In chapter 3 (example 4) of the book "Advanced Mathematical Methods for Scientists and Engineers", by Bender and Orszag, I want to get the approximate solution for $+\infty$ for the parabolic cylinder differential equation:
$$y'' + (\nu + 1/2 -x^2/4)y = 0.$$
First I get rid of the singularity at infinite using the following transformation:
$$y = \mathrm e^{S(x)}$$ yielding
$$S'' + S'^2 + \nu + 1/2 - x^2/4=0.$$
Then I assume the following approximations $S'' \ll S'^2$ and $\nu + 1/2 <\ll1/4 x^2$, which gives,
$$S(x) \approx \pm x^2/4$$ when $x \to \infty$.
This is the controlling factor of the general differential equation. To get the leading behavior I assume that
$$S(x) \approx \pm x^2/4 + C(x)$$ where $C(x) \ll \pm x^2/4 $.
I know that the answer to this problem is given by
$$y \approx C_1 x^{-(\nu+1)} \mathrm e^{x^2/4 }$$ and $$y \approx C_2 x^{\nu} \mathrm e^{-x^2/4 }.$$
However I don't know how to reach such result. The differential equation for $C(x)$ is given as $$(\pm 1/2 + C'') + (\pm x/2 + C')^2 + \nu + 1/2 - x^2/4 = 0,$$
Using the approximation for $C(x)$ I get that $C' \ll\pm x/2$ and $C'' \ll \pm 1/2$, hence, the differential equation would not depend on $C(x)$, which makes no sense.
I have tried to assume only one of these conditions at a time, however, even so, I do not get to the desired result.
By taking
$y=e^{S(x)}$
you have already pulled out the exponential behaviour in the solution.
As a result another exponential correction
$y=e^{S(x)}e^{C(x)}$
will not further elucidate $y$'s behaviour. So no surprise that the differential equation for $C$ becomes independent of it.
Try the following correction instead
$y=e^{S(x)}p(x)=e^{S(x)}e^{ln(p(x))}$
For $S(x)=k \frac{x^2}{4}$
$k=1$ leads to $p''+x p'+(\nu+1)p=0$ which (ignoring $p''$ term) solves as $x^{-(\nu+1)}$
$k=-1$ leads to $p''-x p'+\nu p=0$ which (ignoring $p''$ term) solves as $x^{(\nu)}$