How i can solve limit this ODE question .
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The population P(x) of a species satisfies the logistic differential equation.the initial population P(0)= 3000 and t is the in years. How I can find the lim x -> infinity P(t) ? this is the equation
4 Comments
Jan
on 7 May 2022
Edited: Jan
on 7 May 2022
@Jumanah Al omran: You are funny. The unit does not matter. For ->Inf you get the same result for years or seconds, and even for bananas. :-)
You can determine the equation for x(t) easily. Then it is easy to set x to Inf are determine the result.
A deeper look on the equation for dx/dt helps also. x is growing, -x^2 is shrinking even faster. What dows a curve with a rapidly shrinking derivative do for large values?
Sam Chak
on 7 May 2022
Edited: Sam Chak
on 7 May 2022
Based on your description, P(x) is the population as a function of x (the years...).
And dx/dt?
By the way, I plotted this to help you visualize the rate dx/dt. You see, x grows initially but the growth starts declining after x > 500K. It is still growing but getting slower and slower until dx/dt comes to a halt when it reaches 0.
x = linspace(0, 1e6, 100001);
dx = x/100 - x.^2/1e8;
plot(x, dx)
Answers (3)
Bjorn Gustavsson
on 7 May 2022
This is a Riccati equation. Convert it to a standard linear second-order differential equation and solve it that way, check what transformation to use from: Riccati-equation
Solve the ODE by separation of variables.
At what level will the source and loss-terms on the RHS be equal?
HTH
0 Comments
Torsten
on 7 May 2022
I think you mean
lim (t->oo) P(t)
instead of
lim (x->oo) P(t)
don't you ?
Seems the limit is not 0, but 1e6:
0 Comments
Sam Chak
on 8 May 2022
Looks like the population saturates at 1 million.
t = linspace(0, 1500);
P = (3000000*exp(t/100))./(997 + 3*exp(t/100));
plot(t, P)
2 Comments
Bjorn Gustavsson
on 8 May 2022
Yes. This is obvoius if we rewrite the ODE as:
dxdt = x/100.*(1-x/1e6);
First factor will be positive for all positive x, second factor goes to zero as x->1e6.
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