# Problems On Frobenius Series Solution

**Problems on Frobenius Series Solution:**

**Exampel: 1** Take the first case of;

The right hand side blows up at x = 0 but not too badly. In the notation that we shall use later, there is a right singularities at x = 0. As before we try for a solution of the form:

Looking first at the lowest term, corresponding to n = 0, we see that

But since a_{0} ≠ 0, The only solution is k = α.

Examining now the heigher terms, we have to satiesfy

That is only possible if a_{n} = 0 for n>1, so that the total solution is

**Example 2: **Contrast this with the case of

Where the right hand side blows up a bit faster at x = 0. We call this an irregular singularity. The solution is;

Which has an essential singularities at x = 0 and for which no power series in x is possible in the region. How does this manifest itself in the Frobenius method?

The lowset power of x is x^{k-2 }and is multiplied by αa_{0}. Provided that α≠0, the only solution would require a_{0} = 0. But the value of k was determined by requiring that a_{0} ≠ 0. These two conditions are in mutual contradiction and so there is no power series solution in x.