B U Department of Mathematics
Math 202 Differential Equations
Spring 2005 Final Exam
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1.
Let
L
(
f
(
t
)
)
=
F
(
s
). Show that
L
(
f
(
at
)
)
=
1
a
F
(
s
a
)
,
a >
0
.
Solution:
L
(
f
(
at
)
)
=
∞
0
e

st
f
(
at
)
dt
=
∞
0
1
a
e

su
a
f
(
u
)
du
(
u
=
at
;
du
=
adt
)
=
1
a
F
(
s
a
)
.
2.
Solve the following initial value problem and discuss the interval of existence
(1 +
t
)
x
+
x
= cos
t
;
x
(

π
2
) = 0
.
Solution:
Observe that the left hand side equals
d
dt
((1 +
t
)
x
). Then
(1 +
t
)
x
=
cos
t dt
= sin
t
+
C,
(
C
∈
R
)
.
Inserting the initial condition we get:
0 = sin(

π
2
) +
C.
Hence
C
= 1 and
x
(
t
) =
sin
t
+ 1
1 +
t
.
3.
Given that
y
(
t
) =
e

t
sin
t
is a solution of the constantcoefficient differential equation
9
y
+ 11
y
+ 4
y

14
y
= 0
,
find the general solution of this equation.
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 Winter '11
 gurel
 Equations, Power Series, Fourier Series

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