I think power series are a tedious way to go about solving a relatively easy DE such as this one.
By separating varaibles, it's easy to see the solution is \(\displaystyle 3e^{-2x}\)
\(\displaystyle \L\\\frac{dy}{dx}=-2y\)
\(\displaystyle \L\\\frac{dy}{y}=-2dx\)
Integrate:
\(\displaystyle \L\\ln(y)=-2dx+C\)
Solve for y:
\(\displaystyle y=e^{-2x+C}\Rightarrow{y=C_{1}e^{-2x}}\)
Using the IC, we see that \(\displaystyle C_{1}=3\)
So, we have \(\displaystyle y=3e^{-2x}\)
Now, the infernal power series method. It's so tedious to LaTex, yet would be hard to follow without it because of all the notation.
Letting \(\displaystyle \L\\y=\sum_{n=0}^{\infty}c_{n}x^{n}\)
From our DE, we have:
\(\displaystyle \L\\y'+2y=\underbrace{\sum_{n=1}^{\infty}nc_{n}x^{n-1}}_{\text{k=n-1}}+\underbrace{2\sum_{n=0}^{\infty}c_{n}x^{n}}_{\text{k=n}}\)
\(\displaystyle \L\\=\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}+2\sum_{k=0}^{\infty}c_{k}x^{k}\)
Factor out \(\displaystyle x^{k}\):
\(\displaystyle \L\\\sum_{k=0}^{\infty}\left[(k+1)c_{k+1}+2c_{k}\right]x^{k}=0\)
\(\displaystyle \L\\=(k+1)c_{k+1}+2c_{k}=0\)
\(\displaystyle \L\\c_{k+1}=\frac{-2}{k+1}c_{k}\)
Now, iterate and set up your series. If you need more help later, I will be back after my fingers get the feeling back in them.
