Solving $(4y+2x-5)dx+(6y+4x-1)dy=0$ using 2 methods produced 2 different answers! $$(4y+2x-5)dx+(6y+4x-1)dy=0,y(-1)=2$$
First method: 
$$\frac{dy}{dx}=-\frac{4y+2x-5}{6y+4x-1}$$
let $Y=y-\frac{9}{2}$, $dY=dy$  and $X=x+\frac{13}{2}$, $dX=dx$;
$$\frac{dY}{dX}=-\frac{4Y+2X}{6Y+4X}$$
let $u=\frac{Y}{X}$ , $Y'=u'X+u$;
$$u'X+u=-\frac{4u+2}{6u+4}$$
$$u'X=\frac{-6u^2-8u-2}{6u+4}$$
$$\int\frac{(6u+4)du}{6u^2+8u+2}=\int-\frac{dX}{X}$$
$$\frac{1}{2}\ln|3u^2+4u+1|=-\ln|X|+Constant$$
$$\frac{1}{2}\ln|3(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})^2+4(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})+1|=-\ln|x+\frac{13}{2}|+Constant$$
Solving for $y(-1)=2$, $Constant=0.89587977346$;
$$\frac{1}{2}\ln|3(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})^2+4(\frac{y-\frac{9}{2}}{x+\frac{13}{2}})+1|=-\ln|x+\frac{13}{2}|+0.89587977346$$
Second method:
let $M=4y+2x-5$ , $N=6y+4x-1$
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=4\rightarrow Exact$$
$$\varnothing=4xy+x^2-5x+3y^2-y=Constant$$
Solving for $y(-1)=2$, $Constant=8$;
$$\varnothing=4xy+x^2-5x+3y^2-y=8$$
I plotted those two answers on MATHEMATICA and found that result:

How could the results be different?

The indefinite integration formula
$$
\int \frac{dU}{U} = \ln |U| + c
\tag{1}
$$
(correctly) summarizes two formulas that hold in disjoint intervals,
$$
\left.
\begin{aligned}
\int \frac{dU}{U} &= \ln U + c    &&(U > 0), \\
\int \frac{dU}{U} &= \ln (-U) + c && (U < 0).
\end{aligned}\right\}
\tag{2}
$$
(I'm deliberately using $U$ as a variable since it does not appear in your situation.)
In any particular definite integral (such as in your case, solving an ODE with an initial condition), only one choice of sign in (2) is correct.
As a result, your first method (using (1)) led to an equation of the form $|U| = C$, i.e., $U = \pm C$,  when it should have led (using one equation from (2)) to $U = C$.
As visual confirmation, the dashed curve you expected comprises two branches of the hyperbola $4xy + x^{2} - 5x + 3y^{2} - y = C$; the "extra" portion comprises the two branches of the hyperbola $4xy + x^{2} - 5x + 3y^{2} - y = -C$.
