56_pdfsam_math 54 differential equation solutions odd

56_pdfsam_math 54 differential equation solutions odd - (c)...

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Chapter 2 and so d dy ( e 2 y x ) = e 2 y e 4 y = e 2 y e 2 y x = Z e 2 y dy = e 2 y 2 + C. Solving for x yields x = e 2 y ± e 2 y 2 + C ² = e 4 y 2 + Ce 2 y . 31. (a) On the interval 0 x 2, we have P ( x ) = 1. Thus we are solving the equation dy dx + y = x, y (0) = 1 . The integrating factor is given by µ ( x )=exp ±Z dx ² = e x . Multiplying the equation by the integrating factor, we obtain e x dy dx + e x y = xe x D x [ e x y ]= xe x e x y = Z xe x dx . Calculating this integral by parts and dividing by e x yields y = e x ( xe x e x + C )= x 1+ Ce x . (b) Using the initial condition, y (0) = 1, we see that 1= y (0) = 0 1+ C = 1+ C C =2 . Thus the solution becomes y = x 1+2 e x .
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Unformatted text preview: (c) In the interval x > 2, we have P ( x ) = 3. Therefore, the integrating factor is given by ( x ) = exp Z 3 dx = e 3 x . Multiplying the equation by this factor and solving yields e 3 x dy dx + 3 e 3 x y = xe 3 x D x ( e 3 x y ) = xe 3 x e 3 x y = Z xe 3 x dx . Integrating by parts and dividing by e 3 x gives y = e 3 x 1 3 xe 3 x 1 9 e 3 x + C = x 3 1 9 + Ce 3 x . 52...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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