Dy dx1 dx 2 x 2 2 dx 2 d2 y dy 2d y dx2 dx dx dx

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Unformatted text preview: 1 (a) Let z = ln(x), x > 0. Apply 2chaindy 2 yp = cos =xdx dz cos2 dz + ⇒sinx dx = dz . − =xx x dx 3 3 dy dy dy ￿ ￿ ￿ = dz dy = 1 dy ￿ ￿ ￿ = 2 d dy d 1 dy dz 1 d dzdy ⇒ 1xdy dz . dy ￿ = = dx1 ￿ dx =2 x ￿ 2 − 2 dx 2 d2 y dy 2d y dx2 dx ￿ dx￿ dx ￿ x dz ￿ x dx dz ￿ x dz = 2 + dy = − . d dy d 1 dy cos2 1xd sin x 1 dy ⇒ x d2 y 2 2 3 1 = y 1 dy − 2 dx dz2 dz d = 1 dz d y = 1 dy 2 d2 y dy d2 y dx = dx dx 2 − dx x = 2 x dx 2dz dz x dz ⇒ x2 = − . ￿ 22− ￿ x dx dz x2 dz 1 x dz 2 x dz dx2 dz2 dz 1 dz d2 y 1 dy 1 cos x + 1 . dy 1 dy == eq. (1) above becomes = − Using the bold face expressions, 2 2 − 2 x dx dz 2 x2 dz 3 x dz x dz 2 dface expressions, eq. (1) above becomes dy y dy dy Using left is All that’s the bold − + 5 + 5y = 0 ⇒ d2 y + 4 + 5y = 0 (2) dz 2 dz dz dz 2 dz d2 y dy dy d2 y dy ￿ 1￿ 2 4 − 2 cos x + = (2) y =√λ1 ++ 5 + + 5ysin 0 ⇒ cos2 + + 1 + 5y = 0 x dz . (b) Auxiliary equation Cdz 4λ dz 5 C20. x + = dz 2 3 dz −4 ± 16...
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This homework help was uploaded on 02/22/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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