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 $+\inf$ 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:
$y = e^{S(x)}$ yielding
$S'' + S'^2 + \nu + 1/2 - x^2/4=0$.
Then I assume the following approximations:
$S'' << S'^2$ and $\nu + 1/2 <<1/4 x^2$ which gives
$S(x) \approx \pm x^2/4$ when $x \to \inf$.
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) << \pm x^2/4 $.
I know that the answer to this problem is given by
$y \approx C_1 x^{-(\nu+1)} e^{x^2/4 }$ and $y \approx C_2 x^{\nu} e^{-x^2/4 }$.
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
$C' << \pm x/2$ and $C'' << \pm 1/2$, 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.
I do not know how much this could help you.
If we make $$y=e^{-\frac{x^2}{4}}\, z$$ the equation becomes $$z''-x z'+\nu z=0$$ the solution of which being (try Wolfram Alpha) $$z=c_1 H_{\nu }\left(\frac{x}{\sqrt{2}}\right)+c_2 \, _1F_1\left(-\frac{\nu }{2};\frac{1}{2};\frac{x^2}{2}\right)$$ where appear Hermite polynomials and Kummer confluent hypergeometric function.
For large values of the argument, the expansions are $$H_{\nu }\left(\frac{x}{\sqrt{2}}\right)=x^{\nu } \left(i^{\nu } 2^{\nu /2} \left(\cos \left(\frac{\pi \nu }{2}\right)-i \sqrt{2} \sin \left(\frac{\pi \nu }{2}\right)\right)+O\left(\frac{1}{x^2}\right)\right)$$ $$_1F_1\left(-\frac{\nu }{2};\frac{1}{2};\frac{x^2}{2}\right)=e^{\frac{x^2}{2}} x^{-\nu } \left(\frac{2^{\frac{\nu +1}{2}} \sqrt{\pi }}{\Gamma \left(-\frac{\nu }{2}\right) x}+O\left(\frac{1}{x^2}\right)\right)+x^{\nu } \left(\frac{i^{\nu } 2^{-\frac\nu 2} \sqrt{\pi }}{\Gamma \left(\frac{\nu +1}{2}\right)}+O\left(\frac{1}{x^2}\right)\right)$$