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An integrating factor for the differential equation $\left(1+y^2\right) d x-\left(\tan ^{-1} y-x\right) d y=0$
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The correct answer is:
$e^{\tan ^{-1} y}$
$\begin{aligned} & \left(1+y^2\right) d x-\left(\tan ^{-1} y-x\right) d y=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{1+y^2}{\tan ^{-1} y-x} \Rightarrow \frac{d x}{d y}=\frac{\tan ^{-1} y}{1+y^2}-\frac{x}{1+y^2} \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}\end{aligned}$
This is equation of the form $\frac{d x}{d y}+P x=Q$
So, I.F. $=e^{\int P d y}=e^{\int \frac{1}{1+y^2} \cdot d y}=e^{\tan ^{-1} y}$.
This is equation of the form $\frac{d x}{d y}+P x=Q$
So, I.F. $=e^{\int P d y}=e^{\int \frac{1}{1+y^2} \cdot d y}=e^{\tan ^{-1} y}$.
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