+ 4)(s−1)(s−1)2=50 + (s+ 12)(s2−2s+ 1)
(s+ 4)(s−1)3
Y=50 + s3−2s2+s+ 12s2−24s+ 12
(s+ 4)(s−1)3=s3+ 10s2−23s+ 62
(s+ 4)(s−1)3
Now, it is time to find y(t); it is time for . . . PARTIAL FRACTION DECOMPOSITION
s3+ 10s2−23s+ 62
(s+ 4)(s−1)3=A
s+ 4 +B
s−1+C
(s−1)2+D
(s−1)3
s3+ 10s2−23s+ 62 = A(s−1)3+B(s+ 4)(s−1)2+C(s+ 4)(s−1) + D(s+ 4)
You can solve for Aand Dby substituting s=−4and s= 1, respectively; then substituting two
more values for sallows you to find Band C.
250 = A· −125 so A=−2 [s=−4]
50 = D·5so D= 10 [s= 1]
62 = −2· −1 + B·4 + C· −4 + 10 ·4or B−C= 5 [s= 0]
64 = −2 + B·6 + C·6 + 10 ·6or B+C= 1 [s= 2]
Using elimination, you find out that B= 3 and C=−2. Thus
Y=−2
s−4+3
s−1+−2
(s−1)2+10
(s−1)3
A table shows that there are elementary functions whose Laplace Transforms are 1
s−4,1
s−1,
1
(s−1)2, and 2
(s−1)3. So Yneeds to be rewritten as:
Y=−2·1
s−4+ 3 ·1
s−1−2·1
(s−1)2+ 5 ·2
(s−1)3,and
y(t) = −2·e4t+ 3 ·et−2·tet+ 5 ·t2et.
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