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

  • #1
Pinedas42
12
0

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→∞) Cnxn+r
y'=Ʃ(0→∞) Cn(n+r)xn+r-1
y''=Ʃ(0→∞) Cn(n+r)(n+r-1)xn+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 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:
 
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  • #2
Bumpity. Does anyone even have perhaps a hint that something is wrong?
 
  • #3
Pinedas42 said:

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

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.
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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