Solve the differential equation: (x dy - y dx) ysin(y/x) = (y dx + x dy) xcos(y/x). Find the…

$(x\,dy - y\,dx)\,y\sin\left(\frac{y}{x}\right) = (y\,dx + x\,dy)\,x\cos\left(\frac{y}{x}\right)$ $\Rightarrow (x\,dy - y\,dx)\frac{\sin(y/x)}{\cos(y/x)} = \frac{(y\,dx + x\,dy)\,x}{y}$ $\Rightarrow \frac{(x\,dy - y\,dx)}{x^2}\tan\left(\frac{y}{x}\right) = \frac{(y\,dx + x\,dy)\,x}{x^2 y}$ $\Rightarrow \frac{(x\,dy - y\,dx)}{x^2}\tan\left(\frac{y}{x}\right) = \frac{(y\,dx + x\,dy)}{xy}$ $\Rightarrow \tan\left(\frac{y}{x}\right) d\left(\frac{y}{x}\right) = \frac{d(xy)}{xy}$, which is variable separable form where variables are $\frac{y}{x}$ and $xy$ Integrating both sides, $\Rightarrow \int \tan\frac{y}{x}\, d\left(\frac{y}{x}\right) = \int \frac{d(xy)}{xy}$ $\Rightarrow \log\left|\sec\left(\frac{y}{x}\right)\right| = \log(xy) + \log c$ $\Rightarrow \sec\left(\frac{y}{x}\right) = c\,(xy)$. Given, when $x = 1$, $y = \pi \Rightarrow c = -\frac{1}{\pi}$. Hence required particular solution is: $\sec\left(\frac{y}{x}\right) = -\frac{1}{\pi}(xy)$.