5 - MATH 138 Winter 2011 Assignment 5 – Solutions 1 Find...

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Unformatted text preview: MATH 138 Winter 2011 Assignment 5 – Solutions 1. Find the general solutions of the following linear DEs: (a) dy dx- 2 xy = 2 x The integrating factor is I ( x ) = e R- 2 xdx = e- x 2 . Then multiplying through by I ( x ) gives: dy dx e- x 2- 2 xye- x 2 = 2 xe- x 2 ⇒ d dx e- x 2 y = 2 xe- x 2 Z d dx e- x 2 y dx = Z 2 xe- x 2 dx e- x 2 y =- e- x 2 + C (by substituting u = x 2 on the RHS) y =- 1 + Ce x 2 (b) dy dx + y cos x = 1 2 sin 2 x The integrating factor is I ( x ) = e R cos x = e sin x . Then we have: d dx ( e sin x y ) = 1 2 e sin x sin 2 x Z d dx ( e sin x y ) dx = Z 1 2 e sin x sin 2 x dx It helps to use the identity sin 2 x = 2 sin x cos x so that the RHS integral be- comes: Z e sin x sin x cos x dx which can be evaluated using the substitution u = sin x , so du = cos x dx and Z e sin x sin x cos x dx = Z ue u du = ue u- e u + C (by parts) Going back to x , we are left with e sin x y = e sin x sin x- e sin x + C y = sin x- 1 + Ce- sin x 1 2. Find the solutions of the following initial value problems: (a) dy dx- y = e x , y (0) = 1 The integrating factor is: I ( x ) = e R- 1 dx = e- x . d dx ( e- x y ) = 1 Z d dx ( e- x y ) dx = Z 1 dx e- x y = x + C y = xe x + Ce x Now we use the initial condition y (0) = 1 to obtain C = 1 , and therefore the solution to the IVP is y = ( x + 1) e x (b) 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 2 (c) x dy...
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This note was uploaded on 07/24/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.

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5 - MATH 138 Winter 2011 Assignment 5 – Solutions 1 Find...

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