$(t^2-3t)y'' + ty' - (t+3)y=0$ with $y(1)=2, y'(1)=1$ - Maximal interval 
Find the maximal interval in which the initial value problem
  $$(t^2-3t)y'' + ty' - (t+3)y=0$$ $$y(1)=2, y'(1)=1$$ exists.

According to a well-known theorem, we have $p(t)=\frac{1}{t-3}$, $q(t)=-\frac{t+3}{t(t-3)}$ and $g(t)=0$. So the discontinuity points are $t=0$ and $t=3$. 
I am only $9$ years old, and often I need help to understand certain concepts. Is anyone could explain in details why the maximal interval is $0< t < 3$? I know it is probably trivial, but I blocked on the resolution of the problem.
 A: Your $p$, $q$ and $g$ are correct; and so are your points of disconinuity.
Based on these points, $t=0$ and $t=3$, there are three intervals in which a solution may exist. These intervals are $(-\infty,0)$, $(0,3)$, and $(3,\infty)$.
Now, you have to pay attention to your initial conditions, both of which are given at $t_0 = 1$. This means that whatever the solution to the differential equation is, it must exist at time $t=1$, and have the given value $y_0=2$.
Out of the three choices that you have for the intervals of existence, which one contains $t=1$? Surely it is the interval $(0,3)$.
That fact that your interval (in $t$) around $t=1$ can be extended all the way to $(0,3)$ as compared to, say $(0.5,2)$, is a direct consequence of the existence theorem (which says your solutions exists on the same interval over which $p,\,q$ and $g$ are defined).
A: There are several aspects to this.


*

*There is a solution on $(0,3)$, since this is a linear differential equation and (after dividing by $t^2-3t$) the coefficients are continuous on that interval.

*The solution can't be extended through the singularities to a larger interval.  That is not so obvious (although I think it's true in this case).  For example, the differential equation $t^2 y'' -t y' + y = 0$ has a singularity at $0$, but $y=t$ is a solution on the whole real line.

