81_pdfsam_math 54 differential equation solutions odd

81_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.5 Solving these equations simultaneously, we obtain n =1and m =1 . So , µ ( x, y )= xy. With these choices for n and m we obtain the exact equation (2 xy 3 6 x 2 y 2 ) dx +(3 x 2 y 2 4 x 3 y ) dy =0 . Solving this equation, we have F ( x, y )= Z (2 xy 3 6 x 2 y 2 ) dx = x 2 y 3 2 x 3 y 2 + g ( y ) F y ( x, y )=3 x 2 y 2 4 x 3 y + g 0 ( y )= N ( x, y )=3 x 2 y 2 4 x 3 y. Therefore, g 0 ( y ) = 0. Since the constant of integration can be incorporated into the constant C of the solution, we can pick g ( y ) 0. Thus, we have F ( x, y )= x 2 y 3 2 x 3 y 2 and the solution becomes x 2 y 3 2 x 3 y 2 = C. Since we have multiplied the original equation by
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