Question

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

Answer 1

We have,
\[\left( 1 + y^2 \right) + \left( x - e^{tan^{- 1} y} \right)\frac{dy}{dx} = 0\]
\[ \Rightarrow \left( x - e^{tan^{- 1} y} \right)\frac{dy}{dx} = - \left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{\left( 1 + y^2 \right)}{\left( x - e^{tan^{- 1} y} \right)}\]
\[ \Rightarrow \frac{dx}{dy} = - \frac{x - e^{tan^{- 1} y}}{1 + y^2}\]
\[ \Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{tan^{- 1} y}}{1 + y^2} . . . . . \left(1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dx}{dy} + Px = Q\]
where
\[P = \frac{1}{1 + y^2}\]
\[Q = \frac{e^{tan^{- 1} y}}{1 + y^2} \]
\[ \therefore I.F. = e^{\int P\ dy} \]
\[ = e^{\int\frac{1}{1 + y^2} dy} \]
\[ = e^{tan^{- 1} y} \]
\[\text{Multiplying both sides of }\left( 1 \right)\text{ by }e^{tan^{- 1} y} ,\text{ we get }\]
\[ e^{tan^{- 1} y} \left( \frac{dx}{dy} + \frac{x}{1 + y^2} \right) = e^{tan^{- 1} y} \frac{e^{tan^{- 1} y}}{1 + y^2}\]
\[ \Rightarrow e^{tan^{- 1} y} \frac{dx}{dy} + \frac{x\ e^{tan^{- 1} x}}{1 + y^2} = \frac{e^{2 \tan^{- 1} y}}{1 + y^2}\]
Integrating both sides with respect to y, we get
\[ x\ e^{tan^{- 1} y} = \int\frac{e^{2 \tan^{- 1} y}}{1 + y^2} dy + C\]
\[ \Rightarrow x e^{tan^{- 1} y} = I + C . . . . .. . . . . \left( 2 \right)\]
Here, 
\[I = \int\frac{e^{2 \tan^{- 1} y}}{1 + y^2} dy\]
\[\text{Putting }\tan^{- 1} y = t,\text{ we get }\]
\[\frac{1}{1 + y^2}dy = dt\]
\[ \therefore I = \int e^{2t} dt\]
\[ = \frac{e^{2t}}{2}\]
\[ = \frac{e^{2 \tan^{- 1} y}}{2}\]
\[\text{Putting the value of I in }\left( 2 \right),\text{ we get }\]
\[x\ e^{tan^{- 1} y} = \frac{e^{2 \tan^{- 1} y}}{2} + C\]
\[ \Rightarrow 2x\ e^{tan^{- 1} y} = e^{2 \tan^{- 1} y} + 2C\]
\[ \Rightarrow 2x\ e^{tan^{- 1} y} = e^{2 \tan^{- 1} y} + k ...............\left( \text{where }k = 2C \right)\]
\[\text{ Hence, }2x\ e^{tan^{- 1} y} = e^{2 \tan^{- 1} y} + k\text{ is the required solution.} \]