We have,
$$
\begin{aligned}
& (2 x-3 y+5) d x+(9 y-6 x-7) d y=0 \\
& \frac{d y}{d x}=\frac{2 x-3 y+5}{3(2 x-3 y)+7}
\end{aligned}
$$
Put $\quad 2 x-3 y=z$
$$
\Rightarrow \quad 2-\frac{3 d y}{d x}=\frac{d z}{d x}
$$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1}{3}\left(2-\frac{d z}{d x}\right)$
$\therefore \quad \frac{1}{3}\left(2-\frac{d z}{d x}\right)=\frac{z+5}{3 z+7}$
$\Rightarrow \quad 2-\frac{d z}{d x}=\frac{3 z+15}{3 z+7}$
$\Rightarrow \quad \frac{d z}{d x}=2-\frac{3 z+15}{3 z+7}$
$\Rightarrow \quad \frac{d z}{d x}=\frac{3 z-1}{3 z+7}$


$$
\begin{aligned}
& \Rightarrow \frac{3 z+7}{3 z-1} d z=d x \\
& \Rightarrow \int\left(\frac{3 z-1}{3 z-1}+\frac{8}{3 z-1}\right) d z=d x \\
& =z+\frac{8}{3} \log |3 z-1|=x+c \\
& =3 z+8 \log (3 z-1)=3 x+c \\
& =3 x-9 y+8 \log (6 x-9 y-1)=c
\end{aligned}
$$