Logistic Equation

Logistic Equation - MATH 42 - Solving the Logistic Equation...

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MATH 42 — Solving the Logistic Equation Here is how to solve the logistic equation y 0 ( t ) = k ± 1 - y ( t ) K ² y ( t ) using the method of separable equations. First, use the general ideas (move everything except the k over to the left hand side, integrate both sides from 0 to T , change variables on the left hand side) as well as partial fractions to get kT = Z y ( T ) y (0) du (1 - u/K ) u = Z y ( T ) y (0) ± 1 u + 1 K - u ² du = ln ³ ³ ³ ³ u K - u ³ ³ ³ ³ ³ ³ ³ ³ y ( T ) y (0) = ln ³ ³ ³ ³ y ( T )( K - y (0)) ( K - y ( T )) y (0) ³ ³ ³ ³ . Next set A = ( K - y (0)) /y (0) for convenience and take the exponential of both sides, which gives e kT = Ay ( T ) K - y ( T ) . Next we’ll isolate y ( T ) by clearing denominators and solving the resulting linear equation for y ( T ). Note there’s a bit of sloppiness taking place here. Read this paragraph only if you’re losing sleep because we’ve lost the absolute value sign! This happened for the following reasons. .. Technically speaking we need to assume y (0) 6 = K and y (0) 6 = 0 for the expression above to make sense. (And
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This note was uploaded on 04/07/2008 for the course MATH 51 taught by Professor Staff during the Winter '07 term at Stanford.

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Logistic Equation - MATH 42 - Solving the Logistic Equation...

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