Question

The general solution of the differential equation $y-\frac{dy}{dx}x=y^2\cos x(1-\sin x)$ is

• A. $y(\sin x+C)=\tan x+\sec x$
• B. $y(\sin x+C)=\tan x-\sec x$
• C. $\frac{y}{x}=\sin x+\frac{1}{4}\cos 2x+C$
• D. $\frac{x}{y}=\sin x+\frac{1}{4}\cos 2x+C$

Answer 1

Answer: (D)

$y-\frac{dy}{dx}\cdot x=y^2\cos x(1-\sin x)$
$\Rightarrow \frac{ydx-xdy}{y^2}=\cos x(1-\sin x)dx$
$\Rightarrow \int d\left(\frac{x}{y}\right)=\int \left(\cos x-\frac{\sin 2x}{2}\right)dx$
$\Rightarrow \frac{x}{y}=\sin x+\frac{\cos 2x}{4}+\text{ C }$