This might be a stupid question, but I'm curious if you can find physicallty expected boundaries without explicitly specifying them in numerical solutions. To explain what I mean, imagine a very boring reaction-diffusion equation in 1D (assuming steady state for now), something like $\frac{\partial^2 P}{\partial z^2}-2=0$. Let's say at $z = 0$, $P=1$ always and at some unknown boundary $z_{n}$, there's both no flux ($\frac{\partial P}{\partial z} = 0$) and zero concentration ($P(z_{n}) = 0$).
Even though we don't know the boundary, we can easily solve this equation analytically, $p(z) = z^2 -2z(z_{n}) + z_{n}^2 $. From this, you can explictly deduce that $z_{n}= 1$, and you get this kind of plot of the function.
Now, imagine we wanted to solve this numerically for whatever reason - every method I've seen says we have to pre-specify the boundary conditions, but what if we only knew the source ($z=0$) condition, but knew there was some $z_{n}$ which satisfied the problem? In principle I wonder if you could use a finite difference scheme to approximate the lapacian and explicitly calculate each step until you came to a space where $P(z)=0$ without pre-specifying it? Or is there formal reasons this wouldn't work? Apologies if this is an ignorant question, just one I'm curious about.
Such variable length problems can be reduced to constant length problems by writing $y(s)=P(sT)$ where $s\in[0,1]$. Then you could formally add $T$ as another component of the state with derivative zero. Some BVP solvers have the facility to separately declare such constant parameters. But still each such parameter adds a slot in the boundary conditions.
So in your example, $y''(s)=T^2P''(sT)=2T^2$ with boundary conditions $y(0)=1$, $y(1)=y'(1)=0$.