Question

Prove that:
$y^2-x^2\frac{dy}{dx}=xy\frac{dy}{dx}$

Answer 1

$y^2-x^2\frac{dy}{dx}=xd\left(\frac{dy}{dx}\right)$

$\frac{y}{2}-\frac{x}{y}\left(\frac{dy}{dx}\right)=\frac{dy}{dx}$

$\frac{y}{2}-(\frac{x}{y}+1)\frac{dy}{dx}=0$

Let $\frac{y}{x}=u$

$y=xu$

$\frac{dy}{dx}=x.\frac{dv}{dx}+v$

Now

$u-(\frac{1}{u}+1)(x\frac{du}{dx}+v)=0$

$u=(\frac{1}{u}+1)(x.\frac{du}{dx}+v)$

$v=\frac{x}{u}.\frac{du}{dx}+1+x\frac{du}{dx}+u$

$v-u(\frac{x}{u}+x)\frac{du}{dx}+1$

$x(\frac{1}{u}+1)\frac{du}{dx}=-1$

$(\frac{1}{v}+1)du=\frac{-dx}{x}$

$(\frac{1}{u}+1)du=\int \frac{-dx}{x}$

$\log u+v=-\log x+c$

$\log u+\log e^u=\log \frac{c}{x}$

$\log v{e}^u=\frac{c}{x}$

As $u=\frac{y}{x}$

$\frac{y}{x}{e}^{\frac{y}{x}}=\frac{c}{x}$

$\therefore y{e}^{\frac{y}{x}}=c$