Need help with the following differential equations. $$xy'=y \cos \bigg(\ln{y\over x}\bigg)$$ 
In the first I tried a substitution to no avail.
 A: Hint: Let $z = \frac{y}{x}$ then $y = zx \implies y' = z'x + z$
$$y' = \frac{y}{x} \cos \bigg(\ln \frac{y}{x}\bigg) \implies z'x + z = z\cos (\ln z)  $$
It's separable. 
Edit:
Spoiler: 

 $$\int \frac{1}{z\cos (\ln z)  - z} dz = \cot \bigg(\frac{\ln z}{2}\bigg) + C'$$

Then 

 $$\cot \bigg(\frac{\ln z}{2}\bigg) = \ln x + C \implies z = e^{2 \ arccot ({\ln x+ C})}$$ 

A: i am going to make a change of variable $$\frac{y}{x} = e^u, y = xe^u, \frac{dy}{dx} = xe^u \frac{du}{dx} + e^u.$$ the differential equation $xy' = y\cos \left(\ln (\frac yx)\right)$ turns into a simpler $$ xe^u \frac{du}{dx} + e^u = e^u\left(\cos u\right)$$ that is $$x\frac{du}{dx} + 1= \cos u$$ which is separable and $$\frac{du}{1-\cos u} =\frac 12 \csc^2 u/2 \, du= -\frac{dx}{x}$$ which intgerates to 
$$\cot(u/2) = \ln(x) + \ln C $$
A: Rewrite the equation as follows:
$$\frac{dy}{y} = \cos{(\log{y}-\log{x})} \frac{dx}{x} $$
Let $u=\log{y}$ and $v=\log{x}$.  Then
$$\frac{du}{dv} = \cos{(u-v)}$$
Let $u=q+v$; then
$$\frac{dq}{dv} = \cos{q}-1 = -2 \sin^2{\left ( \frac{q}{2} \right )}$$
Then
$$-\int dq \csc^2{\left ( \frac{q}{2} \right )} = 2 \cot{\left ( \frac{q}{2} \right )} = v + C$$
where $C$ is a constant of integration. Backsub and get
$$2 \cot{\left [ \frac12 \log{\left (\frac{y}{x}\right )} \right ]} = \log{x} + C$$
A: $y=zx$, $y'=z'x+z$, $z'x^2+zx=zx cos(ln(z))$
Separation of variables: $\frac{dz}{z(cos(ln(z))-1)}= \frac{dx}{x}$.
Hint for the Integral (Integratin by parts with first factor $z^{-1}$ and second factor $(cos(ln(z))-1)^{-1}$): $\int \frac{dz}{z(cos(ln(z))-1)}= \frac{ln(z)}{cos(ln(z))-1}- \int \frac{ln(z)sin(ln(z))dz}{z(cos(ln(z))-1)^2}$
