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Solve : $e^x y dx=(e^{2x}+1) dy \quad y \cos =1$

$(a)\;y=e^{\tan ^{-1}x} \\ (b)\;y= e^{\tan ^{-1}(1/x)-\pi/4} \\ (c)\;y=e^{\tan ^{-1} (1/x)} \\ (d)\;y= e^{\tan ^{-1}ex -\pi/4} $

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