x^{2}(x dx+ydy) +2y(xdy - ydx)=0 this equation can be written as x^{2}(xdy+ydx)+ 2x(xdy+ydx)-2(x^{2}+y^{2})dx=0 by adding & subtracting 2x^2dx (xdx+ydx)(x^{2}+2x)-2(x^{2}+y^{2})dx=0 Here xdx can be written as 1/2x^{2} & ydx as 1/2 dy^{2} We have 1/2(dx^{2}+d)(x^{2}+2x)-2(x^{2}+y^{2})dx=0 or \frac{1}{2} d(x^{2}+y^{2})(x^{2}+2x)-2(x^{2}+y^{2})dx=0 \frac{d\left ( x^{2}+y^{2} \right )}{x^{2}+y^{2}} = 4 \frac{dx}{x^{2}+2x} \frac{d\left ( x^{2}+y^{2} \right )}{x^{2}+y^{2}}= \frac{4dx}{x\left ( x+2 \right )} By partial differentiation \frac{4}{x\left( x+2 \right )} = 2\left [ \frac{1}{x}- \frac{1}{x+2} \right ] put 2 in 1 \frac{d(x^{2}+y^{2}) }{x^{2}+y^{2}} + 2\left [\frac{dx}{x+2} - \frac{dx}{x}\right ]=0 by integrating we get log (x^{2} + y^{2}) + 2 log (x+2) - 2 log x = log c \left (x^{2}+ y^{2}\right )+ 2log \left ( \frac{x+2}{x} \right )=log c x or log (x2+y2)+ log \left ( \frac{x+2}{x} \right )^{2} = log c or log \left [ \left ( x^{2}+y^{2} \right )\left ( \frac{x+2}{x}^{2} \right ) \right ]