The general solution of the differential equation \(\frac{d y}{d x}=\frac{x+2 y-1}{x+2 y+1}\) is

\(
\frac{d y}{d x}=\frac{x+2 y-1}{x+2 y+1}
\)
Put \(x+2 y=t \Rightarrow x+2 y=t \Rightarrow 1+2 \frac{d y}{d x}=\) \(\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=\frac{\left(\frac{d t}{d x}-1\right)}{2}\)
\(\therefore \frac{\left(\frac{d t}{d x}-1\right)}{2}=\frac{t-1}{t+1} \Rightarrow \frac{d t}{d x}-1=\frac{2 t-2}{t+1} \)
\( \therefore \frac{d t}{d x}=\frac{2 t-2}{t+1}+1=\frac{3 t-1}{t+1}\)
\( \therefore \int \frac{t+1}{3 t-1} d t=\int d x \)
\( \therefore \frac{1}{3} \int \frac{3(t+1)}{3 t-1} d t=\int d x \Rightarrow \frac{1}{3} \int \frac{3 t-1+4}{3 t-1} d t=\) \(\int d x \)
\( \therefore \frac{1}{3} \int d t+\frac{4}{3} \int \frac{d t}{3 t-1}=\int d x \)
\( \therefore \frac{t}{3}+\frac{4}{3} \frac{\log |3 t-1|}{3}=x+c_1 \)
\( \therefore \frac{x+2 y}{3}+\frac{4}{3} \frac{\log |3(x+2 y)-1|}{3}=x+c_1 \)
\( \therefore 3 x+6 y+4 \log |3 x+6 y-1|=9 x+9 c_1 \)
\( \therefore 6(-x+y)+4 \log |3 x+6 y-1|=K\)