Question

Solution of the differential equation $\cos x \: dy = \: y \: (\sin x - y) dx, 0 < x < \frac {\pi} {2}$ is

• A. $y\, \tan x = \sec x + c$
• B. $\tan x \,=\, (\sec x + c) y$
• C. $\sec x\, =\, (\tan x + c)y$
• D. $y \sec x\, =\, \tan x + c.$

Answer 1

Answer: (C)

$cos\, x\,dy= y (sin\,x -y) \,dx$
$\Rightarrow cos\,x \frac{dy}{dx} = y\,sin\,x-y^{2}$
$\Rightarrow \frac{dy}{dx} = y \, tanx-y^{2}\,sec\,x$
$\Rightarrow \frac{dy}{dx} - y\,tan\,x= -y^{2}\,sec\,x$
$\Rightarrow y^{-2} \frac{dy}{dx}-y^{-1}\,tan\,x = -sec\,x$
Put $y^{-1}= z$
$\therefore -y^{-2}\, \frac{dy}{dx} = \frac{dz}{dx}$
$\therefore -\frac{dz}{dx} - z\,tan\,x=-sec\,x$
which is linear in $z$
$I. F. = e^{\int\,tan\,xdx} = e^{log\,secx} = sec\,x$.
$\therefore $ sol is. $z \, sec\,x=\int\,sec^{2}\,x\,dx+C=tan\,x+C$
$\Rightarrow \frac{sec\,x}{y} = tan\,x+C$
$\Rightarrow sec\,x = \left(tan\,x+C\right)\,y$