Question

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

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

Answer 1

Answer: (D)

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