AMath350.F11.A5.Sol

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online