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Unformatted text preview: nce we’ve assumed that k  > 4, this means that C1 = 0, from which
C2 = 0. Therefore the only solution is the trivial solution, y = 0.
Case II: k  = 4
This is really two cases, but they’re so similar that I’ll discuss them both
at once: if k = ±4, then m = ⌥4, and so y = C1 e⌥x + C2 xe⌥x . The
boundary conditions require that
0 = C1
and (using the fact that C1 = 0) 0 = C 2 e ⌥4 . This gives C2 = 0, so again, the only solution is the trivial solution y = 0.
Case III: k  < 4 p
Under this assumption m = k ± 16 k 2 i,
h
i
p
p
kx
2 x + C sin
2 x . Enforcing the boundso y = e
C1 cos 16 k
16 k
2
ary conditions, we ﬁnd that
0 = C1
p
and
0 = C2 e k sin 16 k 2 .
We conclude that either C2 = 0 (which gives us the trivial solution again)
p
or sin 16 k 2 = 0. This occurs if
p
16 k 2 = n⇡ ,
p
where n is any integer. Solving gives k = ± p
16 n2 ⇡ 2 , and we can see
that there are only two eigenvalues: k = ± 16 ⇡ 2 (choosing n = 0
violates our assumption that k  < 4, so we can have only n = 1). The
corresponding eigenfunctions are
y = Ce⌥ p 16 ⇡ 2 x sin(⇡ x)....
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 Fall '12
 DavidHamsworth

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