TITLE: How can you comment this calculation step to avoid points deduction by teacher?
QUESTION [9 upvotes]: Solve the initial value problem: $y'+2y=e^{-x}$ with $y(0)=4$

Our teacher gave me a mistake / unclear mark at the part which I will mark red here. But I don't understand why and I no longer have the opportunity to ask him.
This is an extract of my solution (it leads to the correct solution):
$$e^{2x} \cdot \frac{dy}{dx} + e^{2x} \cdot 2y = e^{2x} \cdot e^{-x} \Leftrightarrow$$
$$\Leftrightarrow \color{red}{\frac{d}{dx} \cdot e^{2x} \cdot y} = e^{x} \Leftrightarrow$$
$$\int{\frac{d}{dx} \cdot e^{2x} \cdot y} \text{ } dx = \int{e^x} \text{ } dx$$
$$...$$
I don't understand, is an explanation here really required? So I took that $y$ from $\frac{dy}{dx}$ and wrote it at the end of the term and now we got $\frac{d}{dx}$ at the front. $\frac{d}{dx}$ means "the derivative of.." (whatever comes after). And taking that derivative, we will get what we had before (the line before the red marked line), so they are equal to each other. 
But how would you explain that (on paper) in a reasonable way? I don't know if my current explanation is fine.
Edit: When I handed this in, I used brackets too. The reason I got point deduction is "how do you get from here to there?" (first line to red line).

REPLY [12 votes]: It took me a minute or two to "decipher" how you'd gotten from the first line to the second line, especially since I wasn't sure whether you were applying an operation to both sides of the equation, or just manipulating the two sides independently.
For an answer written in "quick" style, I think this would be a fine way to write this manipulation:

$$e^{2x} \cdot \frac{dy}{dx} + e^{2x} \cdot 2y = e^{2x} \cdot e^{-x}$$
Rearrange LHS and simplify RHS:
$$e^{2x} \cdot \frac{dy}{dx} + 2 e^{2x} \cdot y = e^x$$
Apply chain rule and product rule in reverse:
$$e^{2x} \cdot \frac{dy}{dx} + \frac{d}{dx} (e^{2x}) \cdot y = e^x$$
  $$\frac{d}{dx} (e^{2x} \cdot y) = e^x$$