103_pdfsam_math 54 differential equation solutions odd

103_pdfsam_math 54 differential equation solutions odd - (...

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Review Problems Hence, the answer is y x + x 2 2 = 7 2 y = x 3 2 + 7 x 2 . 33. Choosing x as the dependent variable, we transform the equation to dx dt + x = ( t +3) . This equation is linear, P ( t ) 1. So, µ ( t )=exp (R dt ) = e t and d ( e t x ) dt = ( t +3) e t e t x = Z ( t +3) e t dt = ( t +3) e t + Z e t dt = ( t +2) e t + C x = ( t +2)+ Ce t . Using the initial condition, x (0) = 1, we Fnd that 1= x (0) = (0 + 2) + Ce 0 C =3 , and so x = t 2+3 e t . 35. ±or M ( x, y )=2 y 2 +4 x 2 and N ( x, y )= xy , we compute ∂M ∂y =4 y, ∂N ∂x = y ∂M/∂y ∂N/∂x N = 4 y ( y ) xy = 5 x , which is a function of x only. Using (8) on page 70 of the text, we Fnd an integrating factor
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Unformatted text preview: ( x ) = x 5 and multiply the equation by ( x ) to get an exact equation, x 5 ( 2 y 2 + 4 x 2 ) dx x 4 y dy = 0 . Hence, F ( x, y ) = Z ( x 4 y ) dy = x 4 y 2 2 + h ( x ) F x = 4 x 5 y 2 2 + h ( x ) = x 5 M ( x, y ) = 2 x 5 y 2 + 4 x 3 h ( x ) = 4 x 3 h ( x ) = 2 x 2 99...
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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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