Trying to solve an ordinary differential equation of the type: $y^{\prime}+P(x) y=Q(x)$ Wikipedia states: "To derive this, let $M_{(x)}$ be the integrating factor of a first order linear differential equation such that multiplication by $M_{(x)}$ transforms a partial derivative into a total derivative:
$M(x)(\underbrace{y^{\prime}+P(x) y}_{\text {partial derivative }}) = \underbrace{M(x) y^{\prime}+M^{\prime}(x) y}_{\text {total derivative }}$ . "
I cannot understand in what way is $y^{\prime}+P(x) y$ a partial derivative and $M(x) y^{\prime}+M^{\prime}(x) y$ a total derivative.
Could someone try to explain this? thank you!
The terminology is a little confusing. "Partial derivative" as used here does not mean the same thing as the much better known "partial derivative" of multivariate calculus.
Basically, $My' + M'y$ is a "total derivative" because there is a function $f(M, y)$ (i.e., one not directly dependent on $x$) such that $$\dfrac{d}{dx}f(M, y) = M\dfrac{dy}{dx} + \dfrac{dM}{dx}y$$
In this case, obviously $f(M,y) = My$. I'm not completely sure of the reason for the particular terminology. But note that if you "multiply through by $dx$" you get the differential equation $$df = Mdy + ydM$$ which does not include $dx$ anymore. When such a function $f$ exists, the expression on the right ($Mdy + ydM$ in this case) is called an "exact differential".
On the other hand, the similar expression for the other side would be $dg = dy + Pydx$. Such a $g$ must explicitly depend on $x$, not just $y$ and $P$. This is why that side is only "partial".
The great thing here is that you can find a function $M$ that can be multiplied by the "partial" derivative to make it a "total" derivative.