Lecture Notes in Differential Equations - Bruce E. Shapiro

296 LESSON 30. THE METHOD OF FROBENIUS
By the indicial equation, α 2 + α(p 0 − 1) + q 0 = 0, so that
α = 1 2
[1 − p 0 ± √ (1 − p 0 ) 2 − 4q 0
]
(30.126)
Designate the two roots by α + and α − and define
∆ = α + − α − = √ (1 − p 0 ) 2 − 4q 0 (30.127)
If they are real, then the larger root satisfies 2α + ≥ 1 − p 0 , so that
n − 1 + 2α + + p 0 ≥ n − 1 + 1 − p 0 + p 0 = n > 0 (30.128)
Therefore (30.124) gives a solution for the larger of the two roots.
other root gives
The
n − 1 + 2α − + p 0 = n − 1 + 1 − p 0 − √ (p 0 − 1) 2 − 4q 0 + p 0
(30.129)
This is never zero unless the two roots differ by an integer. Thus the smaller
root also gives a solution, so long as the roots are different and do not differ
by an integer.
If the roots are complex, then
where ∆ ≠ 0 and therefore
α = 1 2 [1 − p 0 ± i∆] (30.130)
n − 1 + 2α + p 0 = n − 1 + 1 − p 0 ± i∆ + p 0 = n ± i∆ (30.131)
which can never be zero. Hence both complex roots lead to solutions.
When the difference between the two roots of the indicial equation is an
integer, a second solution can be found by reduction of order.
Theorem 30.2. (Second Frobenius solution.) Suppose that p(t) and q(t)
are both analytic with some radius of convergence r, and that α 1 ≥ α 2 are
two (possibly distinct) real roots of the indicial equation
α 2 + (p 0 − 1)α + q 0 = 0 (30.132)
and let ∆ = α 1 − α 2 . Then for some set of constants {c 0 , c 1 , ...}
y 1 (t) = (t − t 0 ) α1
∞ ∑
k=0
c k (t − t 0 ) k (30.133)