TITLE: Initial Value Problem with Laplace Transform
QUESTION [5 upvotes]: How do you solve the following with Laplace Transform?
$$
{\rm y}''\left(t\right) - 10\,{\rm y}'\left(t\right) + 25\,{\rm y}\left(t\right)
=
24\,t\,{\rm e}^{-2t}\,;
\qquad\qquad
{\rm y}\left(0\right) = -2\,,\quad {\rm y}'\left(0\right) = -10
$$

REPLY [4 votes]: Hints: 

$\mathcal{L} (y''(t)) = s^2y(s) -s y(0) -y'(0)$
$\mathcal{L} (-10y'(t)) = -10(sy(s) -y(0))$
$\mathcal{L} (25y(t)) = 25y(s)$
$\mathcal{L} (24te^{-2t}) = \dfrac{24}{(s + 2)^2}$

Now, substitute the ICs, isolate $y(s)$ on the LHS, everything else on the RHS, do a partial fraction expansion and then find the Inverse Laplace Transform. You should end up with the following.
Spoiler

 $y(t)=\dfrac{2}{343} e^{-2t} (84 t+e^{7 t} (84 t-367)+24)$