Find the roots of the indicial equation

Question

Find the roots of the indicial equation of the differential equation

(x²-1)²y"+(x+1)y'-y=0

At x=-1

Answer 1

The differential equation can be written in the canonical form as

y'' +P(x)y'+Q(x)y=0

where we have defined P(x)=(x+1)/(x²-1)² and Q(x)=-1/(x²-1)²

You can obtain this by simply imposing x not equal to 1 or -1and then divide the whole equation by (x²-1)²

It can be shown that the indicial equation at the singular regular point x=x₀ is given by

λ²+(p₀-1)λ+q₀=0

where

p₀ is the limit as x to x₀ of (x-x₀)P(x) and

q₀ is the limit as x to x₀ of (x-x₀)²Q(x)

Regarding the indicial equation I used the letter λ instead the usual letter ρ to avoid confusion since we have the variable p and q.

We now need to evaluate p₀ and q₀ . In this case we have

p₀ is the limit as x to -1 of (x-(-1))P(x)=(x+1)²/((x+1)²(x-1)²)=1/4

q₀ is the limit as x to -1 of (x-(-1))²Q(x)=-(x+1)²/((x+1)²(x-1)²)=-1/4

We can substitute these values in the indicial equation and solve for λ,

λ²+(p₀-1)λ+q₀=0 ---> λ² -(3/4)λ-1/4=0 which can be rewritten as 4λ²-3λ-1=0

It is then straight forward to evaluate the roots λ=1 and λ=-1/4

Best,

Davide