Help needed solving logistic differential equation with initial conditions

Question

Given information: Solve $$\frac{dP}{dt} = P(10 - 2P)$$ with initial conditions of $P(0) = 1$.

I have separated the variables, integrated, and simplified so I am now at $\ln(p) - \ln(p-5) = 10t + c$

I need help determining firstly, if that´s correct thus far, and secondly, where to go from here as everything else I have tried has not worked.

Yes, that looks correct. Another view of the logistic equation is that it is Bernoulli. Thus substitute $y=1-5P^{-1}$, $y'=-10y$,... — Lutz Lehmann, Feb 11 '22 at 18:33
@DanDoe that was after I separated variables and integrated. Also multiplied by 10. — Asset, Feb 11 '22 at 18:37
@LutzLehmann are you starting from the beginning again? I have not been taught Bernoulli and have no clue what that means. — Asset, Feb 11 '22 at 18:38
Yes, that is an alternative solution path. Comparing multiple approaches can be used to check for errors. Bernoulli equations are of the form $y'+py=qy^\alpha$. They are named because they have a simple solution trick that transforms them into linear DE. — Lutz Lehmann, Feb 11 '22 at 18:41
@LutzLehmann do you have any suggestions for where to go to learn that — Asset, Feb 11 '22 at 18:42

score 3 · Answer 1 · answered Feb 11 '22 at 18:54

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

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

Exponentiate
$\left(\frac {P}{5-P}\right) = \frac {e^{10t}}{4}$

And what is left is algebra
$4P = e^{10t}(5-P)\\ (4+e^{10t})P = 5e^{10t}\\ P = \frac {5e^{10t}}{4+e^{10t}}$

1 Answers1