Help needed solving logistic differential equation with initial conditionsdPdt=P(10−2P)with initial conditions of P(0)=1.

Get the answer to "Help needed solving logistic differential equation with initial..."

Nancy Richmond

2022-03-24

Help needed solving logistic differential equation with initial conditions

\frac{d P}{d t} = P (10 - 2 P)

with initial conditions of

P (0) = 1 .

Theodore Davila

Beginner2022-03-25Added 14 answers

Step 1

\ln | P | - \ln | P - 5 | = 10 t + c

Inspecting the set-up and initial conditions we can reason that

P < 5

. As P approaches

5, P^{'}

goes to 0 and P cannot grow beyond 5.
This lets us resolve the absolute value brackets.

\ln P - \ln (5 - P) = 10 t + c

We might as well resolve our intial conditions.

\ln (1) - \ln 4 = c

C = - \ln 4

Using the rules of logarithms

\ln (\frac{P}{5 - P}) = 10 t - \ln 4

Exponentiate

(\frac{P}{5 - P}) = \frac{e^{10 t}}{4}

And what is left is algebra

4 P = e^{10 t} (5 - P)

(4 + e^{10 t}) P = 5 e^{10 t}

P = \frac{5 e^{10 t}}{4 + e^{10 t}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?