<s>[INST] Regarding solutions of ordinary differential equations Got into a problem with the solution of this one:
$$
x\cos\left(\frac{y}{x}\right)(x\text{d}x+y\text{d}y)=y\sin\left(\frac{y}{x}\right)(x\text{d}y-y\text{d}x)
$$
I can identify differentials of $xy$ and $\frac{x}{y}$ and also of $\cos$ and $\sin$ functions but how should I manipulate the equation into a solvable form?
 [/INST] We have:
$x \cos (\frac{y}{x})(x dx+ydy) = y \sin(\frac{y}{x})(xdy-ydx)$
$\Rightarrow x \cos (\frac{y}{x})(x + y\frac{dy}{dx}) = y \sin(\frac{y}{x})(x\frac{dy}{dx} - y)$
Let $u = \frac{y}{x}$
Then $\frac{du}{dx} = \frac{x\frac{dy}{dx} - y}{x^2} = \frac{\frac{dy}{dx} - u}{x}$
$\Rightarrow \frac{dy}{dx} = x\frac{du}{dx} + u$
Then we substitute this in:
$x\cos u [x+ux(x\frac{du}{dx} +u)] = ux\sin u(x^{2}\frac{du}{dx}+ux-ux)$
$\Rightarrow x^2\cos u [1+u(x\frac{du}{dx} +u)] = ux^3\sin u \frac{du}{dx}$
$\Rightarrow ux\sin u \frac{du}{dx} -\cos u - ux\cos u \frac{du}{dx}-u^2\cos u = 0$
$\Rightarrow ux(\sin u - \cos u)\frac{du}{dx}-(u^2+1)\cos u = 0$
$\Rightarrow ux(\sin u - \cos u)\frac{du}{dx}=(u^2+1)\cos u $
$\Rightarrow \int \frac{u(\sin u - \cos u)}{(u^2+1)\cos u}du = \int\frac{1}{x} dx$
Can you continue from there?
Edit: lol it appears I can't solve this integral after trying, perhaps someone could comment where I've gone wrong?
</s>