Question

Find one-parameter families of solution curves of the following differential equation:-

\[\frac{dy}{dx} - \frac{2xy}{1 + x^2} = x^2 + 2\]

Solve the following differential equation:-

\[\frac{dy}{dx} - \frac{2xy}{1 + x^2} = x^2 + 2\]

Answer 1

We have, 
\[\frac{dy}{dx} - \frac{2xy}{1 + x^2} = x^2 + 2 . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = - \frac{2x}{1 + x^2} \]
\[Q = x^2 + 2\]
\[ \therefore I.F. = e^{\int P\ dx} \]
\[ = e^{- \int\frac{2x}{1 + x^2} dx} \]
\[ = e^{- \log\left| 1 + x^2 \right|} \]
\[ = \frac{1}{1 + x^2}\]
\[\text{ Multiplying both sides of }\left( 1 \right) \text{ by }\frac{1}{1 + x^2},\text{ we get }\]
\[\frac{1}{1 + x^2} \left( \frac{dy}{dx} - \frac{2xy}{1 + x^2} \right) = \frac{1}{1 + x^2}\left( x^2 + 2 \right)\]
\[ \Rightarrow \frac{1}{1 + x^2}\frac{dy}{dx} - \frac{2xy}{\left( 1 + x^2 \right)^2} = \frac{x^2 + 2}{x^2 + 1}\]
Integrating both sides with respect to x, we get
\[\frac{1}{1 + x^2}y = \int\frac{x^2 + 2}{x^2 + 1} dx + C\]
\[ \Rightarrow \frac{1}{1 + x^2}y = \int\frac{x^2 + 1 + 1}{x^2 + 1} dx + C\]
\[ \Rightarrow \frac{1}{1 + x^2}y = \int dx + \int\frac{1}{x^2 + 1} dx + C\]
\[ \Rightarrow \frac{1}{1 + x^2}y = x + \tan^{- 1} x + C\]
\[ \Rightarrow y = \left( 1 + x^2 \right)\left( x + \tan^{- 1} x + C \right)\]
\[\text{ Hence, }y = \left( 1 + x^2 \right)\left( x + \tan^{- 1} x + C \right)\text{ is the required solution.}\]