Laplace Transform solution verification: $\ddot{y} + 2y = 2e^t\implies \frac13\cos(\sqrt{2}t)-\frac{2}{3\sqrt{2}}\sin(\sqrt{2}t)+\frac23e^t\,\text{?}$ Does
$$\ddot{y} + 2y = 2e^t\quad y(0)=1,\dot{y}(0)=0$$
Give
$$\frac13\cos(\sqrt{2}t)-\frac{2}{3\sqrt{2}}\sin(\sqrt{2}t)+\frac23e^t\,\text{?}$$ 
This is what I have got, and it seems to go back and forth just fine. I just wanted a quick verification. Thank you very much sirs, and madams.

I am not sure if you used partial fractions and table to solve the problem so I will post a solution using the inverse Mellin transform.
The Laplace transform of the ODE is
\begin{align}
s^2Y(s) - sy(0) - \dot{y}(0) + 2Y(s) &= 2\int_0^{\infty}e^{t}e^{-st}dt\\
Y(s)(s^2 + 2) - s &= \frac{2}{s-1}\\
Y(s) &= \frac{2}{(s-1)(s^2+2)}+\frac{s}{s^2+2}
\end{align}
Then by the inverse Mellin transform, we have that the poles are at $s=\pm \sqrt{2}i,1$
\begin{align}
y(t) &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\biggl[\frac{2e^{st}}{(s-1)(s^2+2)}+\frac{se^{st}}{s^2+2}\biggr]ds\\
&=\sum\text{Res}\\
&=\lim_{s\to 1}(s-1)\frac{2e^{st}}{(s-1)(s^2+2)}+\lim_{s\to \sqrt{2}i}(s-\sqrt{2}i)\frac{2e^{st}}{(s-1)(s^2+2)}\\
&+\lim_{s\to -\sqrt{2}i}(s+\sqrt{2}i)\frac{2e^{st}}{(s-1)(s^2+2)} + \lim_{s\to \sqrt{2}i}(s-\sqrt{2}i)\frac{se^{st}}{s^2+2}\\
&+\lim_{s\to -\sqrt{2}i}(s+\sqrt{2}i)\frac{se^{st}}{s^2+2}\\
&= \frac{2e^t}{3}-\frac{2}{3}\cos(t\sqrt{2})-\frac{\sqrt{2}}{3}\sin(t\sqrt{2})+\cos(t\sqrt{2})\\
&=\frac{2e^t}{3}+\frac{1}{3}\cos(t\sqrt{2})-\frac{\sqrt{2}}{3}\sin(t\sqrt{2})
\end{align}
which is of no surprise the solution you obtained.
