Find the general integral of the homogeneous equation…

Find the general integral of the homogeneous equation $xy'-y=(x+y)\ln\dfrac{x+y}{x}$.

Explanation of Method

Write $y'=\dfrac{y}{x}+\left(1+\dfrac{y}{x}\right)\ln\left(1+\dfrac{y}{x}\right)$, substitute $y=vx$, and integrate with the nested substitution $w=\ln(1+v)$.

Step 1: Substitute $y=vx$

\[x\frac{dv}{dx}=(1+v)\ln(1+v).\]

Step 2: Separate and integrate

With $u=1+v$ then $w=\ln u$,

\[\frac{dv}{(1+v)\ln(1+v)}=\frac{dx}{x} \quad\Longrightarrow\quad \ln|\ln(1+v)|=\ln|x|+C \quad\Longrightarrow\quad \ln(1+v)=Cx.\]

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

\[1+\frac{y}{x}=e^{Cx} \quad\Longrightarrow\quad \boxed{\,y=x\big(e^{Cx}-1\big).}\]

Result

Final answer:

\[\boxed{\,y=x(e^{Cx}-1)\ \ \Longleftrightarrow\ \ \ln\dfrac{x+y}{x}=Cx.}\]