# Solution to a linear Ordinary Differential Equation (ODE) with variable coefficients.

I need to solve a series of linear ODEs that are close'' to the Euler equation , in a certain sense. One example of my equations is given below:

\begin{equation} 8 \xi^3 \frac{d^3 r_\xi}{d \xi^3} + 36 \xi^2 \frac{d^2 r_\xi}{d \xi^2} + 2(12+\omega^2)\xi \frac{d r_\xi}{d \xi} + \omega^2(2 - \omega \sqrt{\xi}) r_\xi ==0 \end{equation}

As you can see, for small values of $\omega$ the equation is close to the Euler equation and thus is solved by applying an ansatz $r_\xi := \xi^r$. That solution reads: \begin{equation} r_\xi = C_\pm \xi^{-1/4} \xi^{\pm \imath \sqrt{\omega^2-1/4}} + C_3 \frac{1}{\xi} \end{equation}

However, in general ,our equation is not an Euler equation so the ansatz does not work. Yet Mathematica gives the solution readily: \begin{equation} r_\xi = C_\pm \xi^{-1/4} \xi^{\pm\imath 1/2\sqrt{\omega^2-1/4}} F_{0,2}\left[{},{5/2\pm 1/2\sqrt{1-4\omega^2},1\pm\sqrt{1-4\omega^2}};\omega^3 \sqrt{\xi}\right] + C_3 F_{0,2}\left[{},{-1/2+1/2\sqrt{1-4\omega^2},-1/2-1/2\sqrt{1-4\omega^2}};\omega^3 \sqrt{\xi}\right] \frac{1}{\omega^6 \xi} \end{equation} where $F_{0,2}$ is the hypergeometric function.

My question is the following. How do I derive this solution? Is it not strange that the solution is expressed through hypergeometric functions which, as we know, are solutions to a second order ODE whereas our ODE is of the third order.

## 1 Answer

Let us use a method of variation of constants''. Since we know that for small values of $\xi$ the solution behaves as: $C_3/\xi$ why do we not make the constant $C_3$ a variable. Thus we conjecture:

\begin{equation} r_\xi := \frac{f_\xi}{\xi} \end{equation}

Substituting the above into the original ODE yields:

\begin{equation} \frac{-1}{\sqrt{\xi}} \omega^3 f_\xi + 2 \omega^2 \frac{d f_\xi}{d \xi} + 12 \xi \frac{d^2 f_\xi}{d \xi^2} + 8 \xi^2 \frac{d^3 f_\xi}{d \xi^3} == 0 \end{equation}

Substituting for $u = \sqrt{\xi}$ yields an even simpler equation: \begin{equation} -\frac{1}{u} \omega^3 f_u + \frac{1}{u} \omega^2 \frac{d f_u}{du} + u \frac{d^3 f_u}{d u^3} == 0 \end{equation}

This equation is readily solved by series expansion methods $f_u := \sum\limits_{n=0}^\infty a_n u^{n+r}$. The details of this procedure are described in the literature so I will not reiterate them here. It is only worthwhile noting that the resulting iteration equation for the coefficients $a_n$ can indeed be solved since it is a recurrence based on two consecutive values only. In other words the values of $a_{n+1}$ will depend on $a_n$ only rather than on some previous values $a_{n-1},a_{n-2}$, etc. Plugging that final equation into Mathematica we get:

\begin{equation} f_u = C_\pm u^{\frac{1}{2}(3\pm \sqrt{1-4 \omega^2})}F_{0,2}\left[{},1\pm\sqrt{1-4 \omega^2},\frac{5}{2} \pm \frac{1}{2}\sqrt{1-4 \omega^2}; \omega^3 u\right] + C_3 F_{0,2}\left[{},-\frac{1}{2}-\frac{1}{2}\sqrt{1-4 \omega^2}, -\frac{1}{2}+\frac{1}{2}\sqrt{1-4 \omega^2}; \omega^3 u\right] \end{equation}

The above solution matches the answer given in the original question.