$(3x - 2y + 1)~dx + (3x - 2y + 3)~dy = 0$
Let
$z = 3x - 2y$
$dz = 3~dx - 2~dy$
$dy = \frac{1}{2}(3~dx - dz)$
Thus,
$(z + 1)~dx + (z + 3)[ \, \frac{1}{2}(3~dx - dz) \, ] = 0$
$(z + 1)~dx + \frac{3}{2}(z + 3)~dx - \frac{1}{2}(z + 3)~dz = 0$
$[ \, (z + 1) + (\frac{3}{2}z + \frac{9}{2}) \, ]~dx - \frac{1}{2}(z + 3)~dz = 0$
$(\frac{5}{2}z + \frac{11}{2})~dx - \frac{1}{2}(z + 3)~dz = 0$
$\frac{1}{2}(5z + 11)~dx - \frac{1}{2}(z + 3)~dz = 0$ → variables separable
Divide by ½(5z + 11)
$dx - \dfrac{\frac{1}{2}(z + 3)~dz}{\frac{1}{2}(5z + 11)} = 0$
$dx - \dfrac{(z + 3)~dz}{5z + 11} = 0$
$dx - \dfrac{[ \, \frac{1}{5}(5z + 11) + \frac{4}{5} \, ] ~dz}{5z + 11} = 0$
$dx - \left[ \dfrac{\frac{1}{5}(5z + 11)~dz}{5z + 11} + \dfrac{\frac{4}{5}~dz}{5z + 11} \right] = 0$
$\displaystyle \int dx - \frac{1}{5} \int dz - \frac{4}{5} \int \dfrac{dz}{5z + 11} = 0$
$\displaystyle \int dx - \frac{1}{5} \int dz - \frac{4}{25} \int \dfrac{5~dz}{5z + 11} = 0$
$x - \frac{1}{5}z - \frac{4}{25}\ln (5z + 11) = -\frac{2}{5}c$
$25x - 5z - 4\ln (5z + 11) = -10c$
$25x - 5(3x - 2y) - 4\ln [ \, 5(3x - 2y) + 11 \, ] = -10c$
$25x - 15x + 10y - 4\ln (15x - 10y + 11) = -10c$
$10x + 10y - 4\ln (15x - 10y + 11) = -10c$
$5x + 5y - 2\ln (15x - 10y + 11) = -5c$
$5x + 5y + 5c = 2\ln (15x - 10y + 11)$
$5(x + y + c) = 2\ln (15x - 10y + 11)$ answer
