Mathematics - Differential Equation Question with Solution | TestHub

The given differential equation can be written as $\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}\dots$ (i) Now, eq. (i) is a linear differential equation of the form $\frac{dx}{dy}+{\mathrm{P}}_{1}x={\mathrm{Q}}_1$ where ${\mathrm{P}}_1=\frac{1}{1+y^2}$ and ${\mathrm{Q}}_1=\frac{{\tan }^{-1}y}{1+y^2}$ 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 $x{e}^{{\tan }^{-1}y}=\int \left(\frac{{\tan }^{-1}y}{1+y^2}\right)e^{{\tan }^{-1}y}\mathrm{dy}+\mathrm{C}\dots$ (ii) Let $\mathrm{I}=\int \left(\frac{{\tan }^{-1}y}{1+y^2}\right)e^{{\tan }^{-1}y}\mathrm{dy}$ On substituting ${\tan }^{-1}y=t$, so that $\left(\frac{1}{1+y^2}\right)\mathrm{dy}=\mathrm{dt}$, 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 $\mathrm{I}=e^{{\tan }^{-1}y}\left({\tan }^{-1}y-1\right)$ On substituting the value of 1 in equation (i1), we get $x\cdot e^{{\tan }^{-1}y}=\cdot e^{{\tan }^{-1}y}\left({\tan }^{-1}y-1\right)+\mathrm{C}$ or $x=\left({\tan }^{-1}y-1\right)+\mathrm{C}{e}^{{\tan }^{-1}y}$ which is the general solution of the given differntial equation.