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« **on:** September 24, 2014, 11:04:15 PM »
Firstly, I'd like to apologize for the really ugly notations and equations.. there is the solution! I rewrote it--V.I.

Find the solution of the given initial value problem.

\begin{gather}

y' - 2y = e^{2t},\label{A}\\

y(0) = 2.\label{B}

\end{gather}

First, we need to find the integrating factor, which is $I = e^{\int -2\,dt}= e^{-2t}$

so we multiple the entire equation by I, thus, giving us

$e^{-2t} y' - 2e^{-2t} y = 1$.

We see that the left side of the equation can be rewritten as $[e^{-2t}y]'$

~~and we see that the right side of the equation is in fact 1~~

therefore: $[e^{-2t} y]' = 1$

we now take the integral of both sides, giving us: $e^{-2t} y = t + C $ where $C$ is a constant. To find $C$, we substitute $y = 2$ when $t$ = 0 ~~(this information is given in the question)~~. We find out that $C = 2$.

rearranging the formula, we come to the solution as

\begin{equation*}

y = (t+2) e^{2t}.

\end{equation*}