Answer: Let \(z=x+y\). Then the required solution is
\(\frac{1}{6}\ln(x+y)+\frac{1}{2}(x-y)-\frac{1}{2}=0\).
So in the form \(a\ln(x+y)+b(x-y)+c=0\),
\(a=\frac{1}{6},\quad b=\frac{1}{2},\quad c=-\frac{1}{2}\).
Use the substitution \(z=x+y\).
Since \(y=z-x\), differentiating gives
\(\frac{dy}{dx}=\frac{dz}{dx}-1\).
Substitute into the differential equation:
\(\frac{dz}{dx}-1=\frac{1+3x+3y}{3x+3y-1}.\)
Because \(x+y=z\), this becomes
\(\frac{dz}{dx}-1=\frac{1+3z}{3z-1}.\)
Hence
\(\frac{dz}{dx}=1+\frac{1+3z}{3z-1}=\frac{6z}{3z-1}.\)
Separate variables:
\(\frac{3z-1}{6z}\,dz=dx.\)
Now
\(\frac{3z-1}{6z}=\frac{1}{2}-\frac{1}{6}z^{-1},\)
so
\(\int \left(\frac{1}{2}-\frac{1}{6}z^{-1}\right)dz=\int 1\,dx.\)
Integrating,
\(\frac{1}{2}z-\frac{1}{6}\ln z=x+C.\)
Substitute back \(z=x+y\):
\(\frac{1}{2}(x+y)-\frac{1}{6}\ln(x+y)=x+C.\)
So
\(-\frac{1}{2}x+\frac{1}{2}y-\frac{1}{6}\ln(x+y)=C.\)
Use \(y=0\) when \(x=1\):
\(-\frac{1}{2}(1)+\frac{1}{2}(0)-\frac{1}{6}\ln(1)=C=-\frac{1}{2}.\)
Therefore
\(-\frac{1}{2}x+\frac{1}{2}y-\frac{1}{6}\ln(x+y)=-\frac{1}{2}.\)
Rearranging into the requested form,
\(\frac{1}{6}\ln(x+y)+\frac{1}{2}(x-y)-\frac{1}{2}=0.\)
