Find the general integral of the homogeneous equation…

Find the general integral of the homogeneous equation $xy'-y=x\tan\dfrac{y}{x}$.

Explanation of Method

Divide by $x$ to see the equation depends only on $y/x$, substitute $y=vx$, and integrate the resulting separable equation using $\int\cot v\,dv=\ln|\sin v|$.

Step 1: Substitute $y=vx$

Since $y'=\dfrac{y}{x}+\tan\dfrac{y}{x}$ and $y=vx\Rightarrow y'=v+xv'$,

\[x\frac{dv}{dx}=\tan v.\]

Step 2: Separate and integrate

\[\cot v\,dv=\frac{dx}{x} \quad\Longrightarrow\quad \ln|\sin v|=\ln|x|+C \quad\Longrightarrow\quad \sin v=Cx.\]

Step 3: Back-substitute $v=y/x$

\[\boxed{\,\sin\frac{y}{x}=Cx\ \ \Longleftrightarrow\ \ y=x\arcsin(Cx),\qquad |Cx|\le1.}\]

The special case $C=0$ gives the lines $y=n\pi x$ ($n\in\mathbb{Z}$), along which $\tan(y/x)=0$.

Result

Final answer:

\[\boxed{\,\sin\dfrac{y}{x}=Cx.}\]