Question

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

Answer 1

We have,
\[\left( 1 + x^2 \right)\frac{dy}{dx} - x = 2 \tan^{- 1} x\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{dy}{dx} = x + 2 \tan^{- 1} x\]
\[ \Rightarrow dy = \left\{ \frac{x}{1 + x^2} + \left( \frac{2}{1 + x^2} \right) \tan^{- 1} x \right\}dx\]
\[ \Rightarrow dy = \left\{ \frac{1}{2} \times \frac{2x}{1 + x^2} + \left( \frac{2}{1 + x^2} \right) \tan^{- 1} x \right\}dx\]
Integrating both sides, we get
\[\int dy = \int\left\{ \frac{1}{2} \times \frac{2x}{1 + x^2} + \left( \frac{2}{1 + x^2} \right) \tan^{- 1} x \right\}dx\]
\[ \Rightarrow y = \frac{1}{2}\int\frac{2x}{1 + x^2}dx + 2\int\left[ \frac{1}{1 + x^2} \tan^{- 1} x \right] dx\]
\[ \Rightarrow y = \frac{1}{2}\log\left| 1 + x^2 \right| + 2\int\left[ \frac{1}{1 + x^2} \tan^{- 1} x \right] dx\]
\[\text{ Putting }\tan^{- 1} x = t\]
\[ \Rightarrow \frac{1}{1 + x^2}dx = dt\]
\[ \therefore y = \frac{1}{2}\log\left| 1 + x^2 \right| + 2\int t dt\]
\[ = \frac{1}{2}\log\left| 1 + x^2 \right| + t^2 + C\]
\[ = \frac{1}{2}\log\left| 1 + x^2 \right| + \left( \tan^{- 1} x \right)^2 + C\]
\[\text{ Hence, }y = \frac{1}{2}\log\left| 1 + x^2 \right| + \left( \tan^{- 1} x \right)^2 +\text{C is the solution to the given differential equation.}\]