# A05S - I x and integrating gives Z d dx e-x y dx = Z e-x...

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Math 138 Assignment 5 Solutions 1. See answers in textbook 2. Find the solutions of the following initial value problems: (a) x 2 dy dx + y = x 2 e 1 /x , y (1) = 3 e This isn’t in quite the correct form, so we divide through by x 2 , noting that x 6 = 0. Then the DE reads: dy dx + 1 x 2 y = e 1 /x Then I ( x ) = e R 1 /x 2 dx = e - 1 /x . d dx ( e - 1 /x y ) = ( e 1 /x )( e - 1 /x ) Z d dx ( e - 1 /x y ) dx = Z 1 dx e - 1 /x y = x + C y = ( x + C ) e 1 /x Applying the IC y (1) = 3 e : 3 e = (1 + C ) e C = 2 The solution is: y = ( x + 2) e 1 /x (b) dy dx - y = cos 3 x, y (0) = 0 The integrating factor is I ( x ) = e R - 1 dx = e - x . Multiplying the DE by
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Unformatted text preview: I ( x ) and integrating gives: Z d dx ( e-x y ) dx = Z e-x cos 3 xdx y = e x Z e-x cos 3 xdx Evaluating the remaining integral by parts yields: Z e-x cos 3 xdx = 1 10 e-x (3 sin 3 x-cos 3 x ) + C which gives y = 1 10 (3 sin 3 x-cos 3 x ) + Ce x Using the initial condition, we obtain 0 =-1 10 + C = ⇒ C = 1 10 Therefore, y = 1 10 (3 sin 3 x-cos 3 x + e x )...
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## This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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