$
(3x+y-2)dx + (3x+y+4)dy = 0
$
$
u = 3x + y
$
$
du = 3dx + dy
$
$
dy = du - 3dx
$
$
(u-2)dx + (u+4)(du - 3dx) = 0
$
$
(u+4)du + (u-2-3u-12)dx = 0
$
$
(u + 4)du + (-2u-14)dx=0
$
$
(u+4)du- (2u+14)dx=0
$
$$
\int{\left(\frac{u+4}{2u+14}\right)du} - \int{dx} =0
$$
$$
\int{\left(\frac{1}{2}-\frac{3}{2u+14}\right)}du -\int{dx} =0
$$
$$
(*)\frac{1}{2}\int{du}-3\int{\frac{du}{2u+14}} -\int{dx} =0
$$
$$
\frac{1}{2}u-\frac{3}{2} ln|2u+14|-x=c
$$
$
u-3ln|2u+14|-2x=c
$
$
(3x+y)-3ln|2(3x+y)+14|-2x=c
$
$ x+y-3ln|6x+2y+14|=c $
QUESTION: In step($*$), what if I factor out $2$ in $2u+14$. Then it will be...
$$
\frac{1}{2}\int{du}-\frac{3}{2}\int{\frac{du}{u+7}} -\int{dx} =0
$$
$$
\frac{1}{2}u-\frac{3}{2} ln|u+7|-x=c
$$
$
u-3ln|u+7|-2x=c
$
$
(3x+y)-3ln|(3x+y)+7|-2x=c
$
$ x+y-3ln|3x+y+7|=c $
Which is . . . $ [x+y-3ln|6x+2y+14|=c] \neq [x+y-3ln|3x+y+7|=c] $
Is there something wrong in my solution? Or the second solution violate something? Please give me reasons. Thank you.