Linear differential equation-> integrating factor $$ \frac{dy}{dx}+Q(x)y=M(x)$$
multiplying with integrating factor
$$f(x)\frac{dy}{dx}+f(x)Q(x)y=f(x)M(x)  $$
we know that
$$\frac{d(f(x)y)}{dx} = f\cdot \frac{dy}{dx}+\frac{df}{dx}\cdot y$$

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*they both have $y'$ and $y$ but does that immediately make the two equation the same all the time?

 A: Remember the basic product rule:
$$(f g)^\prime(x)=(f g^\prime)(x) + (f^\prime g)(x)$$
The general idea behind introducing an integrating factor in a linear DE is to introduce something that looks like the output of the product rule. So for an expression like
$$I\frac{\mathbf dy}{\mathbf dx}+IQy=IM$$
notice that the left hand side is the output of the product rule, if $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$:
$$\underbrace{I}_{f(x)}\;\underbrace{\frac{\mathbf dy}{\mathbf dx}}_{g^\prime(x)}+\underbrace{IQ}_{f^\prime(x)}\;\underbrace{y}_{g(x)}=IM$$
Thus, anti-differentiating both sides gives us:
$$\underbrace{Iy}_{f(x)g(x)}=\int IM\,\mathbf dx$$
I think that's what you were asking: this is why $\displaystyle\frac{\mathbf d}{\mathbf dx}(Iy)=I\frac{\mathbf dy}{\mathbf dx}+IQy$.
Of course, as already mentioned, this is assuming that $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$. However, the nice thing about $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$ is that it is a separable DE of the form $\displaystyle\frac1I\frac{\mathbf dI}{\mathbf dx}=Q$, which we can solve to get $\displaystyle I=\exp\int Q\,\mathbf dx$. In other words, we introduce a function $I$ with the specific property, that $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$, in order to make the above result true.
