Case 2:
λ
=0
In this case we have a double root
r
=0andasweknowfromODEstheso
lut
ionis
X
=
Ax
+
B
The boundary condition at zero yields
X
0
(0) =
A
=0
and the second condition
X
0
(
L
)=
A
=0
This implies that
A
= 0 and therefore we have no restriction on
B
. Thusinth
iscasethe
solution is a constant and we take
X
(
x
)=1
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
DiFerentiate
X
0
=
−
√
λA
1
sin
√
λx
+
√
λB
1
cos
√
λx
Substitute the boundary condition at zero
X
0
(0) =
√
λB
1
Thus
B
1
= 0 and the solution is
X
=
A
1
cos
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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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