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Math Help - Solve DE using Laplace transform

  1. #1
    MHF Contributor alexmahone's Avatar
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    Solve DE using Laplace transform

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

    tx''+(4t-2)x'+(13t-4)x=0

    My attempt:

    \mathcal{L}\{x'(t)\}=sX(s) and \mathcal{L}\{x''(t)\}=s^2X(s)-x'(0)

    -\frac{d}{ds}[s^2X(s)-x'(0)]-4\frac{d}{ds}[sX(s)]-2sX(s)-13\frac{d}{ds}[X(s)]-4X(s)=0

    -[s^2X'(s)+2sX(s)]-4[sX'(s)+X(s)]-2sX(s)-13X'(s)-4X(s)=0

    (-s^2-4s-13)X'(s)-(2s+8)X(s)=0

    (s^2+4s+13)X'(s)+(2s+8)X(s)=0

    \frac{X'(s)}{X(s)}=-\frac{2s+8}{s^2+4s+13}

    \frac{dX}{X}=-\frac{2s+8}{s^2+4s+13}ds

    \ln X=\int -\frac{2s+8}{s^2+4s+13}ds

    =\int -\frac{2s+8}{(s+2)^2+9}ds

    =-2\int \frac{s+4}{(s+2)^2+9}ds

    =-2\int \left[\frac{s+2}{(s+2)^2+9}+\frac{2}{(s+2)^2+9}\right]ds

    =-2\left[\frac{1}{2}\ln [(s+2)^2+9]+\frac{2}{3}\tan^{-1}\left(\frac{s+2}{3}\right)\right]+K

    =-\ln [(s+2)^2+9]-\frac{4}{3}\tan^{-1}\left(\frac{s+2}{3}\right)+K

    The answer given in the textbook is x(t)=Ce^{-2t}(\sin 3t-3t\cos 3t) but I don't seem to be arriving at it. How do I proceed?
    Last edited by alexmahone; November 17th 2011 at 04:29 AM.
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  2. #2
    A Plied Mathematician
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    Re: Solve DE using Laplace transform

    Quote Originally Posted by alexmahone View Post
    Transform the given differential equation to find a nontrivial solution such that x(0)=0.

    tx''+(4t-2)x'+(13t-4)x=0

    My attempt:

    \mathcal{L}\{x'(t)\}=sX(s) and \mathcal{L}\{x''(t)\}=s^2X(s)-x'(0)

    -\frac{d}{ds}[s^2X(s)-x'(0)]-4\frac{d}{ds}[sX(s)]-2sX(s)-13\frac{d}{ds}[X(s)]-4X(s)=0

    -[s^2X'(s)+2sX(s)]-4[sX'(s)+X(s)]-2sX(s)-13X'(s)-4X(s)=0
    Good so far, but you make an algebra mistake.

    (-s^2-4s-13)X'(s)-(2s+8)X(s)=0
    Should be

    (-s^{2}-4s-13)X'-(4s+8)X=0.

    (s^2+4s+13)X'(s)+(2s+8)X(s)=0

    \frac{X'(s)}{X(s)}=-\frac{2s+8}{s^2+4s+13}

    \frac{dX}{X}=-\frac{2s+8}{s^2+4s+13}ds

    \ln X=\int -\frac{2s+8}{s^2+4s+13}ds

    =\int -\frac{2s+8}{(s+2)^2+9}ds

    =-2\int \frac{s+4}{(s+2)^2+9}ds

    =-2\int \left[\frac{s+2}{(s+2)^2+9}+\frac{2}{(s+2)^2+9}\right]ds

    =-2\left[\frac{1}{2}\ln [(s+2)^2+9]+\frac{2}{3}\tan^{-1}\left(\frac{s+2}{3}\right)\right]+K

    =-\ln [(s+2)^2+9]-\frac{4}{3}\tan^{-1}\left(\frac{s+2}{3}\right)+K

    The answer given in the textbook is x(t)=Ce^{-2t}(\sin 3t-3t\cos 3t) but I don't seem to be arriving at it. How do I proceed?
    Carry through the correction above. It will make a fairly big difference, I think.
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