#### Answer

$(0,4)$

#### Work Step by Step

$t(t-4)y''+3ty'+ty=2$
To use Theorem 3.2.1, divide both sides by $t(t-4)$:
$y''+\frac{3}{t-4}y'+\frac{4}{t(t-4)}y=\frac{2}{t(t-4)}$
$p(t)=\frac{3}{t-4}$ which is continuous on $(-\infty,4) \cup(4,\infty) $
$q(t)=\frac{4}{t(t-4)}$ which is continuous on $(-\infty,0) \cup(0,4)\cup (4,\infty) $
$g(t)=\frac{2}{t(t-4)}$ which is continuous on $(-\infty,0) \cup(0,4)\cup (4,\infty)$
Thus, $p(t),q(t),$ and $g(t)$ are all continuous on $(-\infty,0) \cup(0,4)\cup (4,\infty)$
Since $t_0=3$, the solution exists on $(0,4)$