Intervals of Validity

An interval of validity is simply the stretch of $t$ values, around your starting time, where the solution actually exists and makes sense. "Validity" because outside it the formula you found may still spit out numbers, but they no longer belong to the solution you started on. Knowing it tells you how far you can trust a solution, or a simulation, before it stops meaning anything. All we're really asking is: starting from your initial point, how far can you follow the solution before something goes wrong? The theorem in the formula box looks like a mouthful, but it's just a few pieces: $y' + p(t)y = g(t)$ (a linear equation in standard form), $y(t_0) = y_0$ (where you start) and "$p, g$ continuous" (no division by zero, no logs of negatives, nothing that breaks). As long as nothing breaks, exactly one solution keeps going. It looks complex, but it's mostly common sense, so bear with it.

You've already seen a solution with an expiration date. In the separable lesson, $y = \frac{1}{28 - 3t^2}$ only made sense for $|t| < \sqrt{28/3}$. This lesson is about predicting where that happens.

Linear equations are predictable: put the equation in standard form, find where $p$ and $g$ break down, and take the interval between those points that contains $t_0$. It makes sense that this works. The solution formula $y = \frac{1}{\mu}\left(\int\mu g\,dt + C\right)$ is built only from $p$ and $g$, so it can only break where they do.

A common mistake. For $(t - 3)y' + (\ln t)\,y = 2t$, it's tempting to answer "all $t > 0$ except $3$", because those are all the places where $p$ and $g$ are fine. It isn't. A solution has to be one unbroken curve through the starting point, so it can't jump over $t = 3$. You take only the piece that contains $t_0$.

Nonlinear equations can surprise you. $y' = y^2$, $y(0) = 1$ has the solution $y = \frac{1}{1 - t}$, which blows up at $t = 1$ even though the equation looks perfectly smooth. The blow-up time depends on the initial value, and there's an animation further down that shows it moving as you change the start.

Uniqueness can fail too: $y' = \sqrt y$, $y(0) = 0$ has both $y = 0$ and $y = \frac{t^2}{4}$. (The function $\sqrt y$ isn't smooth enough at $y = 0$ for the uniqueness theorem.)

Numerical solvers can't detect a blow-up on their own; knowing the interval of validity tells engineers when a simulation's output stops meaning anything.