Question: If $f$ and $g$ have compact support, then does it follow that $f \* g$ has compact support?

Answer: 
Where you went wrong is in the second line of your reasoning: It should have read (based on your correct first line)
$$
t^2(4a) + t(4b) + 1(4c\color{red}{+2a}) = t^2(1) + t(0) + 1(0)
$$
From this, you can see that $c=-\frac12 a$ and $a=\frac14$ so one solution of the differential equation is 
$$y = \frac14 t^2 -\frac18 $$
The general solution is, of course,
$$
y = \frac14 t^2 -\frac18 + M\sin \left(\frac12 (t+\delta) \right)
$$
First realize that $(t^2)''=2$, hence let:
$$4x(t)=4y(t)-t^2+\frac{1}{2}$$

$$\Rightarrow x''(t)=y''(t)-\frac{1}{2}$$

The equation becomes:

$$x''(t)=-4x(t)$$

Note that this can be seen as a differential equations relating the position($x(t)$) and acceleration

($a(t)=x''(t)$) of an object in harmonic motion, the solution of which:
$$x(t)=Acos(2t + \varphi)$$
with constants $A,\varphi $

substitue back to get the answer