Since weve assumed that k 4 this means that c1 0 from

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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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