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प्रश्न
\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]
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उत्तर
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
\[x \times e^{\tan^{- 1} y} = \int\frac{e^{- \tan^{- 1} y}}{1 + y^2} \times e^{\tan^{- 1} y} dy + C\]
\[ \Rightarrow x \times e^{\tan^{- 1} y} = \int\frac{1}{1 + y^2} dy + C\]
\[ \Rightarrow x e^{\tan^{- 1} y} = \tan^{- 1} y + C\]
