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

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

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

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

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

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

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

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

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

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

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

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