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$$y'-\frac{1-2x}{y}=1$$

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?

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    You cannot use the integrating factor as it is. You may need to change variables. – 2017-01-24
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    Integrating factors are based off the product rule. I don't think the product rule is helping you here. However, I do believe that equation is separable. – 2017-01-24
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    Integrating factor is $e^{-\int p(x)dx}$ – 2017-01-24

2 Answers 2

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$e^{-\int p(x)dx}$ 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.

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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.