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What strategy to use

by Lyuokdea
Tags: strategy
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Lyuokdea
#1
Aug23-05, 03:29 PM
P: 198
I've been working through an equation for awhile and finally reduced it to a differential equation I have to solve, but I'm not sure how to solve it, the equation is:

[tex]y'' + (At + B)y' + (Ct + D)y = 0[/tex]

Where t is a variable and A..D are constants. I attempted to solve this using taylor approximations and found the iterative relationship:

[tex]a_{n+2} = \frac{B(n+1)a_{n+1} + Ca_{n-1} + (An+D)a_n}{(n+1)(n+2)}[/tex]

But I don't know of any analytical functions that look anything like that. I could of course get a numeric approximation, but I need an actual analytic function. Does anybody have any suggestions on any methods which I should use in order to solve this function?

Thanks in advance,

~Lyuokdea
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MalleusScientiarum
#2
Aug23-05, 03:52 PM
P: n/a
What you've found has to be wrong; it's a second order differential equation and you have three constants of integration. I would recommend trying the series solution again.
Lyuokdea
#3
Aug23-05, 05:28 PM
P: 198
I looked through it again, and i'm still not seeing my error:


[tex] y= \sum_{n=0}a_nt^n [/tex]
[tex] y'= \sum_{n=1}na_nt^{n-1} [/tex]
[tex] y''=\sum_{n=2}n(n-1)a_nt^{n-2} [/tex]

[tex] \sum_{n=2}n(n-1)a_nt^{n-2} + \sum_{n=1}na_nt^{n-1}(At+B) + \sum_{n=0}a_nt^n(Ct+D) = 0 [/tex]


Reindexing yields:

[tex]
\sum_{n=0}(n+2)(n+1)a_{n+2}t^{n} + \sum_{n=1}Ana_nt^{n} + \sum_{n=0}B(n+1)a_{n+1}t^n + \sum_{n=1}Ca_{n-1}t^n + \sum_{n=0}Da_{n}t^n = 0 [/tex]

which seems to yield the iterative formula I gave before, is there something wrong with the math here, maybe I'm just screwing something up.

~Lyuokdea

saltydog
#4
Aug23-05, 05:55 PM
Sci Advisor
HW Helper
P: 1,593
What strategy to use

Quote Quote by Lyuokdea
I've been working through an equation for awhile and finally reduced it to a differential equation I have to solve, but I'm not sure how to solve it, the equation is:

[tex]y'' + (At + B)y' + (Ct + D)y = 0[/tex]

Where t is a variable and A..D are constants. I attempted to solve this using taylor approximations and found the iterative relationship:

[tex]a_{n+2} = \frac{B(n+1)a_{n+1} + Ca_{n-1} + (An+D)a_n}{(n+1)(n+2)}[/tex]

But I don't know of any analytical functions that look anything like that. I could of course get a numeric approximation, but I need an actual analytic function. Does anybody have any suggestions on any methods which I should use in order to solve this function?

Thanks in advance,

~Lyuokdea
How about Laplace Transforms? Recall that if:

[tex]\mathcal{L}\left[y(x)\right]=F(s)[/tex]

then:

[tex]\mathcal{L}\left[xy(x)\right]=-F^{'}(s)[/tex]

and:

[tex]\mathcal{L}\left[xy^{'}(x)\right]=-\frac{d}{ds}\mathcal{L}
\left[y^{'}(x)\right]=-\frac{d}{ds}(sF(s)-y(0))[/tex]

rock and roll

Although you'll end up with a first order ODE in F(s) and the integrating factor may be messy so after that it might be tough unless the initial conditions are simple.

Edit: What are A, B, C, and D and the initial conditions?
saltydog
#5
Aug23-05, 06:18 PM
Sci Advisor
HW Helper
P: 1,593
Alright Lyuokdea I've looked at it. The integration becomes too difficult to analyze via Laplace Transform. I wish to change my recommendation: Use power series.
Lyuokdea
#6
Aug23-05, 10:51 PM
P: 198
Quote Quote by MalleusScientiarum
What you've found has to be wrong; it's a second order differential equation and you have three constants of integration. I would recommend trying the series solution again.
what do you mean three constants of integration? there are four indexing terms in the expression and all four constants appear. I'm not exactly sure what you are talking about.

~Lyuokdea
lurflurf
#7
Aug24-05, 12:08 AM
HW Helper
P: 2,264
Quote Quote by Lyuokdea
what do you mean three constants of integration? there are four indexing terms in the expression and all four constants appear. I'm not exactly sure what you are talking about.

~Lyuokdea
Only 3 indexing terms should appear. The initial conditions will give the first two terms, then the subsequent terms are generated from them.
Lyuokdea
#8
Aug24-05, 12:34 AM
P: 198
I think that problem comes directly from the fact that there is a (Ct+D) in the y term. That leaves four different powers of t in the series a t^(n+1) from the y term down to a t^(n-2) from the y'' term. I'm not sure of a way to get it to not come out that way. Is there something special you are supposed to do to the (Ct + D)y to correct for that?

~Lyuokdea
saltydog
#9
Aug24-05, 06:53 AM
Sci Advisor
HW Helper
P: 1,593
Well, when I shift the index to obtain [itex]x^{n-2}[/itex] for all the summations, I get:

[tex]a_0: \quad\text{arbitrary}[/tex]

[tex]a_1: \quad\text{arbitrary}[/tex]

[tex]a_2=-\frac{Ba_1+Da_0}{2}[/tex]

[tex]n\geq 3:\quad a_n=-\frac{Aa_{n-2}(n-2)+Ba_{n-1}(n-1)+Ca_{n-3}+D a_{n-2}}{n(n-1)}[/tex]

Now, Lyuokdea if you want, you can verify that you get this, then plug it into Mathematica with selected values for all the constants, make sure it agrees with numerical results, then finally try and come up with a nice encapsulated expression for a summation if possible. I've already checked it for:

[tex]y^{''}+(t+1)y^{'}+(t+1)y=0;\quad y(0)=0,\quad y^{'}(0)=1[/tex]

Edit: Know what, this is the third time I correct typos in those expressions up there. Don't want to cause grief for anyone. I'm pretty sure it's correct now.
Lyuokdea
#10
Aug24-05, 11:56 PM
P: 198
I think I got it all, thanks for the help everybody.

~Lyuokdea


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