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642:527
SOLUTIONS: ASSIGNMENT 5
FALL 2009
Some of these solutions were written by Professor Dan Ocone.
5.5.1 (a), (d): See solutions in text.
(b)
f
(
t
) =
H
(
t
)
e
−
t

H
(
t

1)
e
−
t
. The frst
H
(
t
) is in a sense not needed, since in discussing
the Laplace trans±orm we always take
t
≥
0. To take the Laplace trans±orm using (9a) we
write
f
(
t
) =
H
(
t
)
e
−
t

e
−
1
H
(
t

1)
e
−
(
t
−
1)
and fnd
F
(
s
) = (1

e
−
1
−
s
)
/
(
s
+ 1).
(c)
f
(
t
) = 2
H
(
t
)

5
H
(
t

5) + 3
H
(
t

7),
F
(
s
) = (2

5
e
−
5
s
+ 3
e
−
7
s
)
/s
.
5 (d) See solution in text.
7 (d) The equation is
x
′′

x
= 10(
H
(
t

5)

H
(
t

7)), so (
s
2

1)
X
(
s
) = 10(
e
−
5
s

e
−
7
s
)
/s
and
X
(
s
) = 10(
e
−
5
s

e
−
7
s
)
p
1
s
(
s
2

1)
P
= 10(
e
−
5
s

e
−
7
s
)
p
1
2(
s

1)

1
s
+
1
2(
s
+ 1)
P
.
We take the inverse Laplace trans±orm with the use o± Appendix C, ±ormulas 1, 2, and 30,
to obtain
x
(
t
) = 5
H
(
t

5)(
e
t
−
5

2 +
e
−
(
t
−
5)
)

5
H
(
t

7)(
e
t
−
7

2 +
e
−
(
t
−
7)
)
5.6.1 (a) See solution in text.
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This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.
 Fall '08
 Staff
 Math, Cone

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