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103_pdfsam_math 54 differential equation solutions odd

# 103_pdfsam_math 54 differential equation solutions odd - µ...

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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 ( dt ) = e t and d ( e t x ) dt = ( t + 3) e t e t x = ( t + 3) e t dt = ( t + 3) e t + e t dt = ( t + 2) e t + C x = ( t + 2) + Ce t . Using the initial condition, x (0) = 1, we find that 1 = x (0) = (0 + 2) + Ce 0 C = 3 , and so x = t 2 + 3 e t . 35. For 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 find an integrating factor
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Unformatted text preview: µ ( x ) = x − 5 and multiply the equation by µ ( 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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