Differential equation: $(x+2y-1)dx-(x+2y-5)dy=0$

$$(x+2y-1)dx-(x+2y-5)dy=0$$

Let $u = x + 2y - 1 \rightarrow du = dx + 2dy$ which gives $dx = du - 2dy$
Substituting to the equation
$$\begin{eqnarray}
u(du - 2dy) - (u - 4)dy &=& 0\\
u du - (3u - 4)dy &=& 0\\
\dfrac{udu}{3u-4} - dy &=& 0\\
\dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)du - dy &=& 0\\
\displaystyle{\int \dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)}du - \int dy &=& C\\
\dfrac{1}{3}u - \dfrac{4}{3} \ln|3u - 4| - y &=& C\\
u - 4 \ln|3u - 4| - 3y &=& C\\
4 \ln|3u - 4| &=& u - 3y + C\\
4 \ln|3(x + 2y - 1) - 4| &=& x + 2y - 3y + C\\
4 \ln|3x + 6y - 7| &=& x - y + C\\
3x + 6y - 7 &=& Ce^{\frac{x-y}{4}} (ANSWER)
\end{eqnarray}$$

P.S. There can also be other forms of the general solution