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Thread: Laplace Transform of a DE

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    Super Member craig's Avatar
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    Laplace Transform of a DE

    First time solving DEs this way so bare with me.

    Consider $\displaystyle \ddot{y} + 2 \dot{y} + 5y = u(t)$ with $\displaystyle y_0 = 1$ and $\displaystyle \dot{y}_0 = -1$.

    Using the Laplace Transform of both sides we get:

    $\displaystyle (s^2 Y(s) - sy_0 - \dot{y}_0) + 2(s Y(s) - y_0) + 5Y(s) - U(s)$

    Isolating $\displaystyle Y(s)$ and using the initial conditions, we get:

    $\displaystyle Y(s) = \frac{U(s) + s + 1}{s^2 + 2s + 5}$

    Now in the answers I've got it says let $\displaystyle u(t) = 2t-1$ and they then proceed to do the Laplace Transform of that and substitute it in.

    Just wondering why they've chosen $\displaystyle u(t) = 2t-1$, is this something I've missed?
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  2. #2
    Super Member craig's Avatar
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    Re: Laplace Transform of a DE

    Actually it doesn't matter, turns out this is just following on from another example!!
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