6
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Are you seeking an analytical or a numerical solution? $\endgroup$ Aug 13, 2018 at 20:01
  • $\begingroup$ Analytical would be better. $\endgroup$
    – phi1123
    Aug 13, 2018 at 22:59

1 Answer 1

2
$\begingroup$

$$z(2+aw)w'+w+z=0$$ 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

$\endgroup$
1
  • $\begingroup$ This appears to be the answer. I actually did this a slightly different way before I saw your answer and got the same thing. Unfortunately doesn't really look like you can progress any further and solve the original 2nd order ODE after solving the transformed 1st order one. Too bad, looks like I will have to mess around with numerical solutions. $\endgroup$
    – phi1123
    Aug 15, 2018 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.