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December 18th, 2016, 03:09 AM  #1 
Newbie Joined: Oct 2015 From: London Posts: 19 Thanks: 0  Difference equations Convert to difference equations 
December 18th, 2016, 03:26 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 855 Thanks: 312 Math Focus: Yet to find out. 
Any particular method?

December 18th, 2016, 03:52 AM  #3 
Newbie Joined: Oct 2015 From: London Posts: 19 Thanks: 0  
December 18th, 2016, 03:53 AM  #4 
Newbie Joined: Oct 2015 From: London Posts: 19 Thanks: 0 
yÎ„>dy/dx

December 18th, 2016, 06:28 PM  #5 
Senior Member Joined: Feb 2016 From: Australia Posts: 855 Thanks: 312 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 . Last edited by Joppy; December 18th, 2016 at 06:48 PM. Reason: swapped x for t from post #4 

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