Simple Homogeneous ODE $$y'-\frac{1-2x}{y}=1$$
If we look at the homogeneous part can we use $y_{h}=e^{-\int p(x)dx}$ or because it is in the form of $y'+p(x)y^{-1}=0$ we can not?
 A: $e^{-\int p(x)dx}$ can be taken as integrating factor for the ODEs on the form :
$$\frac{dy}{dx}+p(x)y(x)=q(x)$$
See the proof in : http://mathworld.wolfram.com/IntegratingFactor.html
Of course, if the ODE has not the above form, $e^{-\int p(x)dx}$ is no longer the integrating factor. It is absurd to think that the same integrating factor be convenient for two ODEs of different kind, one linear as above, the other non-linear as $\frac{dy}{dx}+p(x)\frac{1}{y(x)}=q(x)$
In the case of $$\frac{dy}{dx}-\frac{1-2x}{y}=1$$
$$\frac{dy}{dx}-\left(\frac{1-2x}{y^2}\right)y=1$$
$p=-\frac{1-2x}{y^2}$ is not a function of $x$ only. Thus $\int pdx$ cannot be explicitly expressed because it includes the unknown $y(x)$.
Forget this ready-made idea that $e^{-\int p(x)dx}$ is an integrating factor in all cases.
A: You are mixing two different meanings of "homogeneous". Since the equation is not linear in $y$, the only one that applies is that $y'$ is equal to a quotient of linear terms. And indeed,
$$
y'=\frac{1+y-2x}y
$$
Setting $y=(1-2x)u$ leaves one with
$$
(1-2x)u'-2u=\frac{1+u}{u}\iff \frac{uu'}{1+u+2u^2}=\frac1{1-2x}
$$
which can be solved via
$$
\frac{uu'}{1+u+2u^2}=\frac14\frac{d}{dx}\ln|1+u+2u^2|-\frac14\frac1{1+u+2u^2}
$$
where the last term has an antiderivative based on the arcus tangent.
