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Can someone explain to me what this is and how it's solved?

$$ \frac{dK}{dt}=\lambda(P-K) $$

$$ \frac{1}{K}\frac{dK}{dt}=\lambda(P-K)$$

I mean there is no t on the right hand side and there are two constants, what do I do about them?

From searching I understand this to be an autonomous, linear ODE? It looks sort of like the logistic growth model, but it's just the constant (lambda) in front?

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$\lambda$ is just a proportionality constant. You are correct; the first ODE is an autonomous, linear one while the second one is nonlinear. –  Josué Molina Oct 30 '12 at 17:10
Presumably the $K$ should be $K(t)$ to indicate the 'time' dependence. It is not clear what form $P$ or $\lambda$ take. –  copper.hat Oct 30 '12 at 17:19
First looks very similar to Sturm–Liouville equations, but without the boundary conditions. –  m0nhawk Oct 30 '12 at 17:37

1 Answer 1

up vote 1 down vote accepted

I assume $\lambda$ is a parameter. The second one is the logistic equation $$\frac{{dK}}{{dt}} = \lambda K(P - K)$$ and I think $K = K(t)$ and $P$ is the carrying capacity, but you usually see $K$ and $P$ switched. I do not know why you have them that way, but it can be solved by the separation of variables using partial fractions and the solution is probably some sigmoid function depending on $P$ and $\lambda$. The first one looks almost like a logistic equation. Are you sure it is not $dP/dt$ on the right hand side (of the first equation)? If so, then $$\frac{{dP}}{{dt}} = \lambda (P - K) = - \lambda (K - P) = - \lambda K\left( {1 - \frac{P}{K}} \right)$$ which is again a logistic equation but here $P = P(t)$ and $K$ is the carrying capacity.

Hope this helps.

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