Question

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

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

Answer 1

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

$\frac{dy}{dx}-\frac{1}{x}.y=\frac{1}{x^2}+\frac{1}{x^2}\cos \frac{y}{x}....\left(i\right)$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore$ Eq. (i) becomes
$v=x\frac{dv}{dx}-v=\frac{1}{x^2}+\frac{1}{x^2}\cos v$
$\Rightarrow \frac{dv}{1+\cos v}=\frac{dx}{x^3}\Rightarrow \frac{1}{2}{\sec }^2\frac{v}{2}dv=\frac{-x^{-2}}{2}+C$
$\Rightarrow \tan \frac{v}{2}=-\frac{1}{2x^2}+C$
$\Rightarrow \tan \frac{y}{2x}=C-\frac{1}{2x^2}$.