= Solution
The <Laplace integral method for a differential equation> seeks
$$
y(z)=\int_Ce^{zt}f(t)\,dt.
$$
Differentiating under the integral and substituting into
$$
zy''+y'-zy=0
$$
gives
$$
\int_Ce^{zt}\left[z(t^2-1)f(t)+tf(t)\right]dt=0.
$$
Since $ze^{zt}=\partial_te^{zt}$, integration by parts yields
$$
\left[e^{zt}(t^2-1)f(t)\right]_{\partial C}
-\int_Ce^{zt}\left[(t^2-1)f'(t)+tf(t)\right]dt=0.
$$
It is therefore enough to choose $f$ and $C$ so that
$$
(t^2-1)f'(t)+tf(t)=0
$$
and the boundary term vanishes. Solving the first-order equation on $-1<t<1$ gives
$$
f(t)=\frac{K}{\sqrt{1-t^2}}.
$$
Take $C$ to be the interval $[-1,1]$. Although $f$ has integrable endpoint singularities,
$$
(t^2-1)f(t)=-K\sqrt{1-t^2}
$$
vanishes at both endpoints, so the boundary term is zero. Thus
$$
y(z)=K\int_{-1}^1\frac{e^{zt}}{\sqrt{1-t^2}}\,dt.
$$
At $z=0$, substituting $t=\sin\theta$ gives
$$
\int_{-1}^1\frac{dt}{\sqrt{1-t^2}}=\pi.
$$
The normalization $y(0)=1$ therefore sets $K=1/\pi$. By the defining regularity and normalization of the <modified Bessel function>, this proves the <real integral representation of the modified Bessel function I0>:
$$
\boxed{I_0(z)=\frac1\pi\int_{-1}^1
\frac{e^{zs}}{\sqrt{1-s^2}}\,ds}.
$$
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