Logistic equation: Difference between revisions

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(Created page with "Of the form <math>\begin{cases} y' = y(1 - y)\\ y(0)=y_0 \end{cases}</math> The solution is <math>y(t) = \frac{y_0e^t}{y_0 e^t - (y_0 - 1)}</math> = Analysis = ''y'' is undefined at <math>t = \ln \left( \frac{y_0 - 1}{y_0} \right)</math> If ''0 < y_0 < 1,'' The solution is defined for all ''t.'' ''y(t)'' approaches 1 as t approaches infinity. ''y(t)'' approaches 0 as ''t'' approaches negative infinity. For ''y_0 > 1,'' the blowup time is negative <math>\ln \left(...")
 
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[[Category:Differential Equations]]
Of the form
Of the form



Latest revision as of 19:34, 17 May 2024


Of the form

The solution is

Analysis

y is undefined at

If 0 < y_0 < 1, The solution is defined for all t. y(t) approaches 1 as t approaches infinity. y(t) approaches 0 as t approaches negative infinity.

For y_0 > 1, the blowup time is negative

As t approaches positive infinity, y(t) approaches 1.

For y_0 < 0, as t approaches negative infinity, y(t) approaches 0.