BITSAT 2021 Paper

The given differential equation can be written as $\frac{dx}{dy} + \frac{x}{1+y^2}$ $= \frac{\tan^{-1} y}{1+y^2}$ ... (i) Now, eq. (i) 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, I.F $= e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$ Thus, the solution of the given differential equation is given by $x e^{\tan^{-1} y}$ $= \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + C$ ...(ii) Let $I$ $= \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy$ On 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 $I = e^{\tan^{-1} y} (\tan^{-1} y - 1)$ On substituting the value of 1 in equation (i1), 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 differntial equation.