The particular solution of the differential equation, \(x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x^2+2 y^2\) when \(y(1)=0\) is

\(\begin{aligned}
& \quad x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x^2+2 y^2 \\
& \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{x^2+2 y^2}{x y} ...(i)\\
& \text { Put } y=\mathrm{v} x...(ii)
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{~d} x}...(iii)\)
Substituting (ii) and (iii) in (i), we get
\(\begin{array}{ll}
& \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{x^2+2 \mathrm{v}^2 x^2}{x(\mathrm{v} x)} \\
\therefore \quad & \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{x^2\left(1+2 \mathrm{v}^2\right)}{x^2 \mathrm{v}} \\
\therefore \quad & x \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{1+2 \mathrm{v}^2}{\mathrm{v}}-\mathrm{v}=\frac{1+\mathrm{v}^2}{\mathrm{v}} \\
\therefore \quad & \frac{\mathrm{v}}{1+\mathrm{v}^2} \mathrm{dv}=\frac{1}{x} \mathrm{~d} x
\end{array}\)
Integrating on both sides, we get
\(\begin{aligned}
& \frac{1}{2} \int \frac{2 \mathrm{v}}{1+\mathrm{v}^2} \mathrm{dv}=\int \frac{\mathrm{d} x}{x} \\
\therefore \quad & \frac{1}{2} \log \left|1+\mathrm{v}^2\right|=\log |x|+\log \left|\mathrm{c}_1\right|
\end{aligned}\)
\(\begin{aligned} \therefore \quad \log \left|1+\mathrm{v}^2\right| & =2 \log |x|+2 \log \left|c_1\right| \\ & =\log \left|x^2\right|+\log |c| \ldots\left[\log c_1^2=\log \mathrm{c}\right] \\ \therefore \quad \log \left|1+\mathrm{v}^2\right| & =\log \left|c x^2\right|\end{aligned}\)
\(\begin{array}{ll}
\therefore & 1+v^2=c x^2 \\
\therefore & 1+\frac{y^2}{x^2}=c x^2 \\
\therefore & x^2+y^2=c x^4...(iv)
\end{array}\)
Putting \(x=1\) and \(y=0\), we get
\(\begin{aligned}
& 1+0=c(1) \\
\therefore \quad & c=1
\end{aligned}\)
Substituting \(\mathrm{c}=1\) in (iv), we get
\(x^2+y^2=x^4\)