Substituting this into the de we have xy y 4x3 y 2x

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Unformatted text preview: ￿ = C1Cx +x + 4 −3 + C2 u − ( 1 ( 1)− 1) 3 1 u . −C y x (x +−3 + C Take simplest form: u(x) = (x + 1)−3=Thus 1 3 (= + 1) 1)−3 (x2+ 1) = (x + 1)−2 is a solution 2 for x > −1. y1 and y2 are linearly 3 independent on x > 3 1 by Proposition2 3.31. So the − Take simplest form:DEx) = (= + 1)−+ 1) + c y2 = (x −2 . − (x + 1) = (x + 1)− is a solution general solution of u( is yh x c1 (x . Thus 2 (x + 1) + 1) 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 ￿ 2￿ ￿ 2￿ ￿￿ ￿￿ ￿ ￿￿ i) y1 = sin x =⇒ y1 = 2x cos x , 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...
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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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