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Morework Solutions
Math 213 – Spring 2008
Week 13
5.1#8:
For which values of
λ
does the boundary value problem
y
00

2
y
0
+ (1 +
λ
)
y
= 0
,
y
(0) = 0
,
y
(1) = 0
have a solution.
As usual, we look at the three separate cases:
Case
λ >
0
.
Then the characteristic polynomial of the equation is
r
2

2
r
+
(1 +
λ
) = 0 and the roots are
r
=
2
±
p
4

4(1 +
λ
)
2
=
2
±
√

4
λ
2
= 1
±
iμ
where
μ
=
√
λ
. So the general solution will have the form
y
=
c
1
e
x
cos(
μx
) +
c
2
e
x
sin(
μx
)
.
Our boundary conditions then give the following equations:
0 =
c
1
,
0 =
c
1
e
cos(
μ
) +
c
2
e
sin(
μ
) =
c
2
e
sin(
μ
)
.
So we will have a nontrivial solution exactly when sin(
μ
) = 0, which hap
pens when
μ
=
π,
2
π,
3
π,.
..
(Note that
μ >
0, so 0
,

π,

2
π,.
..
must be
discarded.)
Therefore, the equation has nontrivial solutions when
λ
=
π
2
,
4
π
2
,
9
π
2
,...
.
Case
λ
= 0
.
The characteristic polynomial of the equation is
r
2

2
r
+ 1 =
(
r

1)
2
= 0. So the general solution will have the form
y
=
c
1
e
x
+
c
2
xe
x
.
1
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View Full Document Our boundary conditions then give the following equations:
0 =
c
1
,
0 =
c
1
e
+
c
2
e
=
c
2
e.
We conclude that
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This note was uploaded on 05/09/2008 for the course MATH 2130 taught by Professor Dorais during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 DORAIS
 Math, Calculus

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