# Modeling Population Growth [dP/dt = k P - A P2 - h]

1. May 18, 2013

### NZBRU

Does anyone know how to solve dP/dt = k P - A P2 - h for P. I understand partial fractions are needed and I have already solved dP/dt = k P - A P2. Is anyone able to solve it, Cheers NZBRU.

2. May 18, 2013

### sweet springs

Hi.

dt = dP/(k P - A P＾2 - h) = -1/A dP/[ (P- k/A)^2-{(k/A)^2+h} ] = -1/A dp/[p^2 - {(k/A)^2+h}) ] , p= P- k/A

= -1/A dp/[p - sqrt{(k/A)^2+h} ][p +sqrt {(k/A)^2+h} ]

now you can integrate.

3. May 18, 2013

### NZBRU

If I typed that in correctly the line would be [-1/A [dp]/[[p - sqrt{(k/A)^2+h} ][p +sqrt {(k/A)^2+h} ]]]=dt or would it be: [-1/A [dp] [p +sqrt {(k/A)^2+h} ]/[p - sqrt{(k/A)^2+h} ]]=dt? (I have not used ASCIIMath extensively). Thank you for the fast response.

4. May 18, 2013

### JJacquelin

Hi !
integration gives t(P)
The inverse fonction is P(t)

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