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

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Aicharya · Accepted Answer

$\begin{aligned} & \quad x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x^2+2 y^2 \\ & \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{x^2+2 y^2}{x y} ...(i)\\ & \text { Put } y=\mathrm{v} x...(ii) \end{aligned}$
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 $x^2+y^2=x^4$

The particular solution of the differential equation, $x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x^2+2 y^2$…

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