In the book, Mathematical Methods for Students of Physics and Related Fields, Second Edition by Sadri Hassani, Page 667, the author has stated that, for the following differential equation
$y''(x) - x^2 y(x) \approx 0$,
where $x \to \infty$, one can easily obtain an approximate solution of the form $e^{\pm x^2/2}$.
Is there any approach to obtain this solution, besides solving the exact differential equation $y''(x) - x^2 y(x) = 0$ by the Frobenius method, and then taking the limit of the solutions (Hermite polynomials) as $x \to \infty$?
In general the equation $y''-qy=0$ with $0<q(x)\to\infty$ for $x\to\infty$ has the WKB approximation (see wikipedia, this is a standard example) of basis solutions $$ y(x)=q(x)^{-\frac14}\exp\left(\pm\int\sqrt{q(x)}dx\right) $$ where the first factor is the second order term in the expansion. So indeed, in first order you would get $y(x)=\exp(\pm\frac{x^2}2)$, while in second order, there would be additionally a factor $\frac1{\sqrt{x}}$.
The approach is to set $y=\exp(S)$ with an expansion $S=S_0+S_1+S_2+...$ where $S_0\gg S_1\gg S_2\gg...$ for $x\to\infty$. This scale also translates to the derivatives. Set for shortness $S'=s$ then isolating the components of equal scale in $s'+s^2-q=0$ gives $$ s_0^2=q\\ s_0'+2s_0s_1=0\\\vdots $$ which implies $s_0=\pm\sqrt q$ and $s_1=-\frac{s_0'}{2s_0}\implies S_1=-\frac12\ln|s_0|=-\frac14\ln(q)$.