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Help with x=exp(t)

  1. Oct 27, 2007 #1

    cks

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    1. The problem statement, all variables and given/known data

    [tex] x=exp(t) [/tex]

    2. Relevant equations

    Prove that [tex] x^2\frac{d^2}{dx^2}=(\frac{d}{dt})(\frac{d}{dt}-1) [/tex]

    3. The attempt at a solution

    Let a tested function y
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2007 #2

    Gib Z

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    [tex]\frac{d}{dt}[/tex] makes no sense...
     
  4. Oct 27, 2007 #3

    cks

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    why you say it doesn't make sense. Actually,

    it's the method to solve the linear different equation of the form

    [tex] b_0x^nD^n+b_1x^n^-1D^n^-^1+... +b_n)y=R(x) [/tex]

    it can be solved by letting x=exp(t)

    with [tex] xD=D_t [/tex]

    [tex] x^2D^2=D_t(D_t-1) [/tex]

    but I don't know how to get it.
     
  5. Oct 27, 2007 #4

    Gib Z

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    Derivative of WHAT with respect to t?
     
  6. Oct 27, 2007 #5

    cks

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    It's an operator w.r.t x and we can find another equivalent operator that is w.r.t t
     
  7. Oct 27, 2007 #6

    arildno

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    Well, Let G(x) be a function, F(t) another function, so that G(x=exp(t))=F(t).

    Then we have:
    [tex]\frac{dF}{dt}=\frac{dG}{dx}\frac{dx}{dt}=\frac{dG}{dx}x[/tex]
    [tex]\frac{d^{2}F}{dt^{2}}=\frac{d^{2}G}{dx^{2}}(\frac{dx}{dt})^{2}+\frac{dG}{dx}\frac{dx}{dt}=\frac{d^{2}G}{dx^{2}}x^{2}+\frac{dG}{dx}x=x^{2}\frac{d^{2}G}{dx^{2}}+\frac{dF}{dt}[/tex]
    Rearranging, we get:
    [tex]x^{2}\frac{d^{2}}{dx^{2}}(G)=\frac{d}{dt}(\frac{dF}{dt}-F)=\frac{d}{dt}(\frac{d}{dt}-1)(F)[/tex]
    whereby the operator equality is shown by remembering G(x=exp(t))=F(t).
     
    Last edited: Oct 27, 2007
  8. Oct 27, 2007 #7
    Let us introduce an arbitrary trial function [itex] f(x) [/itex], which can be also written as [itex] f(e^t) [/itex]. Then

    [tex] \frac{dx}{dt} = x [/tex]

    [tex] \frac{d}{dt} f(x) = \frac{df}{dx} \frac{dx}{dt} = x \frac{df}{dx}[/tex]

    [tex] \frac{d^2}{dt^2} f(x) = \frac{d}{dt} (\frac{df}{dx} x) = \frac{d^2f}{dx^2} \frac{dx}{dt} x + \frac{df}{dx} \frac{dx}{dt} = \frac{d^2f}{dx^2} x^2 + \frac{df}{dx} x [/tex]

    [tex] x^2 \frac{d^2f}{dx^2} = \frac{d^2f}{dt^2} - x\frac{df}{dx} = \frac{d^2f}{dt^2} - \frac{df}{dt}[/tex]

    [tex] x^2 \frac{d^2}{dx^2} = \frac{d}{dt} ( \frac{d}{dt} - 1) [/tex]


    Eugene.

    EDIT: OOps! arildno was 5 min. faster than me.
     
    Last edited: Oct 27, 2007
  9. Oct 27, 2007 #8

    Gib Z

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    O damn I suck :( Idiocy conceded =]
     
  10. Oct 27, 2007 #9

    cks

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    I'll read your all answers later, but somehow I still couldn't detect my mistakes.
     
  11. Oct 27, 2007 #10

    Hurkyl

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    x and t aren't independent. You can't just interchange the order of derivatives!

    Try computing
    [tex]\frac{d}{dx} \frac{d}{dt} x[/tex]
    and
    [tex]\frac{d}{dt} \frac{d}{dx} x[/tex]
    for a concrete example!
     
  12. Oct 27, 2007 #11

    arildno

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    Remember, cks, that x and t are related by an EQUATION; they are not independent variables.
     
  13. Oct 27, 2007 #12

    cks

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    ooo, I see. Thank you very much.
     
  14. Oct 28, 2007 #13
    [tex] x^2 \frac{d^2f}{dx^2} = \frac{d^2f}{dt^2} - x\frac{df}{dx} = \frac{d^2f}{dt^2} - \frac{df}{dt}[/tex]

    This has been learning ...

    But i don't get this part ...

    [tex] - x\frac{df}{dx} = - \frac{df}{dt}[/tex]

    Kindly Pellefant ...
     
  15. Oct 28, 2007 #14

    cristo

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    [tex]\frac{d}{dt}=\frac{d}{dx}\frac{dx}{dt}=x\frac{d}{dx}[/tex] from the given relationship between x and t.
     
  16. Oct 28, 2007 #15
    Thank you cristo :)

    Kindly Pellefant!
     
  17. Oct 28, 2007 #16

    cristo

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    You're welcome!
     
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