Frobenius Method - Explanation

Explanation

The Frobenius method tells us that we can seek a power series solution of the form

Differentiating:

Substituting:

\begin{align} & {} \quad z^2\sum_{k=0}^\infty (k+r-1)(k+r)A_kz^{k+r-2} + zp(z) \sum_{k=0}^\infty (k+r)A_kz^{k+r-1} + q(z)\sum_{k=0}^\infty A_kz^{k+r} \\ & = \sum_{k=0}^\infty (k+r-1) (k+r)A_kz^{k+r} + p(z) \sum_{k=0}^\infty (k+r)A_kz^{k+r} + q(z) \sum_{k=0}^\infty A_kz^{k+r} \\ & = \sum_{k=0}^\infty (k+r-1)(k+r) A_kz^{k+r} + p(z) (k+r) A_kz^{k+r} + q(z) A_kz^{k+r} \\ & = \sum_{k=0}^\infty \left A_kz^{k+r} \\ & = \left A_0z^r+\sum_{k=1}^\infty \left A_kz^{k+r} \end{align}

The expression

is known as the indicial polynomial, which is quadratic in r. The general definition of the indicial polynomial is the coefficient of the lowest power of z in the infinite series. In this case it happens to be that this is the rth coefficient but, it is possible for the lowest possible exponent to be r − 2, r − 1 or, something else depending on the given differential equation. This detail is important to keep in mind because one can end up with complicated expressions in the process of synchronizing all the series of the differential equation to start at the same index value which in the above expression is k = 1. However, in solving for the indicial roots attention is focused only on the coefficient of the lowest power of z.

Using this, the general expression of the coefficient of zk + r is

These coefficients must be zero, since they should be solutions of the differential equation, so

The series solution with Ak above,

satisfies

If we choose one of the roots to the indicial polynomial for r in Ur(z), we gain a solution to the differential equation. If the difference between the roots is not an integer, we get another, linearly independent solution in the other root.

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