79_pdfsam_math 54 differential equation solutions odd

79_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.5 Since µ ( x )= x 2 we must check to see if the solution x 0 was either gained or lost. The function x 0 is a solution to the original equation, but is not given by the above implicit solution for any choice of C . Hence, y 2 2 y x +3 x = C and x 0 are solutions. 9. We compute partial derivatives of M ( x, y )=2 y 2 +2 y +4 x 2 and N ( x, y )=2 xy + x . ∂M ∂y = ∂y ( 2 y 2 +2 y +4 x 2 ) =4 y +2 , ∂N ∂x = ∂x (2 xy + x )=2 y +1 . Although the equation is not exact ( ∂M/∂y 6 = ∂N/∂x ), the quotient ∂M/∂y ∂N/∂x N = (4 y +2) (2 y +1) 2 xy + x = 2 y +1 x (2 y +1) = 1 x depends on x only, and so the equation has an integrating factor, which can be found by applying formula (8) on page 70 of the text. Namely, µ ( x )=exp ±Z 1 x dx ² =exp( ln | x | )= | x | .
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