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using logistic differential equatin (reproducing Ti Nspire CAS code)

asked 2016-07-29 09:43:11 +0100

etb gravatar image

updated 2016-07-29 09:55:38 +0100

I’m trying to reproduce some results from Ti Nspire CAS to use a differential equation. Only secondarily am I interested to actually solve the differential equation.

My Ti Nspire codes is as follows

deSolve(y’ = 2*10^(-5)*y(1500-y) and y(0) = 50, t, y)

and yields the following result

y = 1500*e^((3*t)/100) / (e^((3*t)/100) + 29)

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Can anyone give me a hit as to how I can do this in Sagemath?

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answered 2016-07-29 10:56:16 +0100

mforets gravatar image

updated 2016-07-29 10:57:04 +0100

t = var('t')
y = function('y')(t)
ye = desolve(diff(y,t) == 2*10^(-5)*y*(1500-y), y, ics=[0,50])

ye = ye*3/100
yt = solve(ye.simplify_log(), y)
yt

$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left[y\left(t\right) = \frac{1500 e^{\left(\frac{3}{100} t\right)}}{e^{\left(\frac{3}{100} t\right)} + 29}\right] $$

Remark. Here I multiplied by $3/100$, since otherwise the simplify_log() was not factoring the terms containing $y(t)$. For instance,

f = 2*log(x) - 2*log(x-1)
f.simplify_log()

gives $\log(x^2/(x-1)^2)$, but

f = 1/3*log(x) - 1/3*log(x-1)
f.simplify_log()

gives $-\frac{1}{3}\log(x-1) + \frac{1}{3}\log(x)$, instead of $\frac{1}{3}\log(x/(x-1))$ (and in the first case I would prefer to get rather $2\log(x/(x-1))$; neither factor() nor simplify_full() do it -- why?).

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Thanks a lot. I cannot answer your question, but I'll get back to you here if I find an answer to that.

etb gravatar imageetb ( 2016-07-29 11:01:44 +0100 )edit

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Asked: 2016-07-29 09:43:11 +0100

Seen: 583 times

Last updated: Jul 29 '16