Question

Find the particular solution of the differential equation `(1+y^2)+(x-e^(tan-1 )y)dy/dx=` given that y = 0 when x = 1.

 

Answer 1

We have

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

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

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

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

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

\[\text { Clearly, it is a linear differential equation of the form } \]

\[\frac{dx}{dy} + Px = Q\]

\[\text { where }, \]

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

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

\[ \therefore I . F . = e^\int P dy \]

\[ = e^\int\frac{1}{1 + y^2} dy \]

\[ = e^{\tan^{- 1}} y \]

\[\text { Multiplying both sides of } \left( 1 \right) by e^{\tan^{- 1}
}y , we get\]

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

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

\[\text { Integrating both sides with respect to y, we get }\]

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

\[ \Rightarrow x e^{\tan^{- 1} } y = I + C . . . . . \left( 2 \right)\]

\[\text { Here }, \]

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

\[\text { Putting } \tan^{- 1} y = t, \text { we get }\]

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

\[ \therefore I = \int e^{2t} dt\]

\[ = \frac{e^{2t}}{2}\]

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

\[\text { Putting the value of I in } \left( 2 \right), \text { we get }\]

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

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

\[ \Rightarrow 2x e^{\tan^{- 1} } y = e^{2 \tan^{- 1} y} + k \left( \text { where }k = 2C \right)\]

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

\[2x e^{\tan^{- 1}} y = e^{2 \tan^{- 1} y} + k \]

\[\text { To fiind the particular solution we have to put the values of x and y as 1 and 0 respectively } . \]

\[2 e^{\tan^{- 1} 0} = e^{2 \tan^{- 1} 0} + k\]

\[ \Rightarrow 2 = 1 + k\]

\[ \Rightarrow k = 1\]

\[\text{ So, the particular solution is } 2x e^{\tan^{- 1}} y = e^{2 \tan^{- 1} y} + 1 . \]