= Solution
The intended integral representation of the <modified Bessel function> is
$$
I_0(x)=\frac1\pi\int_0^\pi e^{x\cos\theta}\,d\theta.
$$
<Differentiation under the integral sign> gives
$$
I_0'(x)=\frac1\pi\int_0^\pi\cos\theta\,e^{x\cos\theta}\,d\theta,
\qquad
I_0''(x)=\frac1\pi\int_0^\pi\cos^2\theta\,e^{x\cos\theta}\,d\theta.
$$
Since
$$
\frac d{d\theta}\left(\sin\theta\,e^{x\cos\theta}\right)
=\left(\cos\theta-x\sin^2\theta\right)e^{x\cos\theta},
$$
<integration by parts> and the vanishing endpoint term yield $xI_0''+I_0'-xI_0=0$.
For large positive $x$, <Laplace's method> localizes the integral near the endpoint $\theta=0$. Put $\theta=u/\sqrt{x}$ and use
$$
x\cos\theta=x-\frac{u^2}{2}+\frac{u^4}{24x}+O(x^{-2}u^6).
$$
Extending the exponentially small tail to infinity and evaluating the resulting <Gaussian integrals> gives
$$
I_0(x)=\frac{e^x}{\pi\sqrt{x}}
\left(\int_0^\infty e^{-u^2/2}du
+\frac1{24x}\int_0^\infty u^4e^{-u^2/2}du+O(x^{-2})\right),
$$
and therefore
$$
\boxed{I_0(x)\sim\frac{e^x}{\sqrt{2\pi x}}
\left(1+\frac1{8x}+O(x^{-2})\right)}.
$$
Now write $y=x^{-1/2}w$. Direct differentiation gives
$$
y'=x^{-1/2}w'-\frac12x^{-3/2}w,
\qquad
y''=x^{-1/2}w''-x^{-3/2}w'+\frac34x^{-5/2}w,
$$
so substitution into the <ordinary differential equation> gives
$$
\boxed{w''+\left(\frac1{4x^2}-1\right)w=0}.
$$
For the <Liouville–Green approximation>, put $Q(x)=1-1/(4x^2)$. The two approximate solutions are
$$
w_\pm(x)\sim Q(x)^{-1/4}
\exp\left(\pm\int^x\sqrt{Q(s)}\,ds\right),
$$
so at leading order $y_\pm(x)\sim x^{-1/2}e^{\pm x}$. The growing solution agrees with the preceding <asymptotic expansion>, including its leading normalization $I_0(x)\sim e^x/\sqrt{2\pi x}$.
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