Question

The solution of the differential equation $x^2\frac{dy}{dx}-xy=1+\cos \frac{y}{x}$ is

• A. $\cos \frac{y}{x}=1+\frac{c}{x}$
• B. $x^2=\left(c+x^2\right)\tan \frac{y}{x}$
• C. $\tan \frac{y}{2x}=c-\frac{1}{2x^2}$
• D. $\tan \frac{y}{x}=c+\frac{1}{x}$

Answer 1

Answer: (C)

We have, $x^2\frac{dy}{dx}-xy=1+\cos \frac{y}{x}=2{\cos }^2\frac{y}{2x}.$
$\Rightarrow {\sec }^2\left(\frac{y}{2x}\right)\cdot \left[x^2\frac{dy}{dx}-xy\right]=2$
$\Rightarrow \frac{1}{2}{\sec }^2\left(\frac{y}{2x}\right)\cdot \frac{x\frac{dy}{dx}-y}{x^2}=\frac{1}{x^3}$
$\Rightarrow \frac{d}{dx}\left(\tan \frac{y}{2x}\right)=\frac{1}{x^3}$
Integrating, we get $\tan \frac{y}{2x}=c-\frac{1}{2x^2}$,
which is the required solution,