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December 18th, 2016, 03:09 AM   #1
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Difference equations



Convert to difference equations
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December 18th, 2016, 03:26 AM   #2
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Math Focus: Yet to find out.
Any particular method?
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December 18th, 2016, 03:52 AM   #3
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Like these
Forward and Backward Euler Methods
https://en.wikipedia.org/wiki/Euler_method
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December 18th, 2016, 03:53 AM   #4
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y΄->dy/dx
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December 18th, 2016, 06:28 PM   #5
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Math Focus: Yet to find out.
Using the Euler backward method,

$\displaystyle \frac{y(k + 1) - y(k)}{x_{k + 1} - x_k} = - a y(k), k \in \mathbb{Z}$

Where $T = x_{k + 1} - x_k$ is the sampling period.

$y(k + 1) = y(k)(1 - Ta)$

The first few values of $k$ gives,

$y(1) = y(0)(1 - Ta) = (1 - Ta)$
$y(2) = y(1)(1 - Ta) = (1 - Ta)^2$
${\vdots}$
$y(k + 1) = y(k)(1 - Ta)^{k + 1}$

Not sure about this last part though , maybe wait to see what the experts say .
Thanks from pass92

Last edited by Joppy; December 18th, 2016 at 06:48 PM. Reason: swapped x for t from post #4
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