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Thread: Using Laplace Transform to solve Differential equations

  1. #1
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    Using Laplace Transform to solve Differential equations

    Hi
    The following question i am having trouble solving
    Use Laplace Transforms to solve each system of differential equations

    $\displaystyle \frac{dx}{dt} = -2y+1$
    $\displaystyle \frac{dy}{dt} = 2x-t$

    given x(0) = -1, y(0) = 1

    This is what i have done:

    $\displaystyle sX+2Y = 0$
    $\displaystyle sY-2X=1-\frac{1}{s^2}$

    I used cramers rule to find X and Y
    $\displaystyle Y = \frac{s^2-1}{s(s^2+4)}$
    $\displaystyle Y = \frac{s^2}{s(s^2+4)}-\frac{1}{s(s^2+4)}$
    $\displaystyle Y = cos(2t)-\frac{1}{4}+\frac{cos(2t)}{4}$

    $\displaystyle X = \frac{-2s^2+2}{s^2(s^2+4)}$
    $\displaystyle X = \frac{-2s^2}{s^2(s^2+4)}+\frac{2}{s^2(s^2+4)}$
    $\displaystyle X = -sin(2t) + \frac{2t}{4}-\frac{2sin(2t)}{8}$

    Answer:

    $\displaystyle X = \frac{1}{2}(t - \frac{3sin(2t)}{2})-cos(2t)$
    $\displaystyle Y = \frac{1}{4}(1+3cos(2t))-sin(2t)$

    P.S
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  2. #2
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    You don't need Laplace Transforms...

    Take the second derivative of the first DE to get

    $\displaystyle \displaystyle \frac{d^2x}{dt^2} = -2\frac{dy}{dt}$, which means

    $\displaystyle \displaystyle \frac{d^2x}{dt^2} = -2\left(2x - t\right)$

    $\displaystyle \displaystyle \frac{d^2x}{dt^2} = -4x + 2t$

    $\displaystyle \displaystyle \frac{d^2x}{dt^2} + 4x = 2t$.

    This is a second order linear non homogeneous constant coefficient ODE. You should know how to solve this.
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  3. #3
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    yes i know you can use that method but the question specify the use of Laplace Transform
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