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December 14th, 2017, 06:19 AM   #1
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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.
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December 14th, 2017, 07:16 AM   #2
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$z = ce^t$ for any constant $c$.
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January 13th, 2018, 04:05 AM   #3
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Quote:
Originally Posted by SDK View Post
$z = ce^t$ 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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