(t^2-9)y'+(t-1)y = 3ln(t) .
We want to find an interval over which the solution is defined.
That’s tough because we can’t actually solve this analytically.
If we try to use method of integrating factors, that will fail.
If we try to use Bernoulli’s method, that also won’t work.
But let’s think about the problem qualitatively without an explicit solution.
Solving for y’, we have y’ = [3ln(t)]/(t^2 – 9) – [(t – 1)/(t^2 – 9)] y
Notice that y’ is undefined for t = -3, 3.
In fact, it will blow up to either positive infinity or negative infinity.
Those are the only times at which y’ is undefined.
For all other times, y’ is defined.
So qualitatively, around t = 2, we can be pretty sure that
a solution y(t) will be defined for all t in -3 < t < 3, and
this is an open interval containing t = 2, so this is our answer.