\(y^2 d x+\left(x y+x^2\right) d y=0\)
\(\left(x y+x^2\right) d y=-y^2 d x\)
\(\frac{d y}{d x}=\frac{-y^2}{x y+x^2}\) ...(1)
Put \(y=v x, \frac{d y}{d x}=v .1+x \frac{d v}{d x}\)
From (1), \(v+x \frac{d v}{d x}=-\frac{v^2 x^2}{x \cdot v x+x^2}\)
\(\Rightarrow v+x \frac{d v}{d x}=\frac{-v^2 x^2}{x^2(v+1)}\)
\(\Rightarrow x \frac{d v}{d x}=\frac{-v^2}{v+1}-v\)
\(\Rightarrow \frac{v+1}{2 v^2+v} d v=-\frac{d x}{x}\)
\(\Rightarrow \int \frac{v+1}{v(2 v+1)} d v=-\int \frac{d x}{x}\)
\(\Rightarrow \int \frac{v+1}{2 v^2+v} d v=-\int \frac{d x}{x}\)
\(\frac{v+1}{v(2 v+1)}=\frac{A}{v}+\frac{B}{2 v+1}\)
\(\Rightarrow \frac{v+1}{v(2 v+1)}=\frac{A(2 v+1)+B v}{v}\)
\(\Rightarrow v+1=A(2 v+1)+B v\)
\(\text {for } v=0\)
\(1=A(1), A=1\)
\(\text { for } 2 v+1=0, v=-\frac{1}{2}\)
\(\frac{1}{2}=B \cdot\left(-\frac{1}{2}\right), B=-1\)
From (1), \(\int\left(\frac{A}{v}+\frac{B}{2 v+1}\right) d v=-\int \frac{d x}{x}\)
\(\Rightarrow 1 \log v-1 \frac{\log (2 v+1)}{2}=-\log x+\log c\)
\(\Rightarrow \log \frac{y^2}{x^2}-\log \left(\frac{2 y+x}{x}\right)=-\log x^2+\log c^2\)
\(\Rightarrow \frac{y^2}{x^2+2 x y}=\frac{c^2}{x^2} \)
\(\Rightarrow x^2 y^2=c^2\left(x^2+2 x y\right)\)
where c is constant of integration.
