A5 Solutions

D dy d 1 dy 1 d dy 1xdy d2 y dx1 dx dz2 x dz 2 2

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Unformatted text preview: 2. dx dx dy dz rule: 1 dy 1 dy dy 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 ￿ ￿ ￿ = . d dy d 1 dy 1 d dy ⇒ 1xdy d2 y ￿ = = dx1 ￿ dx dz2 x dz 2 − 2 dx dz = d2 y dy d2 y dx2 dx ￿ dx￿ dx ￿ x dz ￿ x dx ￿ dz ￿x x dz = 2 + − . ⇒ x2 2 = d dy d 1 dy cos2 1xd sin d2 y 2 3 1 = y 1 dy − 12 dy dx dz2 dz d dy = 1 dz d y = 1 dy 2 2 − dx2 = dx dx 2 − dx x = 2 x dx 2dz dz x dz ⇒ x2 d y = d y − dy . ￿ 22 ￿ x dx dz x2 dz 1 x dz 2 x dz dx2 dz2 dz 1 dz d2 y 1 dy 1 d y x 1 dy = − 2 == 2cos2 − + 1 becomes Using the bold face expressions, eq. (1) above . 2 2 dz 3 x dz x dx dz x dz x dy d2 y Using left is 2 y dy (1) All that’s the bolddface expressions, eq. = 0 above becomes4 dy + 5y = 0 − + 5 + 5y ⇒ + (2) 2 dz dz dz dz 2 dz 2 d2 y dy dy 1 ￿ d y + 4 dy ￿ 5y = 0 2 − 2 cos x + = (2) y =√λ1 ++ 5 + + 5ysin 0 ⇒ cos2 x + 1 + (b) Auxiliary equation Cdz 4λ dz 5 C20. x + = dz 2 dz . 3 dz −4 ±...
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This note was uploaded on 05/29/2013 for the course AMATH 350 taught by Professor Noidea during the Fall '11 term at Waterloo.

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