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Differential Equations

TG EAPCET 2025 (Online) 4th May Evening Shift

MCQ+1 / -02025

If the general solution of $\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$ is $x=f(y)+c e^{-\tan ^{-1} y}$, then $f(y)=$


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