Unformatted text preview: Continuous dynamic systems 51 Hence
1
a dx
+
x dx
=
a−x k dt + c0 1
[ln x − ln a − x] = kt + c0
a
x
1
ln
= kt + c0
a
a−x
ln x
= akt + ac0
a−x Taking antilogs, we have
x
= eakt+ac0 = eac0 eakt = ceakt
a−x
where c = eac0 . Substituting for the initial condition, i.e., t = t0 then x = x0 , we
can solve for the constant c, as follows
x0
= ceakt0
a − x0
x0
c=
e−akt0
a − x0
Substituting, then
x
=
a−x
= x0
e−akt0 eakt
a − x0
x0
eak(t−t0 )
a − x0 Solving for x
x0
eak(t−t0 )
a − x0
x=
x0
1+
eak(t−t0 )
a − x0
a Which can be further expressed6
x= ax0
(a − x0 )e−ak(t−t0 ) + x0 From the logistic equation (2.21) we can readily establish the following results,
assuming that x0 is less than a:
1.
2. 6 For t = t0 then x = x0
As t → ∞ then x → a The logistic growth equation is a particular example of the Bernoulli function and can be solved in
a totally different way using a simple transformation. See n. 2 and exercise 6. (2.21) ...
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This note was uploaded on 12/14/2011 for the course ECO 3023 taught by Professor Dr.gwartney during the Fall '11 term at FSU.
 Fall '11
 Dr.Gwartney
 Economics

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