Question

Solve:
$(x\frac{dy}{dx}-y){\tan }^{-1}\frac{y}{x}=x$

Answer 1

Given, $(\frac{xdy}{dx}-y)ta{n}^{-1}\frac{y}{x}=x$

$\Rightarrow \left(\frac{xdy-ydx}{dx}\right)ta{n}^{-1}\frac{y}{x}=x$

$\frac{-\frac{x^{2}d|y|x}{dx}}{dx}.ta{n}^{-1}\frac{y}{x}=x$ $[\because \frac{d\left(y\right)x}{dx}=\frac{ydx-xdy}{x^2}]$

$-ta{n}^{-1}\left(\frac{y}{x}\right)d\left(\frac{y}{x}\right)=\frac{dx}{x}$

Integrating both sides

$\Rightarrow -\int ta{n}^{-1}\left(\frac{y}{x}\right)d\left(\frac{y}{x}\right)=\int \frac{dx}{x}$

$-\left[ta{n}^{-1}\right(\frac{y}{x}).\int d(\frac{y}{x})-\int \frac{1}{(\frac{y}{x}{)}^2+1}\int d(\frac{y}{x}\left)d\right(\frac{y}{x}\left)\right]=logx+c$ [using by parts ]

$-\left[\right(ta{n}^{-1}\left(\frac{y}{x}\right)\left)\right(\frac{y}{x})-\frac{1}{2}log(\frac{y}{x}{)}^2+1\left]\right]=logx+c$

$-\frac{y}{x}ta{n}^{-1}\left(\frac{y}{x}\right)+\frac{1}{2}log\left(\right(\frac{y}{x}{)}^2)+1]=logx+c$