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Find the solution for the equation $\left[x-y \tan ^{-1}\left(\frac{y}{x}\right)\right] \mathrm{d} x+x \tan ^{-1}\left(\frac{y}{x}\right) \mathrm{d} y=0$

(1) $\left[x-y \tan ^{-1}\left(\frac{y}{x}\right)\right] \mathrm{d} x+x \tan ^{-1}\left(\frac{y}{x}\right) \mathrm{d} y=0$

$\frac{d y}{d x}=\frac{-\left[x-y \tan ^{-1}\left(\frac{y}{x}\right)\right]}{x \tan ^{-1}\left(\frac{y}{x}\right)} \quad \cdots \cdots (1)$

(2) $y=v x \quad \Rightarrow (3) \: \frac{d y}{d x}=\frac{d v}{d x} x+v \quad \cdots \cdots(2)$

(4) $(2)=(1) \Rightarrow x \frac{d v}{d x}+v=\frac{-\left[x-y \tan^{-1} \left(\frac{y}{x}\right)\right]}{x \tan ^{-1}\left(\frac{y}{x}\right)}$

$x \frac{d v}{d x}=\frac{-\left[x-y \tan ^{-1}\left(y/{x}\right)\right]}{x \tan ^{-1}\left(\frac{y}{x}\right)}-v$

$x \frac{d v}{d x}=\frac{-x[1-v \tan ^{-1}\left(y/{x}\right)]}{x \tan ^{-1}(\frac{y}{x})}-v$

$x \frac{d v}{d x}=\frac{-\left[1-v \tan ^{-1}(v)\right]}{\tan ^{-1}(v)}-v$

$\frac{x d v}{d x}=\frac{-1+v \tan ^{-1}(v)}{\tan ^{-1}(v)}-v$

$\frac{x d v}{d x}=\frac{-1+v \tan ^{-1}(v)}{\tan ^{-1}(v)}-v \cdot \frac {\tan^{-1}(v)}{\tan^{-1}(v)}$

$x \frac{d v}{d x}=\frac{-1+v \tan ^{-1}(v)-v \tan ^{-1}(v)}{\tan ^{-1}(v)}$

$x \frac{d v}{d x}=\frac{-1}{\tan ^{-1}(v)}$

$d x \cdot x \frac{d v}{d x}=\frac{-1}{\tan ^{-1}(v)} \cdot d x$

$\Rightarrow x d v=\frac{-1}{\tan ^{-1}(v)} d x \quad \cdot \frac{\tan ^{-1}(v)}{x}$

$\tan ^{-1}(v) d v=\frac{-1}{x} d x \Rightarrow \frac{d x}{x}=-\tan ^{-1}(v) d v$