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and the second condition
X
(
π
)=
Aπ
=0
This implies that
A
= 0 and therefore we again have a trivial solution.
Case 3:
λ>
0
In this case the two roots are imaginary
r
=
±
i
√
λ
Thus the solution is a combination of sine and cosine
X
=
A
1
cos
√
λx
+
B
1
sin
√
λx
Substitute the boundary condition at zero
X
(0) =
A
1
Thus
A
1
= 0 and the solution is
X
=
B
1
sin
√
λx
Now use the condition at
π
X
(
π
)=
B
1
sin
√
λπ
=0
If we take
B
1
= 0, we get a trivial solution, but we have another choice, namely
sin
√
λπ
=0
This implies that the argument of the sine function is a multiple of
π
±
λ
n
π
=
nπ
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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