Question

The solution of the differential equation ${\left(\frac{dy}{dx}\right)}^2+2y\cos x\frac{dy}{dx}-y^2=0$ is

• A. $y=-k(1+\cos x)$
• B. $y=-k(1-\cos x)$
• C. $y=k(1-\cos x)$
• D. None of these

Answer 1

Answer: (B)

$y=-k(1-\cos x)$

From the given equation treating as a quadratic in $\frac{dy}{dx},$ we have
$\frac{dy}{dx}=\frac{-2y\cot x\pm \sqrt{4y^2+{\cot }^{2}x+4y^2}}{2}$
$=y[-\cot x\pm cosecx]$ $\frac{dy}{dx}=y[-\cot x+cosecx]$ or $-y(\cot x+cosecc)$
$\Rightarrow \frac{dy}{y}=[cosecx-\cot x]dx$ or $-(cosecx+\cot x)dx$

Integrating,$\log y=\log \tan \frac{x}{2}+\log \sin x\log k$ or $-[\log \tan \frac{x}{2}+\log \sin x]+\log k$
$\log \frac{y}{k}=\log \frac{\tan \frac{x}{2}}{\sin x}=\log \frac{\tan \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\log \frac{1}{2{\cos }^2\frac{x}{2}}$
$\therefore y=\frac{k}{1+\cos x}$
The of other result gives $\frac{y}{k}=-\sin x\tan \frac{x}{2}=-2\sin \frac{x}{2}\cos \frac{x}{2}.\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
or $\frac{y}{k}=-2{\sin }^2\frac{x}{2}=\frac{x}{2}-(1-\cos x)$ or $y=-k(1-\cos x)$