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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 isnt 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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 Winter '07
 Anoymous
 Math

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