Question
Solve the following differential equations : $xy \frac{d y}{d x}= x ^2+2 y ^2$

Answer

$
\begin{array}{r}
x y \frac{d y}{d x}=x^2+2 y^2 \\
\therefore \frac{d y}{d x}=\frac{x^2+2 y^2}{x y}
\end{array}
$
Put $y=v x$. Then $\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=\frac{x^2+2 v^2 x^2}{x \cdot v x}=\frac{1+2 v^2}{v}$
$
\therefore x \frac{d v}{d x}=\frac{1+2 v^2}{v}-v=\frac{1+2 v^2-v^2}{v}
$
$\therefore x \frac{d v}{d x}=1+v^2$
$\therefore x \frac{d v}{d x}=\frac{1+v^2}{v}$
$
\therefore \frac{v}{1+v^2} d v=\frac{1}{x} d x
$
Integrating, we get
$
\begin{aligned}
& \int \frac{v}{1+v^2} d v=\int \frac{1}{x} d x \\
\therefore & \frac{1}{2} \int \frac{2 v}{1+v^2} d v=\int \frac{1}{x} d x+\log c_1 \\
\therefore & \frac{1}{2} \log \left|1+v^2\right|=\log |x|+\log c_1 \\
\therefore & \log \left|1+v^2\right|=2 \log |x|+2 \log c_1 \\
\therefore & \log \left|1+v^2\right|=\log \left|x^2\right|+\log c_1{ }^2 \\
\therefore & \log \left|1+v^2\right|=\log \left|c x^2\right|, \text { where } c=c_1{ }^2 \\
\therefore & 1+v^2=c x^2
\end{aligned}
$
$
\begin{aligned}
& \therefore 1+\frac{y^2}{x^2}=c x^2 \\
& \therefore \frac{x^2+y^2}{x^2}=c x^2 \\
& \therefore x^2+y^2=c x^4
\end{aligned}
$
This is the general solution.

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