Question

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

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

Answer 1

Answer: (D)

$secx=(\tan x+c)y$

$\cos xdy$ $=y(\sin x-y)$ dx

$\Rightarrow \frac{dy}{dx}=y\tan x-y^2$ $\sec x$

$\Rightarrow \frac{1dy}{y^{2}dx}-\frac{1}{y}$ tanx $=-\sec x$

Let $\frac{-1}{y}=t\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dt}{dx}+t\left(tanx\right)=-\sec x=\frac{dt}{dx}+\left(\tan x\right)t=\sec x.$

$I.F=e^{\int \tan xdx}=\sec x$

Solution is $t(I.F)=\int (I.F)$ secx dx
$\frac{1}{y}$ secx $=$ tanx $+c$
Hence, option 'D' is correct.