I have to reduce the following differential equation to the Bessel equation.
$$y''+\lambda xy = 0$$
Upon seeing articles on the internet, I found that this is an Airy differential equation, which can be converted to a Bessel equation by using the substitution given in the below answer.
Is there any general method by which we can predict this substitution, or do we have to brute force different substitutions to see which one works?
The solutions of the Bessel equation have, in first order, an asymptotic behavior at infinity of $e^{\pm ix}$, or its trigonometric versions.
The given equation has, by the WKB method, asymptotics of $\exp(\pm i\int\sqrt{λx}dx)=\exp(\pm i\frac23\sqrt{λ}x^{3/2})$. This indicates a substitution $$ s=\frac23\sqrt{λ}x^{3/2},~~~u(s)=y(x) $$ Then the derivatives are $$ y'(x)=\sqrt{λx}u'(s)\\ y''(x)=λxu''(s)+\frac{\sqrt{λ}}{2\sqrt{x}}u'(s) $$ giving the differential equation $$ su''(s)+\frac13u'(s)+su(s)=0. $$ At $s=0$ the equation for the leading order is $r(r-1)+\frac13r=0$ giving $r=0$ and $r=\frac23$. To center these powers around zero use $u(s)=s^{1/3}v(s)$, $$ u'(s)=s^{1/3}v'(s)+\frac13s^{-2/3}v(s)\\ u''(s)=s^{1/3}v''(s)+\frac23s^{-2/3}v'(s)-\frac29s^{-5/3}v(s) $$ so that $$ 0=\left[s^2v''(s)+\frac23sv'(s)-\frac29v(s)\right] +\frac13\left[sv'(s)+\frac13v(s)\right]+s^2v(s) \\ 0=s^2v''(s)+sv'(s)+\left[s^2-\frac19\right]v(s) $$ This is now the requested Bessel equation. In total, and adjusting some constants, $$ y(x)=\sqrt{x}\,v\left(\frac23\sqrt{λ}x^{3/2}\right) $$
