Series solution about a regular singular point (x=0) of xy''-xy'-y=0

[Pinedas42]

Homework Statement

Find the indicial equation and find 2 independent series solutions for the DE:
xy''-xy'-y=0 about the regular singular point x=0

Homework Equations

y=Ʃ(0→∞) C_nx^n+r
y'=Ʃ(0→∞) C_n(n+r)x^n+r-1
y''=Ʃ(0→∞) C_n(n+r)(n+r-1)x^n+r-2

The Attempt at a Solution

Finding the indicial eq.
Stan. form y''-y'-(1/x)y=0

p(x)=x*(-1)=-x
q(x)=x²*(-1/x)=-x

Making a_o and b_o both zero for

r(r-1)+a_or+b_o=0

so r=0,1

Solving for the equation I finish with (I'll skip a few steps, confident in this portion)

C_or(r-1)x^r-1+Ʃ(0→∞) [C_n+1(n+r+1)(n+r)-C_n(n+r+1)]x^n+r

Inside the brackets = 0 so the recurrence relation is

C_n=C_n+1(n+r) , n=0,1,2,3...

For r=1, C_n=C_n+1(n+1)

C_o=C₁, n=0
C₁=C₂(2)=C_o/2, n=1
C₂=C₃(3)=C_o/2*3, n=2
C₃=C₄(4)=C_o/2*3*4, n=3

I conclude y₁=C_o(1+x+x²/2!+x³/3!...)
which is the series for e^x, though our professor wants this in series form.

For r=0, C_n=C_n+1(n)

C_o=0, n=0
C₁=C₂, n=1
C₂=C₃(2)=C₁/2, n=2
C₃=C₄(3)=C₁/2*3, n=3

I'm not to sure how put this into summing terms, the zero is throwing me off.
I'd like to know if I'm on the right track with this. I feel like I did everything as I was supposed to, but something is giving me gut feeling that some portion is erroneous.

Thanks for the help, if you choose to lend it to this tedious problem lol :zzz:

 

[Pinedas42]

Bumpity. Does anyone even have perhaps a hint that something is wrong?

 

[vela]

Homework Statement

Find the indicial equation and find 2 independent series solutions for the DE:
xy''-xy'-y=0 about the regular singular point x=0

Homework Equations

y=Ʃ(0→∞) C_nx^n+r
y'=Ʃ(0→∞) C_n(n+r)x^n+r-1
y''=Ʃ(0→∞) C_n(n+r)(n+r-1)x^n+r-2

The Attempt at a Solution

Finding the indicial eq.
Stan. form y''-y'-(1/x)y=0

p(x)=x*(-1)=-x
q(x)=x²*(-1/x)=-x

Making a_o and b_o both zero for

r(r-1)+a_or+b_o=0

so r=0,1

Solving for the equation I finish with (I'll skip a few steps, confident in this portion)

C_or(r-1)x^r-1+Ʃ(0→∞) [C_n+1(n+r+1)(n+r)-C_n(n+r+1)]x^n+r

Inside the brackets = 0 so the recurrence relation is

C_n=C_n+1(n+r) , n=0,1,2,3...

For r=1, C_n=C_n+1(n+1)

C_o=C₁, n=0
C₁=C₂(2)=C_o/2, n=1
C₂=C₃(3)=C_o/2*3, n=2
C₃=C₄(4)=C_o/2*3*4, n=3

I conclude y₁=C_o(1+x+x²/2!+x³/3!...)
which is the series for e^x, though our professor wants this in series form.

Your conclusion is wrong. You can see this if you plug e^x into the original differential equation. It's not a solution. Remember you're working on the case where r=1. What's the power of x in the lowest-order term?

For r=0, C_n=C_n+1(n)

C_o=0, n=0
C₁=C₂, n=1
C₂=C₃(2)=C₁/2, n=2
C₃=C₄(3)=C₁/2*3, n=3

I'm not to sure how put this into summing terms, the zero is throwing me off.
I'd like to know if I'm on the right track with this. I feel like I did everything as I was supposed to, but something is giving me gut feeling that some portion is erroneous.

C₀ can't be equal to 0. By definition, it's the coefficient of the lowest-order non-vanishing term. Because the two values of r differ by an integer, this method won't give you a second independent solution. You'll have to find the second solution another way.