Economics Dynamics Problems 67

Economics Dynamics Problems 67 - Continuous dynamic systems...

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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 anti-logs, 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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