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maxsthekat

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## Homework Statement

I have dP/dt = (b - kP)P - hP (population problem with harvesting)

I'm told this is separable into a 1st order DE... so this is what I've tried:

dP/dt = bP - kP^2 - hP

dP/dt + (h-b)P = -kP^2

At this point, I get stuck... It doesn't appear to be a separable DE to me... I've tried approaching this as a Bernoulli equation...

let Z = P^-1

dZ/dt = (-1)(p^-2)(dP/dt)

(-P^2)(dZ/dt) = dP/dt

therefore, substituting back into the 3rd equation:

(-P^2)(dZ/dt) + (h-b)P^-1 = -kP^-2

dZ/dt + (h-b)P = -k

let mu = e^[Integral (h-b)P^-1 dp] = e^[(h-b) ln P] = P^(h-b) (we don't care about the "+c" since it falls out in the next step...)

Multiplying the 1st order DE by mu

P^(h-b)*dZ/dt + [P^(h-b)](h-b)P^-1 = -kP^(h-b)

d/dt [ P^(h-b) * Z] = -kP^(h-b). Integrating both sides gives:

P^(h-b) * Z = -ktP^(h-b). Substituting Z = P^-1 back in:

P^(h-b) * P^-1 = -ktP^(h-b) + C

P^-1 = -kt + C / P^(h-b)

P = 1 / [-kt + C/P^(h-b)]

Is this right? It seems horribly wrong, given this is a logistic equation... Any help in spotting my mistake would be greatly appreciated. I've been told you can just divide both sides of the original equation to get dP/[kP^2 +(h-b)P = dt. However, in order to integrate that, would you need to complete the square on the bottom?

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