Find the general integral of the homogeneous equation…
Find the general integral of the homogeneous equation $xy'=y\cos\ln\dfrac{y}{x}$.
Explanation of Method
Substitute $y=vx$ to get $x\frac{dv}{dx}=v[\cos(\ln v)-1]$, then use $u=\ln v$ to turn the remaining integral into the standard form $\int\frac{du}{\cos u-1}$, evaluated with the half-angle identity $\cos u-1=-2\sin^2(u/2)$.
Step 1: Substitute $y=vx$
\[x\frac{dv}{dx}=v\cos(\ln v)-v=v[\cos(\ln v)-1].\]
Step 2: Reduce and integrate
With $u=\ln v$ (so $dv/v=du$),
\[\frac{dv}{v[\cos(\ln v)-1]}=\frac{dx}{x} \quad\Longrightarrow\quad \int\frac{du}{\cos u-1}=\ln|x|+C.\]
Since $\cos u-1=-2\sin^2\frac{u}{2}$, $\displaystyle\int\frac{du}{\cos u-1}=-\frac12\int\csc^2\frac u2\,du=\cot\frac u2$, so
\[\cot\frac{\ln v}{2}=\ln|x|+C.\]
Step 3: Back-substitute $v=y/x$
\[\boxed{\,\cot\!\left(\frac12\ln\frac{y}{x}\right)=\ln|x|+C.}\]
Result
Final answer:
\[\boxed{\,\cot\!\left(\dfrac12\ln\dfrac{y}{x}\right)=\ln|x|+C.}\]