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JEEMAIN and NEET
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JEEMAIN PAST PAPERS
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2003
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Q)
The solution of the differential equation $(1+y^2) + (x - e^{\tan^{-1} y} )\frac{dy}{dx} = 0 $ is :
( A ) $ x e^{\tan^{-1} y} = \tan^{-1} y +k$
( B ) $(x-2) = ke^{-\tan^{-1} y}$
( C ) $ x e^{2\tan^{-1} y} = e^{\tan^{-1} y} +k$
( D ) $2 x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} +k$
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