Solve the differential equation (tan^-1 y - x) dy = (1 + y^2) dx.

The correct answer is Option (1) → $x = (\tan^{-1} y - 1) + Ce^{-\tan^{-1} y}$ ##

The given differential equation can be written as

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

Now (1) is a linear differential equation of the form $\frac{dx}{dy} + P_1 x = Q_1$,

where, $P_1 = \frac{1}{1 + y^2}$ and $Q_1 = \frac{\tan^{-1} y}{1 + y^2}$.

Therefore, $\text{I.F.} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y}$

Thus, the solution of the given differential equation is

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

Let $I = \int \left( \frac{\tan^{-1} y}{1 + y^2} \right) e^{\tan^{-1} y} dy$

Substituting $\tan^{-1} y = t$ so that $\left( \frac{1}{1 + y^2} \right) dy = dt$, we get

$I = \int t e^t dt = t e^t - \int 1 \cdot e^t dt = t e^t - e^t = e^t (t - 1)$

or $I = e^{\tan^{-1} y} (\tan^{-1} y - 1)$

Substituting the value of $I$ in equation (2), we get

$x \cdot e^{\tan^{-1} y} = e^{\tan^{-1} y} (\tan^{-1} y - 1) + C$

or $x = (\tan^{-1} y - 1) + C e^{-\tan^{-1} y}$

which is the general solution of the given differential equation.