How can I solve this equation?

1. Jun 2, 2012

atomqwerty

Hello,

it goes like this: x^2y'' + 2xy' +(2x+1)y = 1

If I use Laplace Transform... I know that L(y'') = s^2F(s) - sy(0) - y'(0), but... what do I do with the 'x^2' in the L(x^2y'')?

Thanks!

2. Jun 3, 2012

jackmell

How in the world is that a Bessel function? I just don't see it. Was a thread in here recently about that. I don't get it. How about the homogeneous part? Yeah, I'll try little of this, little of that, but then, I'm going to Mathematica's DSolve. That's not cheating. Anyway, it gives me the homogeneous solution in terms of Bessel functions. Can I get that into a Bessel DE then? Don't see it initially. Maybe you can. Alright, suppose we got that solution. Can we then find the non-homogeneous solution via reduction of order or variation of parameter? Yeah, me neither. But I do know trying to find out is the true art of problem solving. :)

Last edited: Jun 3, 2012
3. Jun 3, 2012

atomqwerty

What do you mean?

4. Jun 3, 2012

jackmell

. . . I have utterly failed completely then in my attempt to teach. My fault, not yours.

I can give you some suggestions but there is always a chance of an easier way. I'm no expert. How about this then: let's look only at the homogeneous equation first:

$$x^2 y''+2 x y'+(2x+1)y=0$$

We could solve it via power-series and that looks like a good problem for that if you need practice with that method. Me, I just used Mathematica's DSolve function to solve it just to get a handle on a solution. Mathematica gives the solution in terms of Bessel functions which are solutions to the Bessel DE:

$$t^2 w''+t w'+(t^2-n^2)w=0$$

I'm pretty sure that's the Bessel DE. Look it up to make sure. But does that mean the equation in x can be converted to a Bessel DE in terms of t? I think so.

Ok then, these are starts. That's what it takes in math, starts, little of this, little of this, try this try that and learn to expect that often what you try doesn't work so you try some more: good cooks try again.

Last edited: Jun 3, 2012
5. Jun 5, 2012

JJacquelin

Hi !

Jackmell is right : The ODE if of the Bessel kind, but in fact not in a standard Bessel form.
So, we look if there are solutions on a more general form :
y(x) = g(x)*f( h(x) )
where g might be an elementary function, f a Bessel function and h an elementary function.
If no, we will have to search for something else.
If yes, OK, the problem is solved. Fortunately, that is the case
(In attachment, i will not detail in order to avoid receiving a warning and have my post deleted as it happened recently in a similar case)

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6. Jun 5, 2012

jackmell

Ok, that's beautiful Jacquelin really, but if I'm teachin' it, that just won't do. I mean what's a complex-order Bessel function anyway (I think you need the conjugate form the $-i\sqrt{3}$ in the final expression also). Suppose I'm workin' on a real-life project, a dam, a rocket, a building, whatever with real stuf in it and I got this equation. Surely can't just use an imaginary solution for the length of a beam right? Suppose I have a IVP with real initial conditions, what's the (real) solution then? And that's just the homogeneuous part. Suppose the assignment was to derive an "expression" for the non-homogeneous equation, even if it's ugly, can he?

I'm just sayin' ok. It's a whole new problem good enough for a new thread. All I mean is if we stop here, we're just lookin' at the shadow of the object and not seein' the object itself. :)

Last edited: Jun 5, 2012
7. Jun 5, 2012

JJacquelin

OK, there is a typo in the last formula : the sign - is missing in the order of the second Bessel. The formuia is corrected in next attachment.
Don't be afraid by the notation "imaginary order". In the standard Bessel ODEs, there is a parameter which, centuries ago, has been noted n². So, if this parameter is negative, now we say "imaginary order". But real solutions exist as well (attachment).
If you use softwares such as Matematica, WolframAlpha, or others, you can compute Bessel functions of real or imaginary orders exactly on the same manner.

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8. Jun 6, 2012

jackmell

Aren't all the solutions complex:

$$z^2 \frac{d^2 w}{dz^2}+2z \frac{dw}{dz}+(2z+1)w=0$$

9. Jun 6, 2012

JJacquelin

No. The sum of a complex solution with the conjugate complex solution is a real solution of the equation.

10. Jun 6, 2012

JJacquelin

If one is working on a real-life project, a dam, a rocket, a building, whatever with real stuf in it and got this equation, he will probably use numerical methods. That is the common way in engeenering.
If he is more theoricien, he will analytically solve it and obtain awfull formulas with a lot of beautiful special functions (which is smart in a report). Then, in practice, he will have to compute those functions thanks to numerical means. Well, in any case all ends with numerical computation. Is there an advantage to appeal to special functions ?
They are several. A very important one is that, knowing the name of a convenient special function gives access to a large background well related to the problem.
For example, in the case of the non-homogeneous equation, if I say that a particular solution is related to the Struve functions (attachment), I bet that searching in the litterature related to the Struve functions will show that someone else already had worked on problems similar to the atomqwerty's problem and, may be, had already solved his problem.
Note : I suggest to make the same study in the case x<0. Then, sqrt(8x) is replaced by sqrt(-8x). With some other little changes in the equations. And why not real order sqrt(3) instead of complex order i*sqrt(3) ?
In all cases, x<0 or x>0, of course, the real solutions will be the same solutions than whose obtained by direct numerical solving of the ODE.

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