What is power series solution of y''-xy

[cloud360]

Homework Statement

[PLAIN]http://img196.imageshack.us/img196/5241/recurrenceq.gif

Homework Equations

The Attempt at a Solution

This is my attempted solution:

1) i got a recurrence relation (n+2)(n+1)a_(n+2)=a_(n-1)
2) i also used the matching coefficients method to get a2=0, a5=0, a7=0, but can't find ones in between
3) I tried finding Ueven and Uodd, so i could join them together as a general solution u=Aueven + Auodd
3.1) For Ueven: let a1=0, a0=1 (i assumed this, i don't know if i was right to do that), then got
a0=1
a1=0
a2=0 (from matching coefficients)
a3=a0/6 -->a3=1/6 (from n=1)
a4=a1/12-->a4=0 (from n=2)
a5=a2/20-->a5=0 (from n=3)
a6=a3/30-->a6=1/180 (from n=4)
a7=a4/42-->a7=0 (from n=5)
a8=a5/56-->a8=0 (from n=6)
a9=a6/72-->a9=1/12960 (from n=7)
I don’t know nth term

3.2) Uodd: let a1=0, a0=1
a0=0
a1=1
a2=0 (from matching coefficients)
a3=a0/6 -->a3=0 (from n=1)
a4=a1/12-->a4=1/12 (from n=2)
a5=a2/20-->a5=0 (from n=3)
a6=a3/30-->a6=1/180 (from n=4)
a7=a4/42-->a7=1/504 (from n=5)
I don’t know nth termGeneral solution is AUeven+BUodd?

 

[LCKurtz]

Your solution looks good, but I wouldn't call the two solutions Ueven and Uodd, because they aren't even or odd. Your recurrence gives the each coefficient in terms of the coefficient 3 preceding it. So the powers on x for each solution will increase by 3 each term, giving both even and odd terms in each.

What you might do is, instead of letting a₀ and a₁ be alternatively 1 and 0 is just leave them in the solution, expressing all the other coefficients in terms of them. Then you can factor out the a₀ from the terms containing it and a₁ from the terms containing it, and the remaining factors will be the two independent series you want.

 

[cloud360]

Your solution looks good, but I wouldn't call the two solutions Ueven and Uodd, because they aren't even or odd. Your recurrence gives the each coefficient in terms of the coefficient 3 preceding it. So the powers on x for each solution will increase by 3 each term, giving both even and odd terms in each.

What you might do is, instead of letting a₀ and a₁ be alternatively 1 and 0 is just leave them in the solution, expressing all the other coefficients in terms of them. Then you can factor out the a₀ from the terms containing it and a₁ from the terms containing it, and the remaining factors will be the two independent series you want.

thanks for the reply. but can you tell me how i could find the nth term. i have collected the terms (of the the series i incorrectly called Ueven and Uodd)

Ueven= 1,0,0,1/6 ,0 ,0 ,1/180 ,0 ,0 ,1/12960

Uodd= 0,1,0 ,0 ,1/12 ,0 ,1/180 ,1/504
[PLAIN]http://img684.imageshack.us/img684/9696/sequence.gif

and to me, a6 and a7 on Uodd don't look right, what is your opinion, given the recurrence relation

(n+2)(n+1)a_n+2=a_n-1

 

[cloud360]

Correction: a6=0, so series is actually !

Ueven= 1,0,0,1/6 ,0 ,0 ,1/180 ,0 ,0 ,1/12960

Uodd= 0,1,0 ,0 ,1/12 ,0 ,0 ,1/504

 

[LCKurtz]

thanks for the reply. but can you tell me how i could find the nth term. i have collected the terms (of the the series i incorrectly called Ueven and Uodd)

Ueven= 1,0,0,1/6 ,0 ,0 ,1/180 ,0 ,0 ,1/12960

Uodd= 0,1,0 ,0 ,1/12 ,0 ,1/180 ,1/504
[PLAIN]http://img684.imageshack.us/img684/9696/sequence.gif

and to me, a6 and a7 on Uodd don't look right, what is your opinion, given the recurrence relation

(n+2)(n+1)a_n+2=a_n-1

The secret to finding the pattern for the coefficients, assuming there is a recognizable one is to not multiply the numbers out. For example in this series you have:

a₄ = a₁/(4*3)

a₇ = a₄/(7*6) = a₁/((7*6)(4*3))

You can probably see what a₁₀, a₁₃,... are going to be. You may have to write the general term using the ellipsis ...and ending with the last factor.

 

[cloud360]

The secret to finding the pattern for the coefficients, assuming there is a recognizable one is to not multiply the numbers out. For example in this series you have:

a₄ = a₁/(4*3)

a₇ = a₄/(7*6) = a₁/((7*6)(4*3))

You can probably see what a₁₀, a₁₃,... are going to be. You may have to write the general term using the ellipsis ...and ending with the last factor.

what do you mean by elipses? can you please elaborate a little ?

 

[LCKurtz]

what do you mean by elipses? can you please elaborate a little ?

An ellipsis is the three dots: ... meaning continue the pattern.

You might write the denominator for the "nth" term

(3*4)(6*7)...( )( )

where the last two parentheses contain the formula for the "nth" term, which you have to figure out. Something like (n)(n+1) -- not those -- but whatever the last factors would be in your problem.

 

[cloud360]

An ellipsis is the three dots: ... meaning continue the pattern.

You might write the denominator for the "nth" term

(3*4)(6*7)...( )( )

where the last two parentheses contain the formula for the "nth" term, which you have to figure out. Something like (n)(n+1) -- not those -- but whatever the last factors would be in your problem.

this is already in the recurrence relation as (n+2)(n+1)

but i don't know how to convert a_n-1i.e the full recurrence relation is a_n-1/(n+2)(n+1)

i already got the denominator of the nth term right, its (n+2)(n+1)? but i don't know what the numerator is, please kindly help !i know the denominator of the terms i called Ueven goes like this

(3*2), (3*2)(6*5), (3*2)(6*5)(9*8), (3*2)(6*5)(9*8)(12*11)...

do i have to use the product function i.e that pi symbol thing?

 

[LCKurtz]

I'm not going to do it for you, but I will give you an example. Suppose you were given this sequence of coefficients, which I have made similar to your problem:

a₂ = (1*3)
a₄ = (1*3)(5*7)
a₆ = (1*3)*(5*7)*(9*11)

and so on, and you need to write the general term with even subscript. First you would have to write the subscript itself which you would write as a_2n.

Then you have to figure out what its formula is. You would write it like this:

a_2n=(1*3)*(5*7)*(9*11)...(?*?)

The ... means similar factors up to the end. The problem is what are the last factors for a_2n. Look at the pattern for the last factor. You might notice that the last factor in this example is always twice the subscript on the a less 1. The subscript of our last term is 2n, so twice that minus 1 is 4n-1. And the factor preceding it is two less which is 4n-3. So you would write

a_2n=(1*3)*(5*7)*(9*11)...(4n-3)*(4n-1)

Now check it with, for example, n = 3. You get a₆ and the factors go from 1*3 up to (12-3)*(12-1), so it works. That is what you need to do with your problem to get the general term.

 

[cloud360]

here is my solution, but i am not confident that this is the right answer to what the question asking

[PLAIN]http://img829.imageshack.us/img829/2788/s0ll.gif

 

[LCKurtz]

Very nice. I think you have it unless I have overlooked something minor. And you have done more than the original question asked by figuring out the general term. If you now pick a₀ and a₁ alternately 0 and 1 you will have the two linearly independent solutions required.