$y'-y = 2te^{2t}, y(0) = 1$
Answer:
\begin{align*}
y'-y &= 2te^{2t}\\
\mu &= e^{\int-1dt} = e^{-t}\\
e^{-t}y' - e^{-t}y &= 2te^t\\
e^{-t}y &= \int 2te^t dt\\
e^{-t}y &= 2(te^t - \int e^tdt)\\
e^{-t}y &= 2te^t - 2e^t + c\\
y &= 2te^2t - 2e^2t + ce^t\\
\end{align*}
Since $y(0) = 1, y(0) = 0-2+c = 1$, then $c = 3$
Therefore, $y = 2te^2t - 2e^2t + 3e^t$.
