My Math Forum Equation solution ?

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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
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Quote:
 Originally Posted by SDK $z = ce^t$ for any constant $c$.

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 "differential-difference" 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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