Question

Find the solution of $x\cos \frac{y}{x}(ydx+xdy)=y\sin \frac{y}{x}(xdy-ydx)$

• A. $xy=k\sec \frac{x}{y}$
• B. $x{y}^2=k\sec \frac{y}{x}$
• C. $x^{2}y=k\sec \frac{y}{x}$
• D. $xy=k\sec \frac{y}{x}$

Answer 1

The correct option is D $xy=k\sec \frac{y}{x}$

Given, $x\cos \frac{y}{x}(ydx+xdy)=y\sin \frac{y}{x}(xdy-ydx)$
$ydx+xdy=\frac{y}{x}\tan \frac{y}{x}(xdy-ydx)$
Dividing both sides by $xy$
$\frac{ydx+xdy}{xy}=\tan \frac{y}{x}.\frac{xdy-ydx}{x^2}$
or $\frac{d\left(xy\right)}{xy}=\tan \frac{y}{x}d\left(\frac{y}{x}\right)$
Integrate both sides
$\log xy=\log \sec \frac{y}{x}+\log k$
$\therefore xy=k\sec \frac{y}{x}$