AMath350.W13.A5.Sol

# Solve as separable or linear de assuming x 1 1 dv 4 v

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Unformatted text preview: olve as separable or linear DE. Assuming x > 1, 1 dv 4 = v dx x+1 Z Z 1 4 dv = dx General solution to eq. (1): y (x) = c1 e y (x) = c1 x AMATH 350 2 ln(x) 2 cos(ln(x)) + c2 e 2 ln(x) sin(ln(x)), i.e., cos(ln(x)) + c2 x 2 sin(ln(x)), x > 0. Page 3 Assignment #5 Solutions - Fall 2011 6. (x + 1)2 y 00 + 2(x + 1)y 0 2y = 0 0 00 (a) y1 = x + 1 ) y1 = 1, y1 = 0. Substitute into DE: (x + 1)2 (0) + 2(x + 1)(1) 2(x + 1) = 0. So y1 is a solution of DE on <x< . ii) Find second solution: 0 00 (b) Let y2 = u(x)y1 = u(x)(x + 1). Then y2 = (x + 1)u0 + u and y2 = (x + 1)u00 + 2u0. Substitute into DE: (x + 1)2 [(x + 1)u00 + 2u0] + 2(x + 1)[(x + 1)u0 + u] 2(x + 1)u = 0 (x + 1)3 u00 + 4(x + 1)2 u0 = 0 Let v = u0 , v 0 = u00 to get ﬁrst order DE (x + 1)3 v 0 + 4(x + 1)2 v = 0. Solve as separable or linear DE. Assuming x > 1, 1 dv v dx Z 1 dv v ln |v | v u0 4 x+1 Z 4 dx x+1 4 ln(x + 1) + C0 C1 (x + 1) 4 C1 (x + 1) 4 1 C1 (x + 1) 3 + C2 3 = = = = = u= Take simplest form: u(x) = (x + 1) 3...
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## This document was uploaded on 02/09/2014.

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