Question

Solve the differential equation $\left(1+x^2\right)\frac{dy}{dx}+\left(1+y^2\right)=0$, given that y = 1, when x = 0.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1+x^2\right)\frac{dy}{dx}+\left(1+y^2\right)=0,y=1\mathrm{when}x=0 \\ \Rightarrow \left(1+x^2\right)\frac{dy}{dx}=-\left(1+y^2\right) \\ \Rightarrow \frac{1}{1+y^2}dy=-\frac{1}{\left(1+x^2\right)}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{1+y^2}dy=-\int \frac{1}{\left(1+x^2\right)}dx \\ \Rightarrow {\tan }^{-1}y=-{\tan }^{-1}x+C \\ \Rightarrow {\tan }^{-1}y+{\tan }^{-1}x=C.....\left(1\right) \\ \mathrm{Given}:x=0,y=1. \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{\mathrm{\pi }}{4}+0=\mathrm{C} \\ \Rightarrow \mathrm{C}=\frac{\mathrm{\pi }}{4} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ {\tan }^{-1}y+{\tan }^{-1}x=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}x+{\tan }^{-1}y=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{x+y}{1-xy}=1 \\ \Rightarrow x+y=1-xy \\ \mathrm{Hence},x+y=1-xy\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$