Question

The solution of differential equation $\left(x\sin \frac{y}{x}\right)dy=\left(y\sin \frac{y}{x}−2x\right)dx$ is

• A. $\sin \left(\frac{y}{x}\right)=2\ln x+c$
• B. $\cos \left(\frac{y}{x}\right)=â„“nx+c$
• C. $\sin \left(\frac{y}{x}\right)=â„“nx+c$
• D. $\cos \left(\frac{y}{x}\right)=2â„“nx+c$

Answer 1

Answer: (D)

$\sin \left(\frac{y}{x}\right)\frac{dy}{dx}=\frac{y}{x}\sin \frac{y}{x}−2$
Let $y=vx⇒\frac{dy}{dx}=v+x\frac{dy}{dx}$
$\sin v\left(v+x\frac{dv}{dx}\right)=v\sin v−2$
$x\sin v\frac{dv}{dx}=−2$
$\sin vdv=−2\frac{dx}{x}$
$−\cos v=−2ℓnx+c⇒\cos v=2ℓnx+c$
$⇒\cos \frac{y}{x}=ℓn{x}^2+c$