Let $\mu(t)$ be the integrating factor. Then,
$$
y' - 2y = e^{2t} \Leftrightarrow y'\mu(t) - 2y\mu(t) = e^{2t}\mu(t).
$$
We require that $\frac{d\mu(t)}{dt} = -2\mu(t)$. Thus,
$$
\frac{d\mu(t)}{dt} = -2\mu(t) \Rightarrow \frac{d\mu(t)}{\mu(t)} = -2dt \Rightarrow \ln|\mu(t)| = -2t + K,
$$
where $K$ is some arbitrary constant that we null to acquire the simplest integrating factor:
$$
\ln|\mu(t)| = -2t \Leftrightarrow \mu(t) = e^{-2t}.
$$
Now, the differential equation is
$$
y'e^{-2t} - 2ye^{-2t} = e^{2t}e^{-2t} = 1.
$$
Applying the product rule gives $\frac{d}{dt}(e^{-2t}y) = 1$, to which the general solution is
$$
e^{-2t}y = t + C \Leftrightarrow y(t) = e^{2t}(t + C).
$$
Using the initial condition, we find that $C=2$. Conclusively, the desired solution is
$$
y(t) = e^{2t}(t+2).
$$
