Solve the differential equation $\;$ $\left(1 + y^2\right) \; dx = \left(\tan^{-1}\left(y\right) - x\right) \; dy$
The given differential equation is $\;$ $\left(1 + y^2\right) \; dx = \left(\tan^{-1}\left(y\right) - x\right) \; dy$
i.e. $\dfrac{dx}{dy} = \dfrac{\tan^{-1}\left(y\right) - x}{1 + y^2}$
i.e. $\dfrac{dx}{dy} + \dfrac{x}{1 + y^2} = \dfrac{\tan^{-1} \left(y\right)}{1 + y^2}$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dx}{dy} + P \left(y\right)x = Q \left(y\right)$
Here $P \left(y\right) = \dfrac{1}{1 + y^2}$ $\;\;\; \cdots \; (2a)$; $\;$ $Q \left(y\right) = \dfrac{\tan^{-1} \left(y\right)}{1 + y^2}$ $\;\;\; \cdots \; (2b)$
Integrating Factor (I.F) $= e^{\int P \left(y\right) \; dy}$
i.e. $I.F = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1}\left(y\right)}$ $\;\;\; \cdots \; (3)$
According to the general solution,
$x \; e^{\int P \left(y\right) \; dy} = \displaystyle \int Q \left(y\right) \; e^{\int P \left(y\right) \; dy} \; dy$
$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is
$x \; e^{\tan^{-1} \left(y\right)} = \displaystyle \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy$ $\;\;\; \cdots \; (4)$
Consider $\displaystyle \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy$ $\;\;\; \cdots \; (5)$
Let $\tan^{-1} \left(y\right) = p$ $\;\;\; \cdots \; (6a)$
Differentiating equation $(6a)$ gives
$\dfrac{dy}{1 + y^2} = dp$ $\;\;\; \cdots \; (6b)$
$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(5)$ becomes
$\begin{aligned}
\int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy & = \int p \; e^{p} \; dp \\\\
& = p \int e^{p} \; dp - \int \left\{\int e^{p} \; dp \times \dfrac{d}{dp} \left(p\right) \right\} \; dp \\\\
& = p \; e^{p} - \int e^{p}\; dp \\\\
& = p \; e^{p} - e^{p} + c \\\\
& = e^{p} \left(p - 1\right) + c \\\\
& = e^{\tan{-1} \left(y\right)} \left(\tan^{-1} \left(y\right) - 1\right) + c \;\;\; \cdots \; (7) \\
& \hspace{8em} \left[\text{In view of equation } (6a)\right]
\end{aligned}$
$\therefore$ $\;$ We have from equations $(4)$ and $(7)$,
$x \; e^{\tan^{-1} \left(y\right)} = e^{\tan^{-1} \left(y\right)} \left(\tan^{-1} \left(y\right) - 1\right) + c$ $\;\;\; \cdots \; (8)$
Equation $(8)$ is the general solution of the given differential equation.