Answer
$t<1$
Work Step by Step
$$(t-1)y''-3ty'+4y=sint,\quad{y(-2)}=2,\quad{y'(-2)}=1$$
For a unique solution that is twice differentiable, the $y''$ term has to exist.
Hence $t=1$ is a limit of the boundary. Since the given solution $t=-2$ lies to the left of this limit, there is a confirmed solution in the range $t<1$.
Hence, the longest open interval is $t<1$.
