Differential-Equations Series Solutions Near Regular SIngular Points r1-r2-is-an-integer, guaranteed and verified 100% Pass

1


Series Solutions Near Regular Singular Points: 𝑟1 − 𝑟2 is an integer


Consider the differential equation:
𝑝(𝑥) 𝑞(𝑥)
𝑦 ′′ + 𝑦′ + 𝑦=0
𝑥 𝑥2

where 𝑝(𝑥) and 𝑞(𝑥) are analytic at 𝑥 = 0 and thus 𝑥 = 0 is a regular singular
point.


We had a theorem that said there are two linearly independent solutions through
a Frobenius series if 𝑟1 ≠ 𝑟2 and 𝑟1 − 𝑟2 is not a positive integer ( 𝑟1 ≥ 𝑟2 are
the real roots of the indicial equation: 𝑟(𝑟 − 1) + 𝑝0 𝑟 + 𝑞0 = 0).



When 𝑟1 = 𝑟2 there can only be one Frobenius series solution.



When 𝑟1 − 𝑟2 = 𝑁, a positive integer, then there may (or may not) be two
linearly independent Frobenius series solutions.



The Nonlogarithmic Case with 𝑟1 = 𝑟2 + 𝑁
Ex. Solve 𝑥𝑦 ′′ + (3 − 𝑥 )𝑦 ′ − 𝑦 = 0.


Dividing by 𝑥 we get:
3−𝑥 𝑥
𝑦 ′′ + 𝑦′ − 𝑦=0
𝑥 𝑥2

So 𝑝(𝑥 ) = 3 − 𝑥 and 𝑝(0) = 3
𝑞 (𝑥 ) = −𝑥 and 𝑞(0) = 0.

, 2


The indicial equation becomes:

𝑟(𝑟 − 1) + 3𝑟 = 0
𝑟 2 + 2𝑟 = 0
𝑟(𝑟 + 2) = 0
𝑟 = 0, −2.
So 𝑟1 = 0, 𝑟2 = −2 and 𝑟1 − 𝑟2 = 2, a positive integer.



Substituting 𝑦 = ∑∞
𝑛=0 𝑐𝑛 𝑥
𝑛+𝑟
into the differential equation:

𝑥𝑦 ′′ + (3 − 𝑥 )𝑦 ′ − 𝑦 = 0


𝑥 ∑∞
𝑛=0(𝑛 + 𝑟)(𝑛 + 𝑟 − 1)𝑐𝑛 𝑥
𝑛+𝑟−2

+(3 − 𝑥 ) ∑∞
𝑛=0(𝑛 + 𝑟)𝑐𝑛 𝑥
𝑛+𝑟−1
− ∑∞
𝑛=0 𝑐𝑛 𝑥
𝑛+𝑟
=0


∑∞
𝑛=0(𝑛 + 𝑟)(𝑛 + 𝑟 − 1)𝑐𝑛 𝑥
𝑛+𝑟−1
+ 3 ∑∞
𝑛=0(𝑛 + 𝑟)𝑐𝑛 𝑥
𝑛+𝑟−1

−𝑥 ∑∞𝑛=0(𝑛 + 𝑟)𝑐𝑛 𝑥
𝑛+𝑟−1
− ∑∞
𝑛=0 𝑐𝑛 𝑥
𝑛+𝑟
=0


∑∞
𝑛=0(𝑛 + 𝑟)(𝑛 + 𝑟 − 1)𝑐𝑛 𝑥
𝑛+𝑟−1
+ ∑∞
𝑛=0 3(𝑛 + 𝑟)𝑐𝑛 𝑥
𝑛+𝑟−1

− ∑∞𝑛=0(𝑛 + 𝑟)𝑐𝑛 𝑥
𝑛+𝑟
− ∑∞
𝑛=0 𝑐𝑛 𝑥
𝑛+𝑟
=0


∑∞
𝑛=0[(𝑛 + 𝑟 )(𝑛 + 𝑟 − 1) + 3(𝑛 + 𝑟)]𝑐𝑛 𝑥
𝑛+𝑟−1

− ∑∞
𝑛=0(𝑛 + 𝑟 + 1)𝑐𝑛 𝑥
𝑛+𝑟
=0


∑∞ 2
𝑛=0[(𝑛 + 𝑟) + 2(𝑛 + 𝑟)]𝑐𝑛 𝑥
𝑛+𝑟−1
− ∑∞
𝑛=1(𝑛 + 𝑟)𝑐𝑛−1 𝑥
𝑛+𝑟−1
= 0.