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ODE - indeterminate coefficients

  1. Jan 7, 2006 #1
    Hi all,

    I can't figure out where I've doing mistake in this quite simple problem:

    [tex]
    y''' - y = x^3 - 1
    [/tex]

    FSS = [itex][e^{x}, e^{-\frac{1}{2}x} \cos \left(\frac{\sqrt{3}}{2}x\right), e^{-\frac{1}{2}x} \sin \left(\frac{\sqrt{3}}{2}x\right)][/itex]

    Then, because on the right side there is quasipolynom, I can solve it using the method of indeterminate coefficients.

    So I look for solution of form

    [tex]
    y(x) = a\sin (0x) + b\cos (0x) = b
    [/tex]

    [tex]
    y' = b'
    [/tex]

    My condition will be

    [tex]
    b' = 0
    [/tex]

    So

    [tex]
    y'' = y''' = 0
    [/tex]

    And I get

    [tex]
    y''' - y = -b = x^3 - 1
    [/tex]

    [tex]
    b = 1 - x^3
    [/tex]

    Anyway, this is not correct...can you see where am I doing the mistake?

    Thank you!
     
  2. jcsd
  3. Jan 7, 2006 #2

    TD

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    Homework Helper

    The RHS is a polynomial of degree 3 so you're looking for a (particular) solution of the form [itex]y = ax^3+bx^2+cx+d[/itex]. Now also find y''' and plug it into your DE, work it out and group in powers of x to identify the coefficients which will allow you to determine a,b,c,d.
     
  4. Jan 7, 2006 #3
    Thank you! I will try that. Anyway, can you see some flaw in my approach? I think it's maybe more general but should work..what do you think?
     
  5. Jan 7, 2006 #4

    TD

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    Well it's not very clear to me why you would propose a lineair combination of sin and cos as a particular solution if your non-homogenous part is a polynomial. Since you take the angles 0, you end up with just a constant b but that will never be able to form the third power you have at your RHS, which is why your general solution should be of the form of a 3rd degree polynomial.
     
  6. Jan 7, 2006 #5
    Well, in school, we had been told this:

    If the right side is so called quasipolynom, ie. function of form

    [tex]
    f(x) = e^{\alpha x}( P_1(x) \cos (\beta x) + P_2(x) \sin (\beta x))\mbox{ , }P_1\mbox{, }P_2\mbox{ polynoms}
    [/tex]

    Then the solution of similar form exists:

    [tex]
    y(x) = x^{m}e^{\alpha x} (Q_1(x)\sin(\beta x) + Q_2(x)\cos (\beta x))
    [/tex]

    where [itex]m[/itex] is multiplicity of root [itex]\alpha + \beta i[/itex] of polynom [itex]P(\lambda)[/itex] ([itex]m = 0[/itex] if [itex]m[/itex] isn't a root)

    [itex]Q_1[/itex] and [itex]Q_2[/itex] are polynoms of degree [itex]\leq \max(\mbox{deg}(P_1), \mbox{deg}(P_2))[/itex]

    I thought that expressing [itex]y[/itex] as [itex]y = b[/itex] would be ok because b can represent polynom of the desired degree. But it seems that the [itex]a[/itex] and [itex]b[/itex] in the way I use them should be really only coefficients..
     
  7. Jan 7, 2006 #6

    TD

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    Oh I see, well if "b" represents a polynomial of the proper degree (here 3), then it's ok but that wasn't clear from your notation. It seemed to me that b was an unknown, just a coefficient. I think it would be better to write the complete polynomial, since as you saw: you treated b as a constant and not as a polynomial.

    As far as I've been taught, the suggested polynomial Q should be of the same degree as P (which seems more logical to me actually...) not 'less or equal'.
     
  8. Jan 7, 2006 #7
    Great, I see it now, I really treat [itex]b[/itex] as if it were a constant, that's the point.

    Well, maybe it means that [itex]\mbox{deg}(Q_1) = \mbox{deg}(P_1)[/itex] and [itex]\mbox{deg}(Q_2) = \mbox{deg}(P_2)[/itex], that's covered by what I wrote - if you understand it in a way that "for both [itex]P_1[/itex] and [itex]P_2[/itex] holds true that their degree is less or equal to [itex]\mbox{max(deg(}P_1\mbox{), deg(}P_2\mbox{))}[/itex]"
     
  9. Jan 7, 2006 #8

    TD

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    Yes that's possible. Bottom line really is: you need to suggest a polynomial of at least the same (but exactly the same is sufficient) degree for a particular solution.

    Did you manage to finish this DE?
     
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