Solutions to linear differential equations can be written using convolutions as $$y \ \ = \ \ y_{\mathrm{IVP}} \ \ + \ \ \Bigl(h(t) \ {\Large \ast} \ f(t)\Bigr)$$ $\qquad \bullet$ $y_{\mathrm{IVP}}$ is the solution to the associated homogeneous differential equation with the given initial values
$\qquad\qquad$ (ignore the forcing function, keep initial values).
$\qquad \bullet$ $h(t)$ is the impulse response
$\qquad\qquad$ (ignore the initial values and forcing function).
$\qquad \bullet$ $f(t)$ is the forcing function.
$\qquad\qquad$ (ignore the initial values and differential equation).

---

Use the form above to write the solution to the differential equation $$y''+2y'-3y = -6t^{2}e^{-3t} \qquad \mathrm{with} \quad y(0) = 12, \quad y'(0) = 4$$

$y \ \ = \ \ $$\quad + \quad \Bigl($$\!\Large\ast$$\Bigr)$

---

If you don't get this in 1 tries, you can get a hint.

Hint: