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December 14th, 2017, 06:19 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 192 Thanks: 23  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 06:24 AM. 
December 14th, 2017, 07:16 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 239 Thanks: 126 Math Focus: Dynamical systems, analytic function theory, numerics 
$z = ce^t$ for any constant $c$.

January 13th, 2018, 04:05 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785  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 Country Boy; January 13th, 2018 at 04:46 AM. 

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