Question

Solve: $xdy+ydx+\frac{xdy-ydx}{x^2+y^2}=0$

Answer 1

$xdy+ydx+\frac{xdy-ydx}{x^2+y^2}=0$

$\Rightarrow \frac{xdy-ydx}{x^2+y^2}+\frac{1}{x^2+y^2}=0$

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

Differentiate $xy$ and $\frac{y}{x}$

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

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

$\Rightarrow x^{2}d\left(\frac{y}{x}\right)=x\frac{dy}{dx}-y$ -------(3)

Put $2$ and $3$ in $1$, we get

$\frac{d\left(xy\right)}{x^{2}d\left(\frac{y}{x}\right)}=-\frac{1}{x^2+y^2}$

$\Rightarrow \frac{d\left(xy\right)}{d\left(\frac{y}{x}\right)}=-\frac{1}{1+(\frac{y}{x}{)}^2}$

$\Rightarrow d\left(xy\right)=-\frac{dt}{1+t^2}$

Put $\frac{y}{x}=t$

Integrating both sides, we get

$xy={\tan }^{-1}t$

$\Rightarrow xy+{\tan }^{-1}\frac{y}{x}=0$