Question

\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]

Answer 1

We have,

\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]

\[\Rightarrow \frac{dx}{dy} = \frac{e^{- \tan^{- 1} y} - x}{1 + y^2}\]

\[ \Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{- \tan^{- 1} y}}{1 + y^2}\]

\[\text{Comparing with }\frac{dx}{dy} + Px = Q,\text{ we get}\]

\[P = \frac{1}{1 + y^2} \]

\[Q = \frac{e^{- \tan^{- 1} y}}{1 + y^2}\]

\[Now, \]

\[I . F . = e^{\int\frac{1}{1 + y^2}dy} = e^{\tan^{- 1} y} \]

So, the solution is given by

\[x \times e^{\tan^{- 1} y} = \int\frac{e^{- \tan^{- 1} y}}{1 + y^2} \times e^{\tan^{- 1} y} dy + C\]

\[ \Rightarrow x \times e^{\tan^{- 1} y} = \int\frac{1}{1 + y^2} dy + C\]

\[ \Rightarrow x e^{\tan^{- 1} y} = \tan^{- 1} y + C\]