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How to solve this non-homogenous second order ODE?

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]y''-2y+y=xe^xlnx[/itex]

    3. 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?
     
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  3. Nov 26, 2011 #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.
     
  4. Nov 26, 2011 #3

    ehild

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

    ehild
     
  5. Nov 26, 2011 #4
    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.

    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?
     
  6. Nov 26, 2011 #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.
     
  7. Nov 26, 2011 #6

    ehild

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    You applied exactly the method of "variation of parameters":smile:

    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
     
    Last edited: Nov 26, 2011
  8. Nov 26, 2011 #7

    ehild

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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
     

    Attached Files:

  9. Nov 26, 2011 #8

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    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". ;)
     
  10. Nov 26, 2011 #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
     
    Last edited: Nov 26, 2011
  11. Nov 26, 2011 #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. Nov 26, 2011 #11

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  13. Nov 26, 2011 #12

    ehild

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  14. Nov 26, 2011 #13

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    Isn't that just the method of Variation of parameters? :confused:
     
  15. Nov 27, 2011 #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
     
  16. Nov 27, 2011 #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.
     
  17. Nov 27, 2011 #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?
     
  18. Nov 27, 2011 #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
     
  19. Nov 27, 2011 #18

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