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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 December 14th, 2017, 05:19 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 708 Thanks: 96 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: 666 Thanks: 437 Math Focus: Dynamical systems, analytic function theory, numerics $z = ce^t$ for any constant $c$. January 13th, 2018, 03:05 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 skipjack; January 18th, 2018 at 07:55 PM. January 18th, 2018, 07:34 PM   #4
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Quote:
 Originally Posted by Country Boy 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.
I'm not sure what you mean by "differential-difference" equation. The usual name for this type of equation is a delay differential equation. This is because the function depends on its derivative at a different moment in time given in this case by a linear lag time.

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. Tags equation, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Differential Equations 1 March 20th, 2017 02:41 AM andrei Algebra 1 March 13th, 2014 06:22 PM mahmut Algebra 9 May 12th, 2013 07:40 PM greg1313 Real Analysis 3 December 5th, 2011 01:13 PM andrei Number Theory 1 December 31st, 1969 04:00 PM

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