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December 14th, 2017, 05:19 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87  Equation solution?
$\displaystyle e\frac{dz}{dt}=z(t+1) \; \; $ where $\displaystyle e$ [euler constant] or equation can be written as: $\displaystyle \; eZ'_{t}  Z_{t+1}=0$ Last edited by idontknow; December 14th, 2017 at 05:24 AM. 
December 14th, 2017, 06:16 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 
$z = ce^t$ for any constant $c$.

January 13th, 2018, 03:05 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Hmmm. If $z= ce^t$ then $z'= ce^t$ and $z(t+ 1)= ce^{t+ 1}= ce e^t$ so the equation becomes $ce^t= ce e^t$ so $e= 1$. No, that's not true. This is a "differentialdifference" equation. Typically, one can assign values to an interval of length 1, such as t = 0 to 1, and then solve for other values. For example, if $z(t) = \sin(t)$ for t = 0 to t = 1, then $z(t+ 1)= e \cos(t)$ so that $z(t)= e \cos(t 1)$ for t= 1 to 2, etc. If, instead, $z(t)= t^2$ for t = 0 to 1, then $z(t+ 1)= 2e t$ so that $z(t)= 2e(t 1)$ for t = 1 to 2, etc. Last edited by skipjack; January 18th, 2018 at 07:55 PM. 
January 18th, 2018, 07:34 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
In any case, you are correct that $z = ce^t$ yields $z' = ce^t$ and also $z(t+1) = ce\cdot e^t$. However, the delay equation is $ez' = z(t+1)$ so we note that $ez' = e\cdot ce^t = ce^{t+1} = z(t+1)$ so it satisfies the equation. Last edited by skipjack; January 18th, 2018 at 07:55 PM.  

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