Question

Solve the differential equation: $x\sin \left(\frac{y}{x}\right)dy=(y\sin \left(\frac{y}{x}\right)-x)dx$

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

Answer 1

The correct option is B $\log x-\cos \left(\frac{y}{x}\right)=k$

Given differential equation is $x\sin \left(\frac{y}{x}\right)dy=(y\sin \left(\frac{y}{x}\right)-x)dx$

$\Rightarrow x\sin \left(\frac{y}{x}\right)\frac{dy}{dx}=y\sin \left(\frac{y}{x}\right)-x$

Substitute $y=vx$

Differentiating w.r.t. x, we get,

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

$\therefore x(x\frac{dv}{dx}+v)\sin v=-x+xv\sin v$

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

Integrating both sides w.r.t $x$

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

$\Rightarrow -\cos v=-\log x+k$

$\Rightarrow \log x-\cos \left(\frac{y}{x}\right)+k$