Question

The solution of the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\], is

पर्याय

• tan⁻¹ x − tan⁻¹ y = tan⁻¹ C

• tan⁻¹ y − tan⁻¹ x = tan⁻¹ C

• tan⁻¹ y ± tan⁻¹ x = tan C

• tan⁻¹ y + tan⁻¹ x = tan⁻¹ C

Answer 1

tan⁻¹y + tan⁻¹x = tan⁻¹C

 

We have,

\[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\]

\[ \Rightarrow \left( 1 + x^2 \right)\frac{dy}{dx} = - \left( 1 + y^2 \right)\]

\[ \Rightarrow \frac{1}{\left( 1 + y^2 \right)}dy = - \frac{1}{\left( 1 + x^2 \right)}dx\]

Integrating both sides we get,

\[\int\frac{1}{\left( 1 + y^2 \right)}dy = - \int\frac{1}{\left( 1 + x^2 \right)}dx\]

\[ \Rightarrow \tan^{- 1} y = - \tan^{- 1} x + \tan^{- 1} C\]

\[ \Rightarrow \tan^{- 1} y + \tan^{- 1} x = \tan^{- 1} C\]