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Q. The solution of the differential equation $e ^{-x} dy (y +1) dy + (Cos^2 x - Sin 2x)y (dx) = 0$ subjected to the condition that $y = 1$ when $x = 0$ is

KCETKCET 2006Differential Equations

Solution:

We have
$e^{-x}(y+1) d y+\left(\cos ^{2} x-\sin 2 x\right) y d x=0 $
$\Rightarrow \frac{(y+1)}{y} d y=-e^{x}\left(\cos ^{2} x-\sin 2 x\right) d x $
$\Rightarrow \left(1+\frac{1}{y}\right) d y=-e^{x}\left(\cos ^{2} x-\sin 2 x\right) d x$
On integrating both sides, we get
$y+\log y=-e^{x} \cos ^{2} x+\int e^{x} \sin 2 x d x $
$-\int e^{x} \sin 2 x d x+c $
$\Rightarrow y+\log y=-e^{x} \cos ^{2} x+c$
At $ x=0, y=1$
$\Rightarrow 1+0=-e^{0} \cos 0+c $
$\Rightarrow c=2$
$\therefore $ Required solution is
$y+\log y=-e^{x} \cos ^{2} x+2$