Question

The solution of the differential equation (x² + 1) \[\frac{dy}{dx}\] + (y² + 1) = 0, is

पर्याय

• y = 2 + x²

• \[y = \frac{1 + x}{1 - x}\]

• y = x (x − 1)

• \[y = \frac{1 - x}{1 + x}\]

Answer 1

\[y = \frac{1 - x}{1 + x}\]

 

We have,

\[\left( x^2 + 1 \right)\frac{dy}{dx} + ( y^2 + 1) = 0\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{dy}{dx} = - \left( y^2 + 1 \right)\]
\[ \Rightarrow \frac{1}{\left( y^2 + 1 \right)}dy = - \frac{1}{\left( x^2 + 1 \right)}dx\]
Integrating both sides, we get
\[\int\frac{1}{\left( y^2 + 1 \right)}dy = - \int\frac{1}{\left( x^2 + 1 \right)}dx\]
\[ \Rightarrow \tan^{- 1} y = - \tan^{- 1} x + \tan^{- 1} C\]
\[ \Rightarrow \tan^{- 1} y + \tan^{- 1} x = \tan^{- 1} C\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x + y}{1 - xy} \right) = \tan^{- 1} C\]
\[ \Rightarrow \frac{x + y}{1 - xy} = C\]
\[ \Rightarrow x + y = 1 - xy\]
\[ \Rightarrow y + xy = 1 - x\]
\[ \Rightarrow y\left( 1 + x \right) = 1 - x\]
\[ \Rightarrow y = \frac{1 - x}{1 + x}\]