= Boundedness selects a Laplace transform at a singular endpoint
{title2=$|Y(p)|\le\|y\|_\infty/p\quad(p>0)$}
A bounded <function> on $[0,\infty)$ has a <Laplace transform> satisfying the displayed bound, by integration of $|y(t)e^{-pt}|$. Consequently its <Laplace transform> cannot have a pole of order two or higher at $p=0$. This can eliminate the arbitrary constant introduced when multiplication by $t$ turns an <ordinary differential equation> into a first-order <differential equation> for its <Laplace transform>.
For example, transforming $ty''+(t-1)y'-y=0$ with $y(0)=1$ gives $p(p+1)Y'+(3p+2)Y=2$. An <integrating factor> $p^2(p+1)$ yields
$$
Y(p)=\frac1{p+1}+\frac{C}{p^2(p+1)}.
$$
The bound forces $C=0$, selecting $y=e^{-t}$. This selection can be justified without assuming unproved endpoint regularity for every solution: $e^{-t}$ and $t-1$ have nonzero <Wronskian> $te^{-t}$ on $(0,\infty)$, so they form a <fundamental system of solutions>. Boundedness removes $t-1$, and the limit at zero fixes the remaining coefficient. The endpoint limit by itself does not select a unique solution.
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