Question

Solve the following initial value problem:-

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

Answer 1

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