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asst5soln

# asst5soln - MATH 138 Assignment 5 Solutions Winter 2010...

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MATH 138 Assignment 5 Solutions Winter 2010 Note: A f rst-order linear di f erential 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 .S inc e 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 =s in( 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

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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 .S inc e 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 < 6 . 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 .S ince Z P ( x ) dx = Z tan xdx =ln | sec x | + k we can choose the integrating factor I ( x )= e ln(sec x ) 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 2 x or d dx (sec x · y 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 =c o s x tan x + C cos x, C < =s i n x + C cos x, C < Apply the initial condition y ( π / 4) = 2 to get 2=s i n ³ π 4 ´ + C cos ³ π 4 ´ 2= 1 2 + C 1 2 2=1+ C and therefore C . The solution to the IVP is y =sin x +cos x. 3

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2 . ( b ) By writing the D.E. x dy dx y = x 3 sin x as dy dx 1 x · y = x 2 sin x, x 6 =0 (5) we see the D.E. is of the form (1) with P ( x )= 1 x .S in ce Z P ( x ) dx = Z 1 x dx = ln | x | + k we can choose the integrating factor I ( x e ln( x ) = e ln ( x 1 ) = 1 x . (Note: I ( x 1 x also works.) Multilply the D.E. (5) by I ( x )=sec x to obtain 1 x dy dx 1 x μ 1 x · y = x sin x or d dx μ 1 x · y = x sin x.
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asst5soln - MATH 138 Assignment 5 Solutions Winter 2010...

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