Population model: dP/dt = k sqrtP

[Tom McCurdy]

Homework Statement

Suppose that when a certain lake is stocked with a population P of fish, the birth and death rates \alpha and \beta are inversely proportional to \sqrt{P} , so that

\frac{dP}{dt} = k \sqrt{p}

Find P(t) if P(0)=C

The Attempt at a Solution

\frac{dP}{\sqrt{p}} = k dtintegrate2 \sqrt{P} = k t + ASub 0 for t C for XA=2 \sqrt{C}

sub back in

2 \sqrt{P} = 2t+4 \sqrt{c}

P = (2 t k+4 \sqrt{C})^2

 

[Kreizhn]

2 \sqrt{P} = 2t+4 \sqrt{c}
P = (2 t k+4 \sqrt{C})^2

There are a few small problems with this, mostly just technical. Such as the second to last line is missing a k, and the we substitute C for P; there is no X.

However, not so trivial is where did the coefficient of 2 for \sqrt{P} go? How did the 2 in 2kt show up randomly? Your mistakes are algebraic and occur after your line "sub back in"

Thus your solutions should be

P(t) = \displaystyle\left( \frac{1}{2} kt + \sqrt(C) \right) ^2

 

[Tom McCurdy]

I ended up solving it correctly

I ended up solving it correctly (2 min before it was due)

the answer is

\frac{(kt+2\sqrt{C})^2}{4}

 

[HallsofIvy]

Homework Statement

Suppose that when a certain lake is stocked with a population P of fish, the birth and death rates \alpha and \beta are inversely proportional to \sqrt{P} , so that

\frac{dP}{dt} = k \sqrt{p}

Find P(t) if P(0)=C

The Attempt at a Solution

\frac{dP}{\sqrt{p}} = k dt


integrate


2 \sqrt{P} = k t + A


Sub 0 for t C for X


A=2 \sqrt{C}

You were fine up to here.

sub back in

2 \sqrt{P} = 2t+4 \sqrt{c}

No, you just said 2\sqrt{P}= kt+ A. If, now, A= 2\sqrt{C}, then 2\sqrt{P}= kt+ 2\sqrt{C}.
Square both sides: 4 P= (kt+ 2\sqrt{C})^2
P= \frac{kt+ 2\sqrt{C})^2}{4}
which is what you say you eventually got

P = (2 t k+4 \sqrt{C})^2