71_pdfsam_math 54 differential equation solutions odd

71_pdfsam_math 54 differential equation solutions odd - x 2...

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Exercises 2.4 29. (a) We have M ( x, y )= y 2 +2 xy and N ( x, y )= x 2 . Therefore M y ( x, y )=2 y +2 x and N x ( x, y )= 2 x .Thu s M y ( x, y ) 6 = N x ( x, y ), so the diFerential equation is not exact. (b) If we multiply ( y 2 +2 xy ) dx x 2 dy =0by y 2 ,weobta in ± 1+ 2 x y ² dx x 2 y 2 dy =0 . In this equation we have M ( x, y )=1+2 xy 1 and N ( x, y )= x 2 y 2 . Therefore, ∂M ( x, y ) ∂y = 2 x y 2 = ∂N ( x, y ) ∂x . So the new diFerential equation is exact. (c) ±ollowing the method for solving exact equations we integrate M ( x, y )inpart(b)w ith respect to x to obtain F ( x, y )= Z ± 1+2 x y ² dx = x + x 2 y + g ( y ) . To determine g ( y ), take the partial derivative of both sides of the above equation with respect to y to obtain ∂F
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Unformatted text preview: x 2 y 2 + g ( y ) . Substituting N ( x, y ) (given in part (b)) for F/y , we can now solve for g ( y ) to obtain N ( x, y ) = x 2 y 2 = x 2 y 2 + g ( y ) g ( y ) = 0 . The integral of g ( y ) will yield a constant and the choice of the constant of integration is not important so we can take g ( y ) = 0. Hence we have F ( x, y ) = x + x 2 /y and the solution to the equation is given implicitly by x + x 2 y = C . Solving the above equation for y , we obtain y = x 2 C x . 67...
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