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The solution of the differential equation $\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$, is
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$2 x e^{\tan ^{-1} y}=\mathrm{e}^{2} \tan ^{-1} y+K$
$\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$
$\left(1+y^{2}\right) \frac{d x}{d y}+x=e^{\tan ^{-1} y}$
$\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{\left(1+y^{2}\right)}$
$I F=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$
$\Rightarrow x . e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}{1+y^{2}} \cdot e^{\tan ^{-1} y} \cdot d y}{}$
$\Rightarrow x\left(e^{\tan ^{-1} y}\right)=\frac{e^{2 \tan ^{-1} y}}{2}+c$
$\therefore 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+K$
$\left(1+y^{2}\right) \frac{d x}{d y}+x=e^{\tan ^{-1} y}$
$\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{\left(1+y^{2}\right)}$
$I F=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$
$\Rightarrow x . e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}{1+y^{2}} \cdot e^{\tan ^{-1} y} \cdot d y}{}$
$\Rightarrow x\left(e^{\tan ^{-1} y}\right)=\frac{e^{2 \tan ^{-1} y}}{2}+c$
$\therefore 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+K$
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