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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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$\large\frac{e^x}{e^{2x}+1}$$dx=\int \large\frac{dy}{y}$
$u=e^x$
$du=e^x dx$
$\int \large\frac{du}{u^2+1}=\int \large\frac{dy}{y}$
$\tan^{-1}u+c =\log y$
Using condition (0,1)
$y= e^{\tan ^{-1} (e^x) -\pi/4}$
Hence d is the correct answer.
answered Feb 6, 2014 by meena.p
 

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