Power series solution method for differential equations

Description of the method

This method aims to find power series for the solution functions to a differential equation. The general idea is as follows:

1. Assume that the solution function has a power series that converges to it. It is typically helpful to write the power series as an exponential generating function, i.e., in the form $\sum _{k=0}^{\infty }a_{k}{\frac {x^{k}}{k!}}$.
2. Compute the power series for the derivatives of the function, and then plug in the differential equation to obtain a recurrence relation between the $a_{k}$s.
3. Solve this recurrence relation. In general, we expect that all the coefficients are determined by the first few, where the number of initial coefficients that we are free to specify equals the order of the differential equation. For an initial value problem, these first few coefficients can be determined.

Examples

Consider the differential equation where the independent variable is $x$ and the dependent variable is $y$:

$y'''=y$

Suppose $y$ has a power series:

$y=\sum _{k=0}^{\infty }a_{k}{\frac {x^{k}}{k!}}$

It is easy to see that:

$y'''=\sum _{k=0}^{\infty }a_{k+3}{\frac {x^{k}}{k!}}$

The equation $y'''=y$ means that these power series are equal coefficient-wise, so we get:

$a_{k}=a_{k+3}\ \forall \ k\in \{0,1,2,\dots \}$

We thus see that $a_{0},a_{1},a_{2}$ can be varied freely, and once these are determined, all the coefficients are determined. Explicitly, we have $a_{0}=a_{3}=a_{6}=\dots$, separately $a_{1}=a_{4}=a_{7}=\dots$, and separately $a_{2}=a_{5}=a_{8}=\dots$. Thus, the power series can be written as:

$a_{0}\left(1+{\frac {x^{3}}{3!}}+{\frac {x^{6}}{6!}}+\dots \right)+a_{1}\left(x+{\frac {x^{4}}{4!}}+{\frac {x^{7}}{7!}}+\dots \right)+a_{2}\left({\frac {x^{2}}{2!}}+{\frac {x^{5}}{5!}}+{\frac {x^{8}}{8!}}+\dots \right),\qquad a_{0},a_{1},a_{2}\in \mathbb {R}$

Comparison with the other method for solving this differential equation: Fill this in later