2 p k we must y c1 sin x2 c2 cos x2 4 a y 00

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Unformatted text preview: 2 p k , we must y = C1 sin x2 + C2 cos x2 . 4. a) y 00 + ky = 0, y (0) = 0, y (1) = 1 Since the characteristic equation is m2 + k = 0, so m = ± consider three cases. AMATH 350 Page 4 Assignment #5 Solutions - Fall 2013 Case I: k < 0 If k < 0, then y = C1 e require that p kx p + C2 e kx . The given boundary conditions C1 + C2 = 0 and C1 e p k + C2 e p k = 1. Therefore C1 e p k C1 e p k =1 =) C1 = e p 1 k e p k = e e2 p p k k 1 . This is a unique number for each k < 0, so we have a unique solution for each k < 0: p p e k(1+x) e k(1 x) p y = 2p k . e 1 e2 k 1 Case II: k = 0 If k = 0, then the DE is simply y 00 = 0, which we can integrate twice to find y = C1 x + C2 . The boundary conditions require that C2 = 0 C 1 + C 2 = 1, and so the (unique) solution is y = x. Case III: k > 0 If k > 0, then y = C1 cos tions require that ⇣p ⌘ ⇣p ⌘ kx + C2 sin kx . The boundary cond...
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