Answer
$t>0$
Work Step by Step
$$ty''+3y=t,\quad{y(1)}=1,\quad{y'(1)}=2$$
For a unique solution that is twice differentiable, the $y''$ term has to exist.
Hence $t=0$ is a limit of the boundary. Since the given solution $t=1$ lies to the right of this limit, there is a confirmed solution in the range $t>0$.
Hence, the longest open interval is $t>0$.
