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