# sol5 - MATH 138 Physics-based Section 007 Assignment 5...

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Unformatted text preview: MATH 138 Physics-based Section 007 Assignment 5 Solutions Winter 2010 Note: A first-order linear differential equation (D.E.) is of the form dy dx + P ( x ) y = Q ( x ) (1) and the integrating factor for such a D.E. is I ( x ) = e R P ( x ) dx . 1 . ( a ) The D.E. dy dx + y = cos ( e x ) (2) is of the form (1) with P ( x ) = 1. Since Z P ( x ) dx = Z 1 dx = x + k we can choose the integrating factor I ( x ) = e x . Multilply the D.E. (2) by I ( x ) = e x to obtain e x dy dx + e x y = e x cos ( e x ) or d dx ( e x y ) = e x cos ( e x ) . Integrate both sides with respect to x to obtain Z d dx ( e x y ) dx = Z e x cos ( e x ) dx or e x y = sin ( e x ) + C, C ∈ < since d dx sin ( e x ) = e x cos ( e x ) by the Chain Rule. Therefore the general solution of the D.E. (2) is y = e- x sin ( e x ) + Ce- x , C ∈ < . 1 1 . ( b ) By writing the D.E. dy dx = x- y x , x 6 = 0 as dy dx + 1 x y = x (3) we see the D.E. is of the form (1) with P ( x ) = 1 x . Since Z P ( x ) dx = Z 1 x dx = ln | x | + k we can choose the integrating factor I ( x ) = e ln x = x. (Note: I ( x ) =- x also works.) Multilply the D.E. (3) by I ( x ) = x to obtain x dy dx + xy = x 2 or d dx ( xy ) = x 2 . Integrate both sides with respect to x to obtain Z d dx ( xy ) dx = Z x 2 dx or xy = 1 3 x 3 + C, C ∈ < . Therefore the general solutions of the D.E. (3) is y = 1 3 x 2 + C x , C ∈ < , x 6 = 0 . 2 2 . ( a ) By writing the D.E. cos x dy dx + y sin x = 1 as dy dx + sin x cos x · y = 1 cos x or dy dx + tan x · y = sec x (4) we see the D.E. is of the form (1) with P ( x ) = tan x . Since Z P ( x ) dx = Z tan xdx = ln | sec x | + k we can choose the integrating factor I ( x ) = e ln(sec x ) = sec x. (Note: I ( x ) =- sec x also works.) Multilply the D.E. (4) by I ( x ) = sec x to obtain sec x dy dx + sec x tan x · y = sec 2 x or d dx (sec x · y ) = sec 2 x. Integrate both sides with respect to x to obtain Z d dx (sec x · y ) dx = Z sec 2 xdx or sec x · y = tan x + C, C ∈ < . Therefore the general solution of the D.E. (4) is y = cos x tan x + C cos x, C ∈ < = sin x + C cos x, C ∈ < Apply the initial condition y ( π/ 4) = √ 2 to get √ 2 = sin π 4 + C cos π 4 √ 2 = 1 √ 2 + C 1 √ 2 2 = 1 + C and therefore C = 1 ....
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## This note was uploaded on 01/06/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.

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sol5 - MATH 138 Physics-based Section 007 Assignment 5...

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