Pop_compu - at Rearranging we have x C K xe at = Ce at or x ´ K Ce at K µ = Ce at Solving for x we arrive at the explicit form x = KCe at K Ce at

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Some Computations for The Logistic Equation Here are some of the computations involved in ﬁnding the explicit form of the solution for the logistic equation dx dt = a x ± 1 - x K ² , x (0) = x o . Once the variables are separated and the partial fraction decomposition has been made, the integration is simple and leads to ln ( x ) - ln ³ ³ ³ ³ 1 - x K ³ ³ ³ ³ = a t + c , which reduces to x ( 1 - x K ) = C e at . At this stage, one can compute the value of the integration constant C by using the initial condition. Thus, setting t = 0, we ﬁnd that C = Kx o / ( K - x o ). However, for further computations, it is easier to continue to use C . The object is to ﬁnd an explicit form of the solution. To this end, x = Ce at ± 1 - x K ² = Ce at - C K xe
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Unformatted text preview: at . Rearranging, we have x + C K xe at = Ce at , or x ´ K + Ce at K µ = Ce at . Solving for x we arrive at the explicit form x = KCe at K + Ce at = 1 1 C e at + 1 K . We can now substitute for C and simplify to get x ( t ) = 1 ± K-x o Kx o ² e-at + 1 K = Kx o ( K-x o ) e-at + x o . NOTES: From this explicit form, we can see that if x o = 0 then x ( t ) ≡ 0r, while if x o > then lim t →∞ Kx o ( K-x o ) e-at + x o = Kx o x o = K . So we see that the asymptotic behavior predicted from the analysis of the direction ﬁeld is also evident from the form of the solution....
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This note was uploaded on 12/02/2009 for the course MATH 352 taught by Professor Staff during the Spring '08 term at University of Delaware.

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