Integrating Factors Proof Help I am trying to learn the methodology behind this ODE technique.  I feel the source I am using (pic-related) leaves out something between Steps 2 and 3.  How does the original equation transform to that in Step 3 without any algebraic manipulation?  Help!?

 A: We start with the initial value problem
$$\frac{dy}{dx} + 3y = 9$$ $$y(0) = 7$$ and multiply through by an undetermined function $\mu(x)$ to give us $$\mu(x)\frac{dy}{dx} + 3\mu(x)y = 9\mu(x).$$  Now, note that if we differentiate $\mu(x)y$ with respect to $x$ we would obtain $$\frac{d(\mu(x)y)}{dx} = \mu(x)\frac{dy}{dx} + y\frac{d\mu(x)}{dx}.$$  Compare now $$\mu(x)\frac{dy}{dx} + 3\mu(x)y \quad \text{and}\quad\frac{d(\mu(x)y)}{dx} = \mu(x)\frac{dy}{dx} + y\frac{d\mu(x)}{dx},$$ and remember that we can choose $\mu(x)$ to be whatever is convenient.  By observing the similarities in the two, namely the presence of the $\mu(x)\frac{dy}{dx}$ in the first term and the $y$ in the second, we choose $$\frac{d\mu(x)}{dx} = 3\mu(x)$$ so they match up.  If we solve this, we get $$\mu(x) = e^{3x}$$ as our integrating factor.  This allows us to replace $$\mu(x)\frac{dy}{dx} + 3\mu(x)y = 9\mu(x)$$ with $$\frac{d(e^{3x}y)}{dx} = 9e^{3x}.$$
This is now a separable equation, and we can solve it to get 
$$e^{3x}y = 3e^{3x} + c \implies y = 3 + ce^{-3x}.$$  Applying the initial condition gives us the solution $$y = 3 + 4e^{-3x}.$$ 
