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Math Help - Differential Equation Using Laplace Transforms

  1. #1
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    i have stalled in another problem.

    3\frac{dy}{dx} + 4y = 4 + 6x + 4x^2

    Given that y=1 when x=0

    3(sL(y)-1) + 4L(y) = \frac{4}{s} + \frac{6}{s^2} + \frac{8}{s^3}

    L(y) = \frac{4}{(s)(3s+4)} + \frac{6}{(s^2)(3s+4)} + \frac{8}{(s^3)(3s+4)} + \frac{3}{(3s+4)}



    Solving using partial fractions:

    A(3s+4) + B(s) = 4
    A=1 B=-3

    C(3s+4) + D(s^2) = 6
    C = \frac{3}{2} D = \frac{27}{8}

    E(3s+4) + F(s^3)
    E = 2 F=\frac{-27}{8}

    Therefore:

    L(y) = \frac{1}{s} - \frac{3}{3s+4} + \frac{3}{2s^2} + \frac{27}{8(3s+4)} + \frac{2}{s^3} - \frac{27}{8(3s+4)} + \frac{3}{3s+4}

    L(y) = \frac{1}{s} + \frac{3}{2s^2} + \frac{2}{s^3}



    y = 1 + x^2 is the solution. however if i take the inverse of L(y) that does not match the solution.
    can you spot where i have gone wrong ?

    -------------------------------------------------------------------------------------

    I seem to be making the same mistake again with this question:

    \frac{d^2x}{dt^2} + 9x = 2sin2t

    Given that:

    s=0 and \frac{ds}{dt} = 1 when t=0

    Taking Laplace Transforms:

    L(x) = \frac{s^2+8}{(s^2+4)(s^2+9)}

    Now solving using partial fractions:

    \frac{A}{s^2+4} + \frac{B}{s^2+9} = s^2+8

    A(s^2+9) + B(s^2+4) = s^2+8

    As^2 + 9A + Bs^2 +4B = s^2+8

    From which:

    A+B=1 and 9A+4B=8

    A=\frac{4}{5} B=\frac{1}{5}

    Hence:

    L(x) = \frac{4}{5(s^2+4)} + \frac{1}{5(s^2+9)}

    The solution gives x=\frac{1}{15}sin3t + \frac{2}{5}sin2t

    What step am i missing out or where have i gone wrong ?
    Last edited by mr fantastic; April 30th 2009 at 06:32 PM. Reason: Merged posts
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  2. #2
    MHF Contributor Calculus26's Avatar
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    you don't need to use PFD write s^2 + 8 as s^2 + 4 + 4 and separate


    Write as [(s^2+4)/[(s^2+4)(s^2+9)] + 4/[(s^2+4)(s^2+9)]


    L(s) = 1/(s^2+9) + 4/(s^2+4)(s^2+9) both of which are standard transforms

    y(t) = 1/3sin(3t) + 4[-1/5*(1/3sin(3t)-1/2sin(2t)]


    = 1/3sin(3t)-4/15sin(3t) + 2/5sin(2t)

    = 1/15sin(3t) + 2/5sin(2t)
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  3. #3
    MHF Contributor Calculus26's Avatar
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    in general you are using the incorrect PFD

    For an irreducible quadratic 1/(s^2+a) the term in the decomposition

    is (As+B)/(s^2+a) not A/(s^2+a)

    It is for linear factors1/(s+a) you have A/(s+a)

    For linear factor1/[(s-a)^n] include A1/(s-a) +A2/(s-a)^2 + ..........+ An/(s-a)^n

    For example for 1/s^2 you would have a A/s and a B/s^2 in the decomposition
    Last edited by Calculus26; May 6th 2009 at 07:54 AM.
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