Question

The general solution of the differential equation $(x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right))ydx=(y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right))xdy$ is
(where $c$ is constant of integration)

• A. $y^{2}x\sec \left(\frac{y}{x}\right)=\pm c$
• B. $\left|yx\cos \left(\frac{y}{x}\right)\right|=c$
• C. $\left|\frac{y}{x}\right|\cos \left(\frac{y}{x}\right)=c$
• D. $\frac{y}{x}\sec \left(\frac{y}{x}\right)=\pm c$

Answer 1

The correct option is B $\left|yx\cos \left(\frac{y}{x}\right)\right|=c$

$(x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right))ydx=(y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right))xdy$
$\frac{dy}{dx}=\frac{(x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right))y}{(y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right))x}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{(x\cos v+vx\sin v)vx}{(vx\sin v-x\cos v)x}$
$\Rightarrow \frac{xdv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}$
$\Rightarrow \frac{(v\sin v-\cos v)}{2v\cos v}dv=\frac{dx}{x}$
Integrating both sides, we get
$\int \frac{1}{2}\tan vdv-\int \frac{1}{2}\cdot \frac{1}{v}dv=\int \frac{dx}{x}$
$\Rightarrow \frac{1}{2}(\ln |\sec v|-\ln |v|)=\ln |x|+\ln |k|$
$\Rightarrow \ln \left|\frac{\sec v}{v}\right|=\ln (kx{)}^2$
$\Rightarrow \ln \left|\frac{x\sec \left(\frac{y}{x}\right)}{y}\right|=\ln (kx{)}^2$
$\therefore \left|yx\cos \left(\frac{y}{x}\right)\right|=c,$ where $(c=\frac{1}{k^2})$