AMath350.W13.A5.Sol

# The boundary condip c1 0 p c1 cos k c2 sin k 1 p p

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Unformatted text preview: ip C1 = 0 p C1 cos k + C2 sin k = 1, p p = csc k , as long as sin k 6= 0. Hence the unique 1 p sin k solution, for all other positive values of k , is ⇣p ⌘ ⇣p ⌘ p 1 p sin y= kx = csc k sin kx . sin k so 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 5 Assignment #5 Solutions - Fall 2011 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 . Si...
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## This document was uploaded on 02/09/2014.

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