Question

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

• A. $\sec x=\left(\tan x+c\right)y$
• B. $y\sec x=\tan x+c$
• C. $y\mathrm{tanx}=secx+c$
• D. $\tan x=\left(\sec x+c\right)y$

Answer 1

The correct option is A $\sec x=\left(\tan x+c\right)y$

$\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\tan x=-secx$

Let $-\frac{1}{y}=t$

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

$\Rightarrow \frac{dt}{dx}+t\tan x=-secx$

$\Rightarrow t.e^{\int \tan xdx}=\int e^{\int \tan xdx}.-secxdx$

$\Rightarrow t.e^{ln\left(secx\right)}=\int e^{ln\left(secx\right)}.-secxdx$

$\Rightarrow -\frac{1}{y}secx=\int -se{c}^{2}xdx$

$\Rightarrow -\frac{1}{y}secx=-\tan x-c$

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