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

1. May 13, 2012

### Pinedas42

1. The problem statement, all variables and given/known data
Find the indicial equation and find 2 independent series solutions for the DE:
xy''-xy'-y=0 about the regular singular point x=0

2. Relevant equations
y=Ʃ(0→∞) Cnxn+r
y'=Ʃ(0→∞) Cn(n+r)xn+r-1
y''=Ʃ(0→∞) Cn(n+r)(n+r-1)xn+r-2

3. 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 ao and bo both zero for

r(r-1)+aor+bo=0

so r=0,1

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

Cor(r-1)xr-1+Ʃ(0→∞) [Cn+1(n+r+1)(n+r)-Cn(n+r+1)]xn+r

Inside the brackets = 0 so the recurrence relation is

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

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

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

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

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

Co=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:

2. May 13, 2012

### Pinedas42

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

3. May 14, 2012

### vela

Staff Emeritus
Your conclusion is wrong. You can see this if you plug ex 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?

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.

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