We continue our study of the method of Frobenius for finding series solutions of linear second order differential equations, extending to the case where the indicial equation has a repeated real root.

### The Method of Frobenius II

In this section we discuss a method for finding two linearly independent Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated real root. As in the preceding section, we consider equations that can be written as

where . We assume that the indicial equation has a repeated real root . In this case Theorem thmtype:7.5.3 implies that (eq:7.6.1) has one solution of the form but does not provide a second solution such that is a fundamental set of solutions. The following extension of Theorem thmtype:7.5.2 provides a way to find a second solution.

- Proof
- Theorem thmtype:7.5.2 implies (eq:7.6.4). Differentiating formally with respect to in (eq:7.6.3) yields
To prove that satisfies (eq:7.6.6), we view in (eq:7.6.2) as a function of two variables, where the prime indicates partial differentiation with respect to ; thus, With this notation we can use (eq:7.6.2) to rewrite (eq:7.6.4) as

where

- Proof
- Since is a repeated root of , the indicial polynomial can be factored as so which is nonzero if . Therefore the assumptions of Theorem thmtype:7.6.1 hold with , and (eq:7.6.4) implies that . Since it follows that , so (eq:7.6.6) implies that This proves that and are both solutions of . We leave the proof that is a fundamental set as an exercise.

Setting in (eq:7.6.11) and (eq:7.6.12) yields

and Computing recursively with (eq:7.6.13) and (eq:7.6.14) yields and Substituting these coefficients into (eq:7.6.10) yields andSince the recurrence formula (eq:7.6.11) involves three terms, it’s not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of (eq:7.6.9). However, as we saw in the preceding sections, the recurrence formula for involves only two terms if either or in (eq:7.6.1). In this case, it’s often possible to find explicit formulas for the coefficients. The next two examples illustrate this.

To obtain in (eq:7.6.17), we must compute for . We’ll do this by logarithmic differentiation. From (eq:7.6.18), Therefore Differentiating with respect to yields Therefore Setting here and recalling (eq:7.6.19) yields

Since (eq:7.6.20) can be rewritten as Therefore, from (eq:7.6.17),To obtain in (eq:7.6.23), we must compute for . Again we use logarithmic differentiation. From (eq:7.6.24), Taking logarithms yields Differentiating with respect to yields Therefore Setting and recalling (eq:7.6.25) yields

Since (eq:7.6.26) can be rewritten as Substituting this into (eq:7.6.23) yieldsIf the solution of reduces to a finite sum, then there’s a difficulty in using logarithmic differentiation to obtain the coefficients in the second solution. The next example illustrates this difficulty and shows how to overcome it.

To obtain in (eq:7.6.29) we must compute for . Let’s first try logarithmic differentiation. From (eq:7.6.30), so Differentiating with respect to yields Therefore

However, we can’t simply set here if , since the bracketed expression in the sum corresponding to contains the term . In fact, since for , the formula (eq:7.6.31) for is actually an indeterminate form at .We overcome this difficulty as follows. From (eq:7.6.30) with , Therefore so

From (eq:7.6.30) with , where Therefore which implies that if . We leave it to you to verify that Substituting this and (eq:7.6.32) into (eq:7.6.29) yields### Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored and Edited Books & CDs. 8. (CC-BY-NC-SA)