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Question Number 100184 by bobhans last updated on 25/Jun/20

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

Answered by maths mind last updated on 26/Jun/20

$$u=xy\Rightarrow du=ydx+xdy$$$$\iff \frac{du}{1-u^2}+xdx=0$$$$\Rightarrow \frac{ln\left(\frac{1+u}{1-u}\right)}{2}=-\frac{x^2}{2}+c$$$$\Rightarrow \frac{1+u}{1-u}={ke}^{-x^2}$$$$\Rightarrow u=\frac{{ke}^{-x^2}-1}{1+{ke}^{-x^2}}$$$$y=\frac{1}{x}\left(\frac{{ke}^{-x^2}-1}{1+{ke}^{-x^2}}\right)$$

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