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Mathematics · JEE | DIFFERENTIAL EQUATIONS
हिन्दीThe general solution of the linear differential equation \( (\tan^{-1} y - x) \, dy = (1 + y^2) \, dx \) is:
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Correct answer · Option C
\( x = (\tan^{-1} y - 1) + C e^{-\tan^{-1} y} \)
Explanation
Rewriting in standard form \( \frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2} \), the integrating factor is \( e^{\tan^{-1}y} \), leading to the solution \( x = (\tan^{-1}y - 1) + C e^{-\tan^{-1}y} \).
