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Solve : $y' +\tan x.y =\tan x \quad y(\large\frac{\pi}{4})$$=1$

$(a)\;y= e^{\sec ^2 x}\frac{\tan ^2x}{2}+c \\ (b)\;y= e^{\tan ^2 x/2}+c \\ (c)\;y= e^{\sec ^2 x}\frac{\tan ^2x}{4}+c \\ (d)\;y= e^{-\sec ^2 x}\frac{\tan ^2x}{2}+c $

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It is linear equation so
$I_0 f=e^{\int \tan xdx}$
$\qquad= e^{\sec^2 x dx}$
So, $y. e^{\sec^2 x}=\int \tan x . \sec^2 x dx$
$\tan x =t$
$\sec^2 x dx =dt$
$y. e^{\sec^2 x}=\int t dt$
$y. e^{\sec^2 x} =\large\frac{\tan ^2 x}{2}$$+c$
Hence d is the correct answer.
answered Feb 6, 2014 by meena.p
 

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