What is the logistic equation that satisfies the initial condition of (0,4) where the logistic differential equation is $\frac{d y}{d t} = y (1 - \frac{y}{36})$ $dy/dt=y(1-y/36)$ ?

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Answer 1 · 2018-01-21T10:26:52

$y = \frac{36}{1 + 8 e^{- t}}$ $y=36/(1+8e^(-t))$

The general form of the differential equation is:

$\frac{d y}{d t} = k \cdot y (1 - \frac{y}{L})$ $dy/dt=k*y(1-y/L)$

in that form the solution is of the form:

$y = \frac{L}{1 + A e^{- k t}}$ $y=L/(1+Ae^(-kt))$ where $A = \frac{L - y_{0}}{y_{0}}$ $A = (L-y_0)/y_0$ .

(Arrived at through partial fraction decomposition and separation of variables. [See this video for the )

For this problem $k = 1$ $k=1$ , $L = 36$ $L=36$ , and $y_{0} = 4$ $y_0=4$ .

Calculating $A$ $A$ :

$A = \frac{36 - 4}{4} = 8$ $A=(36-4)/4=8$ .

Substituting everything:

$y = \frac{36}{1 + 8 e^{- t}}$ $y=36/(1+8e^(-t))$

What is the logistic equation that satisfies the initial condition of (0,4) where the logistic differential equation is dydt=y(1−y36)dy/dt=y(1-y/36)?