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Question Number 115539 by bemath last updated on 26/Sep/20

$$\frac{dy}{dx}=\frac{e^{{\tan }^{-1}\left(x\right)}-y}{1+x^2}$$

Answered by bobhans last updated on 26/Sep/20

$$\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{e^{{\tan }^{-1}\left(x\right)}}{1+x^2}$$$$Integratingfactor$$$$IF=e^{\int \frac{dx}{1+x^2}}=e^{{\tan }^{-1}\left(x\right)}$$$$solution:y.e^{{\tan }^{-1}\left(x\right)}=\int e^{{\tan }^{-1}\left(x\right)}.\frac{e^{{\tan }^{-1}\left(x\right)}}{1+x^2}dx$$$$y.e^{{\tan }^{-1}\left(x\right)}=\int \frac{e^{{2\tan }^{-1}\left(x\right)}}{1+x^2}dx$$$$set{\tan }^{-1}\left(x\right)=u\Rightarrow x=\tan u$$$$\int \frac{e^{2u}}{1+\tan {}^{2}u}\sec {}^{2}udu=\frac{1}{2}{e}^{2u}+c$$$$\therefore y.e^{{\tan }^{-1}\left(x\right)}=\frac{1}{2}{e}^{{2\tan }^{-1}\left(x\right)}+c$$$$y=\frac{1}{2}{e}^{{\tan }^{-1}\left(x\right)}+C.e^{-{\tan }^{-1}\left(x\right)}.$$

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