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Thread: Delay differential equation

  1. #1
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    Delay differential equation

    The problem is as follows:

    \frac{d}{dt}x(t) = x(t-1) with x(t) = 1 for t \in [-1,0].

    The solution is given by x(t) = \sum_{k = 0}^{n} \frac{t-(k-1))^k}{k!} for n-1 \leq t \leq n.

    Find the solution using the method of steps.

    I have no idea what the method of steps is and I can't seem to find anything about it online. Any help/hints would be much appreciated!
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  2. #2
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    Re: Delay differential equation

    take a look at

    http://www.mathfile.net/hicstat_fde.pdf

    section 1.1.2

    and yell back if that doesn't guide you to the solution
    Thanks from usersmuser
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  3. #3
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    Re: Delay differential equation

    Quote Originally Posted by usersmuser View Post
    The problem is as follows:

    \frac{d}{dt}x(t) = x(t-1) with x(t) = 1 for t \in [-1,0].

    The solution is given by x(t) = \sum_{k = 0}^{n} \frac{t-(k-1))^k}{k!} for n-1 \leq t \leq n.

    Find the solution using the method of steps.

    I have no idea what the method of steps is and I can't seem to find anything about it online. Any help/hints would be much appreciated!
    I don't see why you would need to use a series solution when it can be solved exactly...

    $\displaystyle \begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= x\left( t - 1 \right) \\ \frac{1}{x} \,\frac{\mathrm{d}x}{\mathrm{d}t} &= t - 1 \\ \int{ \frac{1}{x}\,\frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d} t} &= \int{ \left( t - 1 \right) \,\mathrm{d}t} \\ \int{ \frac{1}{x}\,\mathrm{d}x} &= \frac{t^2}{2} - t + C_1 \\ \ln{ \left| x \right| } + C_2 &= \frac{t^2}{2} - t + C_1 \\ \ln{ \left| x \right| } &= \frac{t^2}{2} - t + C_1 - C_2 \\ \left| x \right| &= \mathrm{e}^{ \frac{t^2}{2} - t + C_1 - C_2 } \\ \left| x \right| &= \mathrm{e}^{C_1 - C_2} \,\mathrm{e}^{\frac{t^2}{2} - t} \\ x &= C \,\mathrm{e}^{\frac{t^2}{2} - t} \textrm{ where } C = \pm \mathrm{e}^{C_1 - C_2} \end{align*}$
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  4. #4
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    Re: Delay differential equation

    You misunderstood the question.
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