= Solution
Take \b[any $a\ne0$ and any $b\in\mathbb R$]. Choose $0<h_0<|a|$, so $[a-h_0,a+h_0]$ avoids zero. On
$$
R=[a-h_0,a+h_0]\times[b-1,b+1],
$$
both $f$ and $f_y=(1-2x)/x^2$ are continuous. They are bounded by <compactness>, and the bound on $f_y$ gives the uniform <Lipschitz continuity> in $y$ required by the <Picard-Lindelöf theorem>.
If $M$ is a bound for $|f|$, shrink $h$ so $0<h<h_0$ and $hM<1$. Then the solution has $|y(x)-b|\leq hM<1$, which supplies even strict inequalities at both edges of the requested vertical strip.
At $a=0$, neither $f$ nor the original exponential expression is defined. Thus \b[the theorem applies exactly for $a\ne0$, with no restriction on $b$]. Failure of the hypotheses alone would not prove nonexistence; the explicit solutions below settle that issue.
Back to article page
