Question

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

Answer 1

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