SKEDSOFT
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 a0 ≠ 0, The only solution is k = α.
Examining now the heigher terms, we have to satiesfy
That is only possible if an = 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 xk-2 and is multiplied by αa0. Provided that α≠0, the only solution would require a0 = 0. But the value of k was determined by requiring that a0 ≠ 0. These two conditions are in mutual contradiction and so there is no power series solution in x.
