\(xydy-y^2dx= (x+y)^2 e^(−y/x)\)

\(xy dy/dx-y^2= (x+y)^2 e^(−y/x)\) (1)

Put \(y/x= t => dy/dx= y+x (dt)/dx\)

Now the equation (1) becomes

\(x*xt [t+x (dt)/dx]-x^2t^2= x^2(1+t)^2 e^-t\)

\(t[t+x (dt)/dx]-t^2= (1+t)^2 e^t\)

\(xt (dt)/dx= (1+t)^2 e^t\)

\((e^t tdt)/(1+t)^2= dx/x\)

\((e^t((t+1)-1)dt)/(1+t)^2= dx/x\)

\(e^t (1/(1+t)-1/(1+t)^2)dt= dx/x\) (2)

Integrating on the both sides

\(int e^t(1/(1+t)-1/(1+t)^2)dt= int dx/x\)

Use the standard formula of integration

\(int e^x(f(x)+f'(x))dx= e^x f(x)+C\)

\(e^t/((1+t))= ln(x)+log(C)\)

\(e^(y/x)/(1+y/x)= ln(Cx)\)

\(e^(y/x)= (1+y/x)ln(Cx)\)

\(xy dy/dx-y^2= (x+y)^2 e^(−y/x)\) (1)

Put \(y/x= t => dy/dx= y+x (dt)/dx\)

Now the equation (1) becomes

\(x*xt [t+x (dt)/dx]-x^2t^2= x^2(1+t)^2 e^-t\)

\(t[t+x (dt)/dx]-t^2= (1+t)^2 e^t\)

\(xt (dt)/dx= (1+t)^2 e^t\)

\((e^t tdt)/(1+t)^2= dx/x\)

\((e^t((t+1)-1)dt)/(1+t)^2= dx/x\)

\(e^t (1/(1+t)-1/(1+t)^2)dt= dx/x\) (2)

Integrating on the both sides

\(int e^t(1/(1+t)-1/(1+t)^2)dt= int dx/x\)

Use the standard formula of integration

\(int e^x(f(x)+f'(x))dx= e^x f(x)+C\)

\(e^t/((1+t))= ln(x)+log(C)\)

\(e^(y/x)/(1+y/x)= ln(Cx)\)

\(e^(y/x)= (1+y/x)ln(Cx)\)