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Multiplicative inverse

  1. Nov 26, 2009 #1
    If I know a n-th order differential equation and a particular solution for it, can I easily construct a separate n-th order differential equation if I want the particular solution to be [tex]\frac{1}{y_p}[/tex]?

    Key word is easily
  2. jcsd
  3. Nov 26, 2009 #2


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    Hi epkid08! :smile:
    What's yp? :confused:
  4. Nov 26, 2009 #3
    The differential equation is

    [tex](x+\alpha)^2y'' + (x + \alpha)y' + a^2y = 0[/tex]

    and the particular solution is

    [tex]y_p =c_1\cos(a\ln(x+\alpha)) + c_2\sin(a\ln(x+\alpha)) [/tex]

    So I want to find a differential equation that has a particular solution [tex]\frac{1}{y_p}[/tex].

    I've found an 2nd order differential equation that fits what I want, but It's messy and I want to know if there's a faster way to find it than what I did, and also if it simplifies into something similar to the above differential equation.
    Last edited: Nov 26, 2009
  5. Nov 26, 2009 #4


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    Isn't that the same as A(x + α)a + B(x + α)-a ?

    And are you sure the equation doesn't begin (x + α)2y'' ?
  6. Nov 26, 2009 #5
    OOPS! Yes, my mistake, it does begin with [tex](x+\alpha)^2y''[/tex].
  7. Nov 26, 2009 #6
    I'm not sure, are you saying [tex]y_p = A(x + \alpha)^a + B(x + \alpha)^{-a}[/tex]?

    Anyways, I'll post the differential equation that fits [tex]1/y_p[/tex], and maybe you can help me simplify it, also this is when alpha equals zero and a equals one:

    [tex]x^2\sin(\frac{\pi}{4}+\ln(x))y'' + 2x\cos(\frac{\pi}{4}+\ln(x))y' + [\cos(\frac{\pi}{4}+\ln(x)) - \sin(\frac{\pi}{4}+\ln(x))]y = 0[/tex]

    It's very similar to the original differential equation, though it seems like it could be simplified!
  8. Nov 27, 2009 #7

    Just let [itex]y_p=\frac{1}{Y}[/itex] and form the DE that is satisfies by Y(x).

    Key word is easy!
  9. Nov 27, 2009 #8


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    Hi epkid08! :smile:

    First thing: those are not particular solutions, they are general solutions …

    If P(y) = 0 has solutions Af(x) + Bg(x),

    then P(y) = q(x) has solutions Af(x) + Bg(x) + h(x), where h(x) is any solution of the whole equation.

    In other words, the general solution of the whole equation is the general solution of the "incomplete" equation (with 0 on the RHS), plus any particular solution of the whole equation.
    oops! :redface:

    i left out an i … A(x + α)ia + B(x + α)-ia.

    Apart from that, yes

    any combination of cos and sin is also a combination of e+i… and e-i…
    well, cosπ/4 = sinπ/4, so you can divide by that throughout (after expanding the brackets), and get either coslnx and sinlnx or just xi and x-i.
    :rofl: nice one, matematikawan! :smile:
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