Non exact differential equation - integration factor problem

Question

Can you help me with this one?

I have checked it and it is non-exact:

$$d/dy(-ycos(ln(y/x))=-cos(ln(y/x))+sin(ln(y/x))$$ $$d/dx(x)=1$$ So, I should find integrating factor with solving this:

But I can't figure out how to solve this.

Answer 1

$$xdy-y \cos \left(\log \left(\frac{y}{x}\right)\right)dx=0$$ The differential equation is homogeneous. Substitute $y=t x \implies y'=t'x+t$: $$t'x+t-t \cos \left(\log t\right)=0$$ It's separable: $$xt'=t(\cos \left(\log t\right)-1)$$ $$\int \dfrac {d \log t}{\cos \left(\log t\right)-1}=\int \dfrac {dx}{x}$$

Answer 2

$$x y'-y \cos \left(\log \left(\frac{y}{x}\right)\right)=0$$

Let $y=x e^z$ to get $$x z'-\cos (z)+1=0$$ Now, switch variables $$\frac x{x'}-\cos(z)+1=0\implies \frac {x'}x=\frac 1 {\cos(z)-1}$$ which looks workable.