ODEs- Series Solutions Near a Regular Singular Point

[Roni1985]

Homework Statement

6x²(x+1)²y^''+0.5x(x+2)y^'+y=0

ii) Find all values of r for which there is a series solution of form
x^r\sum(a_nxⁿ,n=0,inf)
a₀ \neq0

Find all values of r for which there is a series solution of form
inf
x^r\suma_n(x-2)ⁿ
n=0
a₀ \neq0

Do not try to solve the problem, just find the values of r without solving for any constants.

Homework Equations

The Attempt at a Solution

What I tried to do was finding the indicial equation. For the first one I got 2 indicial equations, so I got 4 answers. Does it make sense ?

r=1/2
r=1/3
r=0
r=23/24
Also, if I try to reach the Eular equation, I get only the first two, but not the last two. Have I made any mistakes ?

And, I didn't know how to do the second one. Actually, I can do it but it would take me ages, so I believe there is a way to find r's without trying to reach the indicial equation.

Maybe I can try and reach the Eular equation?

Thanks,
Roni.

 

[vela]

Homework Statement

6x²(x+1)²y^''+0.5x(x+2)y^'+y=0

ii) Find all values of r for which there is a series solution of form
x^r\sum(a_nxⁿ,n=0,inf)
a₀ \neq0

Find all values of r for which there is a series solution of form
inf
x^r\suma_n(x-2)ⁿ
n=0
a₀ \neq0

Do not try to solve the problem, just find the values of r without solving for any constants.

Homework Equations

The Attempt at a Solution

What I tried to do was finding the indicial equation. For the first one I got 2 indicial equations, so I got 4 answers. Does it make sense ?

r=1/2
r=1/3
r=0
r=23/24
Also, if I try to reach the Eular equation, I get only the first two, but not the last two. Have I made any mistakes ?

Your second equation isn't an indicial equation; it just tells you how to calculate a₁ in terms of a₀. Only your first equation, the one involving only a₀, dictates what the allowed values of r are.

And, I didn't know how to do the second one. Actually, I can do it but it would take me ages, so I believe there is a way to find r's without trying to reach the indicial equation.

Maybe I can try and reach the Eular equation?

Think about when you need to use the Frobenius method as opposed to the regular old series solution.

 

[Roni1985]

Your second equation isn't an indicial equation; it just tells you how to calculate a₁ in terms of a₀. Only your first equation, the one involving only a₀, dictates what the allowed values of r are.

The thing is that the second equation is also for a₀ and this is confusing me.
((12(r-1)*r+0.5*r)a₀)x^1+r=0

Think about when you need to use the Frobenius method as opposed to the regular old series solution.

We use Frobenius method near singular regular points...
but I don't think I understand how to use the second one.

 

[vela]

The thing is that the second equation is also for a₀ and this is confusing me.
((12(r-1)*r+0.5*r)a₀)x^1+r=0

Your missing another term that contributes to the coefficient of x^r+1.

 

[vela]

We use Frobenius method near singular regular points...
but I don't think I understand how to use the second one.

Ignore my earlier question. :)

Are you sure you have the correct form for the second series? I'm guessing they wanted you to expand about x=2, so the series should be

(x-2)^r \sum_{n=0}^\infty a_n (x-2)^n

whereas you only have x^r out front.

 

[Roni1985]

Your missing another term that contributes to the coefficient of x^r+1.

I went over it like 5 times, and I don't think I'm missing anything when I group a₀

There are two terms but for a₁...


And, how would you do the second one ?

Thanks.

 

[vela]

I went over it like 5 times, and I don't think I'm missing anything when I group a₀

There are two terms but for a₁...

You don't group by a_n. You group by powers of x.

 

[Roni1985]

Ignore my earlier question. :)

Are you sure you have the correct form for the second series? I'm guessing they wanted you to expand about x=2, so the series should be

(x-2)^r \sum_{n=0}^\infty a_n (x-2)^n

whereas you only have x^r out front.

I have only x^r, it might be a typo, though...
Say we have (x-2)^r, should I go the regular Frobenius method ? Is it going to be like very very long for a question on an exam ? is there a shortcut ?

Thanks.

 

[Roni1985]

You don't group by a_n. You group by powers of x.

hmmm... wow... thanks a lot.. didnt know this ...
appreciate it.

 

[vela]

I have only x^r, it might be a typo, though...
Say we have (x-2)^r, should I go the regular Frobenius method? Is it going to be like very very long for a question on an exam ? is there a shortcut?

I'd try the substitution u=x-2, and then go with the regular Frobenius method with the new differential equation.

 

[Roni1985]

I'd try the substitution u=x-2, and then go with the regular Frobenius method with the new differential equation.

substitution... hmm, very smart :\

I'm going to try it...

Thanks.

 

[vanchanh123]

substitution... hmm, very smart :\

I'm going to try it...

Thanks.

because x=2 is reguler point, this eq has solution power serier.
And we can apply Frobinous theorem