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Solution of the general equation : $\large\frac{dy}{dx}$$=\cos (x+y) - \sin (x+y) $

$(a)\;\log (1- \tan (\frac{x+y}{2}))+x=c \\ (b)\;\log (1+\tan (\frac{x+y}{2}))+x=c \\ (c)\;\log (1+\tan (\frac{x+y}{2}))-x=c \\ (d)\;\log (1- \tan (\frac{x+y}{2}))-x=c $
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$\large\frac{dy}{dx}$$=\cos (x+y) -\sin (x+y)$
$x+y=t$
$1+\large\frac{dy}{dx}=\frac{dt}{dx}$
$\large\frac{dt}{dx}$$-1=\cos t -\sin t$
$\large\frac{dt}{dx}$$=\cos t -\sin t+1$
$\large\frac{dt}{\cos t - \sin t +1} $$=dt$
$\large\frac{dt}{\Large\frac{1- \tan ^2 t/2}{1+\tan ^2 t/2}- \Large\frac{2 \tan t/2}{1+\tan ^2 t/2}+1}$$=dx$
$\large\frac{\sec^2 t/2 dt}{1-\tan ^2 t/2 -2 \tan t/2 +1 \tan ^2 t/2}$$dx$
$\int \large\frac{\sec^2 t/2 dt}{2( 1- \tan t/2)}$$=\int dx$
$1- \tan t/2 =m$
$-\large\frac{1}{2} $$\sec^2 \frac{t}{2} dt =dm$
$\int \large\frac{-dm}{m}$$=\int dx$
$-\log m =x+c$
$-\log (1- \tan t/2) =x+c$
$\log \bigg(1- \tan \bigg(\Large\frac{x+4}{2} \bigg) \bigg) $$=-x-c$
$\log (1- \tan (\frac{x+y}{2}))+x=c$
Hence a is the correct answer.
answered Feb 3, 2014 by meena.p
 

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