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2015-02-15T23:19:17+05:30

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I dont know the best way of solving this.  I am able to solve it in the following manner.

        y - x  y'  =  x + y  y'
         (x + y)  y' =  y - x       --- equation (1)

Let  y = g x          =>            y' = g + x g'

    Substitute in equation (1), 
          x g + x² g' + x g² + x² g g' =  g x - x 
             x g' (1 + g) = - (g² + 1) 

               - 1/x = (g + 1) g' / (g² + 1)
            -1/ x  =  g g' / (g² + 1)  +  g' / (g² + 1)
 Integrating we get, 
       Ln 1/x  + Ln K  =  1/2 Ln (1 + g²)  +  tan⁻¹ g
 Substitute g = y/x, and simplify.

Ln\ \frac{K}{x\sqrt{1+g^2}} = tan^{-1}g\\\\Ln\ \frac{Kx}{x^2+y^2}=tan^{-1}\frac{y}{x}\\\\Or,\ x^2+y^2=K_1\ e^{-2\ tan^{-1}\frac{y}{x}}\\

You may verify the solution, by differentiating this equation and obtaining the expression for y'.

2 5 2
thank u sir but isn't there easier way????