Solving non-linear ODEs

Question

I am trying to solve a differential equation of the form:

\begin{equation}x^2y''+2xy'+x^2e^{ay}=0\end{equation}

This arises from calculating the electric potential of ions following the Boltzmann distribution attracted to a spherically symmetric electrode for those who are curious.

Anyway, seeing as this is very similar to #33 from here, I made the substitutions $z=x^2e^{ay}$ and $w=xy'$. I then found

\begin{equation}dz=2xe^{ay}dx+ax^2e^{ay}y'dx=(2+w)z\frac{dx}{x}\end{equation}

\begin{equation}dw=y'dx+xy''dx\end{equation}

I therefore got

\begin{equation}\frac{dw}{dz}=w'=\frac{x^2y''+xy'}{(2+aw)z}\end{equation}

Substituting this and the definition of $z$ into the original ODE gives

\begin{equation}z(2+aw)w'+w+z=0\end{equation}

I thought this would be easier to solve, as it is a 1st order ODE. However, it is non-linear and not separable so far as I can see. In the original from the link there is no 2 on the $y'$ term, which makes the transformed ODE separable.

Can anyone confirm that what I did with the substitution was correct? Also does anyone have any other suggestions for solving either the original or transformed ODE analytically?

Answer 1

$$z(2+aw)w'+w+z=0$$ Let : $\quad u(z)=2+aw(z) \quad;\quad w(z)=\frac{1}{a}(u-2) \quad;\quad w'=\frac{1}{a}u'$ $$\frac{1}{a}zuu'+\frac{1}{a}(u-2)+z=0$$ $$uu'=\frac{2-u}{z}-a$$ Let : $\quad u(z)=\frac{1}{y(z)}$ $$-\frac{y'}{y^3}=\frac{2-\frac{1}{y}}{z}-a$$ $$y'=\frac{az-2}{z}y^3+\frac{1}{z}y^2$$ This is an Abel's differential equation of the first kind : http://mathworld.wolfram.com/AbelsDifferentialEquation.html

Except particular cases, the Abel's equations are not soluble in terms of standard functions and require the definition of non-standard special functions : In fact, in practice, numerical calculus.

For further investigation about solvability see : https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf