A5 Solutions

Then y2 u sin x2 2ux cos x2 and y2 u sin x2

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Unformatted text preview: ; −1. y1 and y2 are linearly 3 independent on x > −1 by Proposition 3.31. So the Take simplest form:DEx) = (= + 1)−+ 1) + c y2 = (x −2 . −3 (x + 1) = (x + 1)−2 is a solution u( is y x c (x . Thus (x + 1) + 1) general solution of h 1 2 for x > −1. y1 and y2 are linearly independent on x > −1 by Proposition 3.31. So the general solution of DE is yh = c1 (x + 1) + c2 (x + 1)−2 . 4 b) xy ￿￿ − y ￿ + 4x3 = 0 4 ￿￿ ￿￿ ￿￿ ￿￿ ￿ ￿￿ i) y1 = sin x2 =⇒ y1 = 2x cos x2 , y1 = 2 cos x2 − 4x2 sin x2 . Substituting this into the DE, we have ￿￿ ￿￿ ￿￿ ￿￿ xy ￿￿ −y ￿ +4x3 y = 2x cos x2 −4x3 sin x2 −2x cos x2 +4x3 s...
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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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