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Thread: Difference equations

  1. #1
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    Difference equations

    Difference equations-diff.png

    Convert to Difference equations
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  2. #2
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    Re: Difference equations

    y΄->dy/dx
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  3. #3
    Senior Member zzephod's Avatar
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    Re: Difference equations

    Quote Originally Posted by jon19 View Post
    Click image for larger version. 

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    Convert to Difference equations
    Please provide more context, like what are you trying to do, what part of your course is this from.

    One of the most common reasons for such a process is related to the numericali ntegration of ODEs, but there is no unique way of doing this.

    The simplest method is to use that approximation:

    $$
    f(x+h)=f(x)+hf'(x)
    $$

    So defining $\Delta_h f(x)=f(x+h)-f(x)=hf'(x)$.

    But these are approximations of a continuous process by discrete.

    As I said before we need more context.

    .
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    Re: Difference equations

    Method Euler for the second equation by step h.
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    Re: Difference equations

    For second equation
    y[n + 2] - 2 y[n + 1] + y[n] == -h^2*y[n], y[0] == 1,
    y'[0] == 0}
    is correct?
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  6. #6
    Senior Member zzephod's Avatar
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    Re: Difference equations

    We want to convert the ODE $y''=-y$ into a difference equation. I will use repeated approximation by a difference quotient:

    $$
    y''(x)=\frac{y'(x+h)-y'(x)}{h}=\frac{\frac{y(x+2h)-y(x+h)}{h}-\frac{y(x+h)-y(x)}{h}}{h}=\frac{y(x+2h)-2y(x+h)+y(x)}{h^2}=-y(x)
    $$

    so:
    $$
    y(x+2h)-2y(x+h)+y(x)=-h^2y(x)
    $$

    with initial conditions $y(0)=0,\ y(h)=h+y(0)=h$


    and the rest is algebra
    Last edited by zzephod; Dec 21st 2016 at 01:27 AM.
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