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. $x^2=(C+x^2)\tan \frac{y}{x}$

Answer 1

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

$x^2\frac{dy}{dx}-xy=1+\cos \frac{y}{x}$
$\frac{dy}{dx}-\frac{y}{x}=\frac{1+\cos \frac{y}{x}}{x^2}$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}-v=\frac{1+\cos v}{x^2}$
$\Rightarrow \int \frac{dv}{1+\cos v}=\int \frac{1}{x^3}dx$
$\Rightarrow \frac{1}{2}\int \frac{dv}{{\cos }^2\frac{v}{2}}=\int \frac{1}{x^3}dx$
$\Rightarrow \frac{1}{2}\int {\sec }^2\frac{v}{2}\cdot dv=\int \frac{1}{x^3}dx$
$\Rightarrow \frac{1}{2}\cdot \tan \frac{v}{2}\cdot 2=\frac{x^{-3+1}}{-3+1}+C$
$\Rightarrow \tan \frac{y}{2x}=C-\frac{1}{2x^2}$