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

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