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Math 293 Prelim #3 Solutions
Spring 2000
1
. (a)
π
2
π
3
π
3
π
π


2
π

0
(b)
π
2
π
3
π
3
π
π


2
π

0
(c)
b
n
=
2
π
R
π
0
sin
x
cos
nx dx
=
1
π
R
π
0
(
sin
(n
+
1
)x

sin
(n

1
)x)dx
=
1
π
(

cos
(n
+
1
)x
n
+
1
+
cos
(n

1
)x
n

1
±
±
π
0
)
=
1
π
(

cos
(n
+
1
)π
n
+
1
+
1
n
+
1
+
cos
(n

1
)π
n

1

1
n

1
)
. This is 0 for
n
odd and
1
π
(
2
n
+
1

2
n

1
)
=

4
π(n
2

1
)
for
n
even. When
n
=
1 this calculation involves dividing by 0, so we
compute
b
1
directly:
b
1
=
2
π
R
π
0
sin
x
cos
xdx
=
2
π
sin
2
x
2
±
±
π
0
=
0. It is not necessary
to calculate
a
0
separately since the calculation for a general value of
n
works when
n
=
0, and gives
a
0
=
4
π
. The ﬁnal answer is
2
π

4
π
∞
X
n
=
1
1
4
n
2

1
cos2
nx
.
2
.
Case
1:
λ
=
α
2
>
0. Then
y
=
c
1
sin
αx
+
c
2
cos
αx
,
y
0
=
αc
1
cos
αx

αc
2
sin
αx
.
We have
y
0
(
0
)
=
αc
1
=
0so
c
1
=
0. Then
y(π)
=
c
2
cos
απ
=
0, which implies
α
=
2
n

1
2
for
n
=
1
,
2
,
···
. Thus we have the eigenvalues
λ
=
(
2
n

1
2
)
2
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This homework help was uploaded on 01/21/2008 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 TERRELL,R
 Math, Differential Equations, Equations

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