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Solve : $y'= 1+y^2$

$(a)\;y= \tan ^2 (x+c) \\ (b)\;y=\tan ^2 x+c \\ (c)\;y= \tan x+c \\ (d)\;y= \tan (x+c) $
Can you answer this question?
 
 

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$\large\frac{dy}{1+y^2}$$=dx$
$\tan ^{-1} y =x+c$
$y= \tan (x+c)$
Hence d is the correct answer.
answered Feb 7, 2014 by meena.p
 

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