Solve y'' + (At + B)y' + (Ct + D)y = 0: Strategies

  • Thread starter Lyuokdea
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In summary: I can't seem to get an analytical function that looks anything like what I've found. If somebody could provide a link or whatever to a website or paper or something that provides an analytic function for this type of equation, that would be much appreciated. Thanks.In summary, you are trying to solve an equation for which you do not have an analytic function. You have tried taylor approximations, but they have not yielded the result you were hoping for. You might try using Laplace transforms, but this may be too difficult. Finally, you may be able to generate the series solution by reindexing the variables.
  • #1
Lyuokdea
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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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  • #2
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.
 
  • #3
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
 
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  • #4
Lyuokdea said:
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?
 
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  • #5
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. :smile:
 
  • #6
MalleusScientiarum said:
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
 
  • #7
Lyuokdea said:
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.
 
  • #8
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
 
  • #9
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.
 
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  • #10
I think I got it all, thanks for the help everybody.

~Lyuokdea
 

1. What is the meaning of the terms in the differential equation?

The terms in the differential equation represent the rate of change of the dependent variable y with respect to the independent variable t. The coefficients A, B, C, and D represent the constants that affect the rate of change.

2. How do I identify the type of differential equation?

The type of differential equation can be identified by the highest order of the derivative present. In this case, the highest order is the second derivative, making it a second-order differential equation.

3. What strategies can I use to solve this type of differential equation?

Some common strategies for solving this type of differential equation include using the method of undetermined coefficients, variation of parameters, and solving for the homogeneous solution and particular solution separately.

4. How do I handle initial conditions in this type of differential equation?

To handle initial conditions, you will need to use the solution to the differential equation and plug in the initial values for t and y. This will give you a system of equations that can be solved for the constants in the general solution.

5. Can I solve this differential equation without using numerical methods?

Yes, it is possible to solve this type of differential equation without using numerical methods. However, it may depend on the specific coefficients and initial conditions given in the problem. In some cases, numerical methods may be necessary to obtain an accurate solution.

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