Sin k so c2 the only values of k for which no

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Unformatted text preview: o C2 = The only values of k for which no solution exists are the squares of the positive multiples of ⇡ (⇡ 2 , 4⇡ 2 , 9⇡ 2 , etc.). AMATH 350 b) Page 2 Assignment #6 Solutions - Fall 2012 y 00 + 2ky 0 + 16y = 0, y (0) = 0, y (1) = 0. The characteristic equation is m2 +2km +16 = 0, or (m + k )2 +(16 k 2 ) = 0, so p m = k ± k 2 16. Case I: |k | > 4 If |k | > 4, then the roots are real, so y = C1 e ( p k + k 2 1 6) x + C2 e ( k p k 2 1 6) x . Enforcing the boundary conditions, we obtain the system of equations 0 = C1 + C2 0 = C1 e ( p k + k 2 1 6) + C2 e ( k p k 2 1 6) . Using the fact that C2 = p C1 from the first equation, and multiplying 2 the second equation by ek+ k 16 we obtain ⇣p ⌘ 2 0 = C 1 e 2 k 16 1 . Since 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 = C...
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