AMath350.F13.A5.Sol

# 1 above 3 all thats left is d2 y dz 2 3 a dy dy

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Unformatted text preview: above . 3 All that’s left is d2 y dz 2 3. a) dy dy + 5 + 5y = 0 dz dz ) 1 ) x2 d2 y dy + 4 + 5y = 0 dz 2 dz d2 y d2 y = dx2 dz2 dy . dz (2) 2 y =p 2 cos + = sin cos x + 1 . (b) Auxiliary equation C1 + 4 x+ 5 C20. x + 3 4 ± 16 20 Roots: = = 2 ± i. 2 General solution to eq. (2): y (z ) = c1 e 2z cos(z ) + c2 e 2z sin(z ). General solution to eq. (1): y (x) = c1 e 2 ln(x) cos(ln(x)) + c2 e 2 ln(x) sin(ln(x)), i.e., 00 x) = (x + 1)2 yy (+ 2 (c1 x 21) y 0 x2y + c20. 2 sin(ln(x)), x > 0. x + cos(ln( )) = x 6. (i) + 1)2 y 00 + 2(x + solution:= 0 x Check given 1)y 0 2y 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< . 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. S...
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## This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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