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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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Verified Answer
The correct answer is:
$2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C$
Given differential equation,
$\begin{aligned}
& \left(1+y^2\right)\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \\
& \frac{d y}{d x}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2} \\
& P=\frac{x}{1+y^2}, \mathrm{Q}=\frac{e^{\tan ^{-1}} y}{1+y^2}
\end{aligned}$
Now, integration factor,
$\mathrm{IF}=e^{\int p d y}=e^{l \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$
The solution of differential equation,
$\begin{aligned}
& x . I F=\int Q . I F d y+C \\
& \Rightarrow x e^{\tan ^{-1} y}=\int \frac{e^{2 \tan ^{-1} y}}{1+y^2} d y+C \\
& \Rightarrow \quad x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{1+y^2}+C \\
& \Rightarrow \quad 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C
\end{aligned}$
$\begin{aligned}
& \left(1+y^2\right)\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \\
& \frac{d y}{d x}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2} \\
& P=\frac{x}{1+y^2}, \mathrm{Q}=\frac{e^{\tan ^{-1}} y}{1+y^2}
\end{aligned}$
Now, integration factor,
$\mathrm{IF}=e^{\int p d y}=e^{l \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$
The solution of differential equation,
$\begin{aligned}
& x . I F=\int Q . I F d y+C \\
& \Rightarrow x e^{\tan ^{-1} y}=\int \frac{e^{2 \tan ^{-1} y}}{1+y^2} d y+C \\
& \Rightarrow \quad x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{1+y^2}+C \\
& \Rightarrow \quad 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C
\end{aligned}$
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