Question

The solution of $[x+y\sin \left(\frac{y}{x}\right)]dx=x\sin \left(\frac{y}{x}\right)dy$ is:

• A. $\cos \left(\frac{y}{x}\right)+\log |x|=c$
• B. $\sin \left(\frac{y}{x}\right)+\log |x|=c$
• C. $x\cos \left(\frac{y}{x}\right)+\log |x|=c$
• D. $x\sin \left(\frac{y}{x}\right)+\log |x|=c$

Answer 1

The correct option is A $\cos \left(\frac{y}{x}\right)+\log |x|=c$

Given $[x+y\sin \left(\frac{y}{x}\right)]dx=x\sin \left(\frac{y}{x}\right)dy$

$\Rightarrow [1+\left(\frac{y}{x}\right)\sin \left(\frac{y}{x}\right)]=\sin \left(\frac{y}{x}\right)\frac{dy}{dx}$

put $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow \frac{1+v.\sin v}{\sin v}=v+x\frac{dv}{dx}$

$\Rightarrow \frac{1}{\sin v}+v=v+x\frac{dv}{dx}$

$\Rightarrow \frac{1}{\sin v}=x\frac{dv}{dx}$

$\Rightarrow \int \frac{dx}{x}=\int \sin vdv+c$

$\log x=-\cos v+c$

$\Rightarrow \cos \left(\frac{y}{x}\right)+\log x=c$