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How to solve this non-homogenous second order ODE?

  • Thread starter AdrianZ
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  • #1
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Homework Statement


[itex]y''-2y+y=xe^xlnx[/itex]

The Attempt at a Solution


I don't know what I should do here because lnx. Is it possible to solve this ODE with undetermined coefficients method? how can I solve it?
 

Answers and Replies

  • #2
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Hi AdrianZ! :smile:

Sounds like a good plan.
Did you try the undetermined coefficients method?

As for the lnx, typically you can integrate that using integration by parts.
 
  • #3
ehild
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Homework Statement


[itex]y''-2y+y=xe^xlnx[/itex]

The Attempt at a Solution


I don't know what I should do here because lnx. Is it possible to solve this ODE with undetermined coefficients method? how can I solve it?
Try variation of parameters. http://en.wikipedia.org/wiki/Variation_of_parameters
It is quite easy in this case.

ehild
 
  • #4
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Hi AdrianZ! :smile:

Sounds like a good plan.
Did you try the undetermined coefficients method?

As for the lnx, typically you can integrate that using integration by parts.
The problem is I don't know what should be the general form that I should consider to determine the coefficients. That's where I'm stuck.

Try variation of parameters. http://en.wikipedia.org/wiki/Variation_of_parameters
It is quite easy in this case.

ehild
I read the wikipedia page, I don't know what that method is, the wikipedia explanations are a bit confusing but I've already solved the ODE using the method where we take yp of the form v1y1+v2y2 and then we find v1 and v2 using a linear system of equations that are dependent to the derivatives of v's.
The thing is that the author has put this problem in the section of U.C method, so I'm wondering if he wants us to solve it using the undetermined coefficients method?
 
  • #5
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Hmm, as I see it both methods are basically the same.
I need a magnifying glass to go over the exact definition to understand the difference.
If I understand correctly the Variation of parameters is more generally put.

So what is it that you are unclear about?
Or what is it that you think you should do?
I don't really understand from what you write.
 
  • #6
ehild
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I read the wikipedia page, I don't know what that method is, the wikipedia explanations are a bit confusing but I've already solved the ODE using the method where we take yp of the form v1y1+v2y2 and then we find v1 and v2 using a linear system of equations that are dependent to the derivatives of v's.
You applied exactly the method of "variation of parameters":smile:

The thing is that the author has put this problem in the section of U.C method, so I'm wondering if he wants us to solve it using the undetermined coefficients method?
Ln(x)has to appear in the solution anyway. You know the solution now, but pretend that you do not know, :smile: and choose a trial function according to it but with unknown constants. But I do not think they meant that.

ehild
 
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  • #7
ehild
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Hmm, as I see it both methods are basically the same.
No, they are basically different.

The method of undetermined coefficients uses trial functions. There are listed such ones for different types of functions on the right hand side in relation with the roots of the characteristic equation of the homogeneous part.See attachment.

ehild
 

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  • #8
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No, they are basically different.

The method of undetermined coefficients uses trial functions. There are listed such ones for different types of functions on the right hand side in relation with the roots of the characteristic equation of the homogeneous part.See attachment.

ehild
Ah, okay.

So for the method of undetermined coefficients, you need to "guess" the particular solution with some undetermined coefficients.
Possibly with the help of a table.

And for the method of Variation of parameters, you consider the coefficient of the homogeneous solution to be a function.

I think that typically, when you work this out, you'll find a particular solution that matches the "guess" that you would otherwise have made.
I guess that with this solution, you can make a very good "educated guess". ;)
 
  • #9
ehild
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Exactly. If the inhomogeneous part is a polynomial or exponential or sine, cosine, (or product of those) the "undetermined coefficient" method is usually easier than the variation of parameters.See the tread as an example https://www.physicsforums.com/showthread.php?t=554093. But I have never seen it applied to a logarithm function.

ehild
 
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  • #10
ehild
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And what is this method called? :tongue:

Let be y=ex Y.
Substituting into the ode, we get the equation Y"=xln(x), so Y=x3(6 ln(x)-5)/36, yp=exx3(6 ln(x)-5)/36.


ehild
 
  • #12
ehild
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  • #13
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And what is this method called? :tongue:

Let be y=ex Y.
Substituting into the ode, we get the equation Y"=xln(x), so Y=x3(6 ln(x)-5)/36, yp=exx3(6 ln(x)-5)/36.


ehild
Isn't that just the method of Variation of parameters? :confused:
 
  • #14
ehild
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In principle, it is. As y1=exp(x) and y2=x exp(x), yp=v1y1+v2y2=exp(x)(Y). In case of equal roots, there is only one parameter to vary.

ehild
 
  • #15
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But y=C ex is the principal homogeneous solution.

So isn't y=Y(x) ex a variation of the parameter? :confused:

Edit: you just beat me to it.
 
  • #16
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Btw, with the homogeneous solution y=C1 ex + C2 x ex, aren't there 2 parameters that you could vary?
 
  • #17
ehild
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excan be factored out. Then [c1(x)+c2(x) x] is a single function of x. I could vary as many parameters as I like, but why if one is enough?

By the way, when we derive the solution with variation of parameters, c1 and c2 are not arbitrary. It is assumed that c1'y1+c2'y2=0.

ehild
 
  • #18
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I see.
 

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