Question

The solution of $\frac{dy}{dx}=\cos x(2-ycosecx)$, where $y=2$, when $x=\frac{\pi }{2}$ is?

• A. $y=\sin x+cosecx$
• B. $y=\tan \frac{x}{2}+\cot \frac{x}{2}$
• C. $y=\frac{1}{\sqrt{2}}\sec \frac{x}{2}+\sqrt{2}\cos \frac{x}{2}$
• D. None of the above

Answer 1

The correct option is A $y=\sin x+cosecx$

Given, $\frac{dy}{dx}=\cos x(2-ycosecx)$
$=\frac{dy}{dx}=2\cos x-y\cot x$
$\Rightarrow \frac{dy}{dx}+y\cot x=2\cos x$
$IF=e^{\int \cot xdx}=\sin x$
$y\sin x=\int 2\cos x\cdot \sin xdx+X$
$y\sin x={\sin }^{2}x+C$ ...........(i)
at $y=2$ and $x=\frac{\pi }{2}$
$\Rightarrow C=1$
$\therefore y\sin x={\sin }^{2}x+1,$ [from Eq. (i)]
$\Rightarrow y=\sin x+cosecx$.