The solution of the following problem
$y'(x) = 100y(x) -101e^{-x}$
$y(0) = 1 $
is $y(x) = e^{-x}$ for $x \in D$
I wonder what the correct set D is.
In my notes it says $D=(0,+\infty)$, But why? I don't see any singularities in the solution to exclude the rest of the real line. Furthermore if the solution is only considered for $x>0$, $y(0)$ should be undefined then I couldn't say $y(0)=1.$ I usually see that in differential equations only one of the half-intervals determined by "a" in the initial condition $y(a)=y_0$ is taken as the domain of the solution (and it is taken open), like $(0, +\infty) $in this case. What is the reason for all this? Would saying the solution is in $D=\mathbb{R}$ or in D=$[0, +\infty)$ make the problem ill-posed?