Consider the following initial value problem:
$$y''+2ty-4y=1, \ \ y(0)=0 \textrm{ and } y'(0)=0. $$
Taking Laplace transform:
$$Y(s) = \frac{1}{s^3} + c\frac{e^{s^2/4}}{s^3} $$
and $\lim_{s\to \infty } Y=0$ shows c=0
However, from what I have remembered,
$\lim_{s\to \infty } Y=0$ applies to y such that y is of exponential order and piecewise continuous function.
I know there is a uniqueness theorem for second order equation, so it's safe to say this (let $c=0$) is the only solution. But the theorem doesn't tell anything about solution itself, the solution can be anything(not of exponential order etc.).
It seems $\lim_{s\to \infty } Y=0$ cannot be used here.
Then how we can just say $c=0$?
If $\lim_{s\to\infty} Y(s) = 0$, then the constant $c$ must be zero since
$$ \frac{e^{s^2/4}}{s^3}\longrightarrow\infty \ \ \textrm{ as }s\longrightarrow\infty. $$