Question

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

$cosxdy=y(sinx-y)dx$
$\Rightarrow cosx\frac{dy}{dx}=ysinx-y^2$
$\Rightarrow \frac{dy}{dx}=ytanx-y^{2}secx$
$\Rightarrow \frac{dy}{dx}-ytanx=-y^{2}secx$
$\Rightarrow y^{-2}\frac{dy}{dx}-y^{-1}tanx=-secx$
Put $y^{-1}=z$
$\therefore -y^{-2}\frac{dy}{dx}=\frac{dz}{dx}$
$\therefore -\frac{dz}{dx}-ztanx=-secx$
which is linear in $z$
$I.F.=e^{\int tanxdx}=e^{logsecx}=secx$.
$\therefore$ sol is. $zsecx=\int se{c}^{2}xdx+C=tanx+C$
$\Rightarrow \frac{secx}{y}=tanx+C$
$\Rightarrow secx=\left(tanx+C\right)y$