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

Answer 1

The correct option is C $\tan \frac{y}{2x}=c-\frac{1}{{2x}^2}$

We have, $x^2\frac{dy}{dx}-xy=1+\cos \frac{y}{x}=2{\cos }^2\frac{y}{2x}$

$\Rightarrow {\sec }^2\left(\frac{y}{2x}\right).[x^2\frac{dy}{dx}-xy]=2$

$\Rightarrow \frac{1}{2}{\sec }^2\left(\frac{y}{2x}\right).\frac{x\frac{dy}{dx}-y}{x^2}=\frac{1}{x^3}$

$\Rightarrow \frac{d}{dx}\left(\tan \frac{y}{2x}\right)=\frac{1}{x^3}$

Integrating, we get

$\Rightarrow \int \frac{d}{dx}\left(\tan \frac{y}{2x}\right)dx=\int \frac{1}{x^3}dx$

$⟹\tan \frac{y}{2x}=c-\frac{1}{{2x}^2}$, which is required solution.