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Math Help - Laplace inverse problem

  1. #1
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    Laplace inverse problem

    Transform the differential equation to find a nontrivial solution such that x(0) = 0.

    tx''+(t-2)x'+x=0

    What I have so far after the laplace transform is:

    -2sX-s^2X'-X'-2X-2sX+X=0
     \frac {X'}{X} = - \frac {4s +1}{s^2+1}

    Then taking the integral, I have lnX = -2ln(s^2+1) - tan^{-1}s
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  2. #2
    Junior Member bleesdan's Avatar
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    Try e to the (what you had)
    The X is found, and then you backtrack to x.
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    Quote Originally Posted by tttcomrader View Post
    Transform the differential equation to find a nontrivial solution such that x(0) = 0.

    tx''+(t-2)x'+x=0

    What I have so far after the laplace transform is:

    -2sX-s^2X'-X'-2X-2sX+X=0
     \frac {X'}{X} = - \frac {4s +1}{s^2+1}

    Then taking the integral, I have lnX = -2ln(s^2+1) - tan^{-1}s
    I have no time now but will take a closer look later.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I have no time now but will take a closer look later.
    Just for the record, Mathematica doesn't like either as a closed form solution. (That doesn't always mean that there isn't one, though.)

    -Dan
    Last edited by topsquark; June 27th 2008 at 04:56 AM.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    Transform the differential equation to find a nontrivial solution such that x(0) = 0.

    tx''+(t-2)x'+x=0

    What I have so far after the laplace transform is:

    -2sX-s^2X'-X'-2X-2sX+X=0
     \frac {X'}{X} = - \frac {4s +1}{s^2+1}

    Then taking the integral, I have lnX = -2ln(s^2+1) - tan^{-1}s
    what you got is incorrect! if, as your notation, \mathcal{L}(x(t))=X(s), then since we are given that x(0)=0, we will get:

    \mathcal{L}(tx''(t))=-s^2 X'(s) - 2sX(s), \ \ \ \mathcal{L}(tx'(t))=-sX'(s)-X(s), \ \ \ \mathcal{L}(x'(t))=sX(s). thus the transformation of your

    differential equation is: -(s^2+s)X'(s) - 4sX(s)=0. that will give you X(s)=\frac{C}{(s+1)^4}, where C is any constant.

    so the general solution is: x(t)=kt^3e^{-t}, where k is any constant. Q.E.D.
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