Answer:
Option B
Explanation:
We have
(2x-3y+5)dx+(9y+6x-7) dy =0
$\frac{dy}{dx}=\frac{2x-3y+5}{3(2x-3y)+7}$
Put 2x-3y=z
$\Rightarrow$ $2-\frac{3dy}{dx}=\frac{dz}{dx}$
$\Rightarrow$ $\frac{dy }{dx}=\frac{1}{3}$ $\left(2-\frac{dz}{dx}\right)$
$\therefore$ $\frac{1}{3}\left(2-\frac{dz}{dx}\right)=\frac{z+5}{3z+7}$
$\Rightarrow$ $2-\frac{dz}{dx}=\frac{3z+15}{3z+7}$
$\Rightarrow$ $\frac{dz}{dx}=2-\frac{3z+15}{3z+7}$
$\Rightarrow$ $\frac{dz}{dx}=\frac{3z+1}{3z+7}$
$\Rightarrow$ $\frac{3z+1}{3z+7}dz= dx$
$\Rightarrow$ $\int\left( \frac{3z-1}{3z-1}+\frac{8}{3z-1}\right)dz= dx$
$=z+\frac{8}{3}log| 3z-1|=x+c$
$=3z+8log( 3z-1)=3x+c$
=3x-9y+8 log (6x-9y-1)=c
