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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 7

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online