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Thread: Help required? Solve the Differential equation by the Laplace transform method"

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    Help required? Solve the Differential equation by the Laplace transform method"

    Hello

    I wonder if you can help? I believe I'm doing this correctly, but iv got caught up in the algebra. Ill list all the question and then as far as I have got.

    Kind regards
    ----------------------------------------------------------------------
    You are required to solve the following differential equation by the Laplace transform method:

    d^2x/dt^2 + 5*dx/dt - 24x = 4t

    with

    x(0) = 2
    x'(0)=9

    (i)
    Take the Laplace transform of the differential equation and solve the resulting equation for X, the Laplace transform of the solution
    Give your answer as a function of only. Omit the "X ="

    d^2x/dt^2 + 5*dx/dt - 24x = 4t
    Taking the Laplace transform

    L(d^2x/dt^2) + 5L(dx/dt) -24L (x) = 4L(t)

    (s^2X - sx(0)-x'(0)) + 5( sX - x(0)) - 24X = 4/s^2
    Sub in values given within questions for x(0) and x'(0)

    (s^2X - 2s - 9) + (5sX - 10) - 24X = 4/s^2
    Expanding brackets out and collecting like terms

    X(s^2+5s-24) = 4/s^2 + 2s +19
    Factorising LHS

    X((s-3)(s+8)) = 4/s^2 + 2s +19

    I can't see where to go next to, iv looked at brining the (s-3)(s+8) to the RHS but I get muddled with all the numbers and it doesn't seem to look correct.

    Thanks again
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    Quote Originally Posted by Davidc2233 View Post
    Hello

    I wonder if you can help? I believe I'm doing this correctly, but iv got caught up in the algebra. Ill list all the question and then as far as I have got.

    Kind regards
    ----------------------------------------------------------------------
    You are required to solve the following differential equation by the Laplace transform method:

    d^2x/dt^2 + 5*dx/dt - 24x = 4t

    with

    x(0) = 2
    x'(0)=9

    (i)
    Take the Laplace transform of the differential equation and solve the resulting equation for X, the Laplace transform of the solution
    Give your answer as a function of only. Omit the "X ="

    d^2x/dt^2 + 5*dx/dt - 24x = 4t
    Taking the Laplace transform

    L(d^2x/dt^2) + 5L(dx/dt) -24L (x) = 4L(t)

    (s^2X - sx(0)-x'(0)) + 5( sX - x(0)) - 24X = 4/s^2
    Sub in values given within questions for x(0) and x'(0)

    (s^2X - 2s - 9) + (5sX - 10) - 24X = 4/s^2
    Expanding brackets out and collecting like terms

    X(s^2+5s-24) = 4/s^2 + 2s +19
    Factorising LHS

    X((s-3)(s+8)) = 4/s^2 + 2s +19

    I can't see where to go next to, iv looked at brining the (s-3)(s+8) to the RHS but I get muddled with all the numbers and it doesn't seem to look correct.

    Thanks again
    So far your algebra looks correct and your idea is sound. You need to isolate $\displaystyle X(s)$

    This gives
    $\displaystyle \displaystyle X(s)=\frac{4}{s^2(s-3)(s+8)}+\frac{2s}{(s-3)(s+8)}+\frac{19}{(s-3)(s+8)}$

    Now you need to use partial fraction decomposition on the RHS side. Then you will be able to find the inverse transform.
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