# How do you find a Power Series solution of a linear differential equation?

To find a solution of a linear ordinary differential equation

${a}_{n} \left(x\right) {y}^{\left(n\right)} + {a}_{n - 1} {y}^{\left(n - 1\right)} \cdots + {a}_{1} \left(x\right) {y}^{p} r i m e + {a}_{0} \left(x\right) y = 0$

around the point ${x}_{0}$, we must first evaluate what power series method we should use.

• If the point ${x}_{0}$ is an ordinary point for the differential equation, that is, all ${a}_{i} \left(x\right)$ are analytic around ${x}_{0}$ (their Taylor Series around ${x}_{0}$ has a non zero convergrence radius), then we can use the ordinary power series method, described below.

• If the point ${x}_{0}$ is a regular singular point for the differential equation, that is, ${x}^{i} {a}_{i} \left(x\right)$ are analytic around ${x}_{0}$, then we should use the Frobenius method (which will not be described in detail, due to it being more complicated).

• If the point ${x}_{0}$ is an irregular singular point, nothing can be said about the solutions of the differential equation.

For the ordinary power series method, start by assuming the solution of the differential equation to be of the form

$y \left(x\right) = {\sum}_{k = 0}^{\infty} {c}_{k} {\left(x - {x}_{0}\right)}^{k}$

Compute the $i$th derivative of $y \left(x\right)$:

${y}^{\left(i\right)} \left(x\right) = {\sum}_{k = 0}^{\infty} {c}_{k} \left(k - i + 1\right) \left(k - i + 2\right) \cdots k {\left(x - {x}_{0}\right)}^{k - i} = {\sum}_{k = i}^{\infty} {c}_{k} \left(k - i + 1\right) \left(k - i + 2\right) \cdots k {\left(x - {x}_{0}\right)}^{k - i}$

Applying the computed derivatives to the differential equation should give a recurrence relation for the coefficients ${c}_{k}$.

Solving that recurrence relation should give at least one solution for the differential equation that lies in the interval $\left({x}_{0} - R , {x}_{0} + R\right)$, where $R$ is the convergence radius of the power series solution found.

This method is generally used for differential equations with polynomial coefficients (that is, ${a}_{i} \left(x\right)$ are polynomials). For ODEs with non-polynomial coefficients, it would be necessary to expand the coeffucient functions ${a}_{i} \left(x\right)$ into their Taylor series around ${x}_{0}$ and multiply them by the Taylor series for ${y}^{\left(i\right)} \left(x\right)$, wich takes considerable effort.

A simple example (generally solved by more elementary methods) to illustrate the recurence relations that appear for the coefficients ${c}_{k}$:

Finding the solution around ${x}_{0} = 0$ of the ODE:

${y}^{p} r i m e \left(x\right) - y \left(x\right) = 0$

Computing the derivatives and applying them to the DE, we get:

${\sum}_{k = 1}^{\infty} k {c}_{k} {x}^{k - 1} - {\sum}_{k = 0}^{\infty} {c}_{k} {x}^{k} = 0$

Changing the index of the first sum using the relation $j = k - 1$ we get:

${\sum}_{j = 0}^{\infty} \left(j + 1\right) {c}_{j + 1} {x}^{j} - {\sum}_{k = 0}^{\infty} {c}_{k} {x}^{k} = 0$

Since $k$ and $j$ are just indexes, we can, using new indexes $l$ rearange the equation above as:

${\sum}_{l = 0}^{\infty} \left[\left(l + 1\right) {c}_{l + 1} - {c}_{l}\right] {x}^{l} = 0$

Which is vallid if and only if

$\left(l + 1\right) {c}_{l + 1} - {c}_{l} = 0 \iff {c}_{l + 1} = {c}_{l} / \left(l + 1\right)$

For the $i$th coefficient ${c}_{i}$:

c_i = c_(i-1)/(i) = c_(i-2)/(i (i-1)) = cdots = c_0/(i!)

Therefore:

y(x) = c_0 sum_(l=0)^(oo) 1/(l!) x^l

$y \left(x\right) = {c}_{0} {e}^{x} ,$

wich is the well known solution for this problem.