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,
$$
y = \frac14 t^2 -\frac18 + M\sin \left(\frac12 (t+\delta) \right)
$$
For
$y''+4y=t^2
$,
suppose
$y
= y(t)
=\sum\_{k=0}^d a\_k t^k
$
with
$a\_d \ne 0$.

Then
$y'
=\sum\_{k=1}^d k a\_k t^{k-1}
=\sum\_{k=0}^{d-1} (k+1) a\_{k+1} t^{k}
$
and
$y''
=\sum\_{k=1}^{d-1} k(k+1) a\_{k+1} t^{k-1}
=\sum\_{k=0}^{d-2} (k+1)(k+2) a\_{k+2} t^{k}
$.

Adding,
and replacing the "$4$"
by "$c$" for generality,

$\begin{array}\\
y''+cy
&=\sum\_{k=0}^{d-2} (k+1)(k+2) a\_{k+2} t^{k}+c\sum\_{k=0}^d a\_k t^k\\
&=\sum\_{k=0}^{d-2} (k+1)(k+2) a\_{k+2} t^{k}+c\sum\_{k=0}^{d-2} a\_k t^k
+ca\_{d-1}t^{d-1}+ca\_d t\_d\\
&=\sum\_{k=0}^{d-2} ((k+1)(k+2) a\_{k+2}+ca\_k) t^k
+ca\_{d-1}t^{d-1}+ca\_d t^d\\
\end{array}
$

If this is $t^2$,
we must have
$d=2$
(or else there are terms
of degree higher than $2$)
and
$ca\_2 = 1$,
so $a\_2 = \frac1{c}$.

Similarly,
we must have
$a\_1 = 0$.

That only leaves the
$k=0$ term in the summation
which gives us
$2a\_2+ca\_0 = 0$,
so
$a\_0
=\frac{-2a\_2}{c}
=\frac{-2}{c^2}
$.

Therefore
$y(t)
=\frac{t^2}{c}-\frac{2}{c^2}
$.

Check:

$y'(t)
=\frac{2t}{c}
$,
$y''(t)
=\frac{2}{c}
$,
so
$y''+cy
=\frac{2}{c}+c(\frac{t^2}{c}-\frac{2}{c^2})
=\frac{2}{c}+t^2-\frac{2}{c}
=t^2
$
as desired.

If $c=4$,
the solution is
$\frac{t^2}{4}-\frac{2}{16}
=\frac{t^2}{4}-\frac{1}{8}
$.