Consider the following initial value problem: $$(t^2 - 14 t + 24 ) \frac { dy }{ dt } = y\quad \mbox{with} \quad y( 7 ) =1.$$

1. Solve the initial value problem (find $y$ as a function of $t$.)
   $y =$
2. On what interval is the solution valid?
   It is valid for $< t <$.
3. Find the limit of the solution as $t$ approaches the left end of the interval.
   Answer:
4. Find the limit of the solution as $t$ approaches the right end of the interval.
   Answer: