Solution of the differential equation \(y^{\prime}=\frac{\left(x^2+y^2\right)}{x y}\), where \(y(1)=-2\) is given by

\(\frac{d y}{d x}=\frac{x^2+y^2}{x y}=\frac{x}{y}+\frac{y}{x} \)
\( \text { Let } \frac{x}{y}=v \Rightarrow y=\frac{v}{x} \)
\( \therefore \frac{d y}{d x}=\frac{v-x\left(\frac{d v}{d x}\right)}{v^2} \)
\( \therefore \frac{v-x\left(\frac{d v}{d x}\right)}{v^2}=v+\frac{1}{v} \Rightarrow \frac{1}{v}-\left(\frac{x}{v^2}\right) \frac{d v}{d x}=v~+\) \(\frac{1}{v} \)
\( \therefore \int \frac{d v}{v^3}=-\int \frac{d x}{x} \)
\( \therefore \frac{-1}{2 v^2}=-\log x-c \quad \Rightarrow \frac{1}{2 v^2}=\log x+c \)
\( \therefore \frac{y^2}{2 x^2}=\log \mathrm{x}+\mathrm{c} \quad \Rightarrow \mathrm{y}^2=\mathrm{x}^2 \log \mathrm{x}^2+2 \mathrm{x}^2 \mathrm{c} \)
\( \text { We have } \mathrm{y}(1)=-2 \)
\( \therefore 4=0+2 c \Rightarrow c=2 \)
\( \therefore \mathrm{y}^2=\mathrm{x}^2 \log \mathrm{x}^2+4 \mathrm{x}^2\)