Question

The solution of the differential equation $\left(1+x^2\right)\frac{dy}{dx}+1+y^2=0$, is
(a) tan⁻¹ x − tan⁻¹ y = tan⁻¹ C
(b) tan⁻¹ y − tan⁻¹ x = tan⁻¹ C
(c) tan⁻¹ y ± tan⁻¹ x = tan C
(d) tan⁻¹ y + tan⁻¹ x = tan⁻¹ C

Answer 1

(d) tan⁻¹y + tan⁻¹x = tan⁻¹C

We have,
$$\begin{gathered} \left(1+x^2\right)\frac{dy}{dx}+1+y^2=0 \\ \Rightarrow \left(1+x^2\right)\frac{dy}{dx}=-\left(1+y^2\right) \\ \Rightarrow \frac{1}{\left(1+y^2\right)}dy=-\frac{1}{\left(1+x^2\right)}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \int \frac{1}{\left(1+y^2\right)}dy=-\int \frac{1}{\left(1+x^2\right)}dx \\ \Rightarrow {\tan }^{-1}y=-{\tan }^{-1}x+{\tan }^{-1}C \\ \Rightarrow {\tan }^{-1}y+{\tan }^{-1}x={\tan }^{-1}C \end{gathered}$$