I'm currently stuck on approximating a solution to a non-linear ODE. I've searched high and low on this site for general methods for approximating non-linear ODE's of the type $$y(x)y''(x) = \frac{A}{x^2}, x > 0, A \in \mathbb{R}$$ with no luck. If you have a link in mind to a previously asked question regarding this, I'd be glad to see it.
It could be that there is something entirely trivial that I am missing in approximating a solution to this equation! My experience with non-linear ODEs is quite limited.
So this question is about finding an approximate solution to the above-mentioned non-linear ODE with BCs $y(0)=\alpha, y'(\infty)=0$, $\alpha \in [0,\pi/2]$. Any ideas? I'd love to hear your thoughts.
The above equation follows by letting $u(x)=xy(x)$ in the equation $$u'''=A/u^2$$ with BCs $u(0)=\varepsilon, u'(0)=\alpha, u''(\infty)=0$ with $\varepsilon \ll 1$ (a small number) and using that $y'(x)$ is small so that $u'=y+xy'=y$ and $u''=y',u'''=y''$.
$$y\frac{d^2y}{dx^2}=\frac{A}{x^2} \tag 1$$ Change of variable : $\quad x=e^t \quad;\quad\frac{dt}{dx}=e^{-t}.\quad$ [Eventually, for more generality $x=c\,e^t$, any $c$]
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=e^{-t}\frac{dy}{dt}$
$\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right)e^{-t}= \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)e^{-2t}$ $$y\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)e^{-2t}=\frac{A}{x^2}=Ae^{-2t}$$ $$y\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)=A$$
This is an autonomous ODE. Change of function : $\quad \frac{dy}{dt}=X(y)$
$\frac{d^2y}{dt^2}=\frac{dX}{dt}=\frac{dX}{dy}\frac{dy}{dt}=X\frac{dX}{dy}$ $$y\left(X\frac{dX}{dy}-X\right)=A$$
$$\left(\frac{dX}{dy}-1\right)X(y)=\frac{A}{y}$$
The ODE has been reduced of the first order. Solving it leads to the solution on the implicit form : $$\frac{2A}{y}\exp\left(-\frac{(X-y)^2}{2A}\right)+\sqrt{2\pi}\text{ erfi}\left(\frac{X-y}{\sqrt{2\pi}}\right)=C$$ The most likely, there is no closed form for the function $X(y)$.
As a consequence, one cannot come back to $t(y)=\int \frac{dy}{X(y)}$, then to the inverse function $y(t)$ and finally to $y(x)$.
This draw to use numerical methods to solve Eq.$(1)$ instead of looking for an exact analytic method.
For approximate solution, it is of interest to first change of function : $\quad y(x)=\sqrt{A}\:Y(x)$ $$Y\,Y''=\frac{1}{x^2}$$ This is an ODE without parameter, which reduces a lot the number of cases to study. Only the initial conditions have to be taken into account. For example : http://m.wolframalpha.com/input/?i=y+y%27%27%3D1%2Fx%5E2