The interval where the solution to the initial value problem
exists and is unique, is:
I began by solving for $y'$ which gave: $$y' = \frac{t}{\ln t} - \frac{1}{(t-3)\ln t}*y$$
by looking at this I would say that $t≠1,3$ since $\ln(1)=0$ and $(3-3)\ln(3)=0$
So by this, the interval should not be $(a)$ or $(b)$ however, the correct answer is $(a)$ and I don't understand how that is possible.
All help is appreciated!
First put the ODE in standard form
$$\frac{dy}{dt} + P(t)y = Q(t)$$
which in your case is
$$\frac{dy}{dt} + \frac{1}{(t-3)\ln(t)}y = \frac{t}{\ln(t)}$$
and then observe the intervals in which $P(t)$ and $Q(t)$ exist and are continuous. Note that both $P(t)$ and $Q(t)$ are undefined for $t=1$. For $$P(t)=\frac{1}{(t-3)\ln(t)}$$
the domain of $t$ is $(0,1)\cup(1,3)\cup(3,\infty)$. For $$Q(t)=\frac{t}{\ln(t)}$$
the domain of $t$ is $(0,1)\cup(1,\infty)$. So, the combined interval of continuity for both $P(t)$ and $Q(t)$ is the restricted domain of $(0,1)\cup(1,3)\cup(3,\infty)$.
The initial condition at $t=2$ is given as $y(2)=0$. This matches one specific part of the combined interval of continuity which is given as the answer $(a)$.