Question

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

Answer 1

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