Question

A model of the population, P, of carp in a lake at time t is given by the DE:
dP/dt=0.25P(1-0.0004P)

A census taken 10 years ago found there were 1000 carp. Estimate the current population.

When I take the integral, just get that P= (.25P^2/2)-(.0001^3/3)+C. What am I doing wrong?

Answer 1

Simply taking the integral of both sides with respect to the same variable will not work here.

What you're doing:
\[\begin{align*}\frac{dP}{dt}&=0.25P(1-0.0004P)\\\\
\int\frac{dP}{dt}dt&=\int 0.25P(1-0.0004P)~dt\end{align*}\]
The right side is like saying you want to take the integral of some unknown function \(f(t)\) with respect to \(t\). However, \(P\) is a function dependent on \(t\), so you can perform this operation unless you know what \(P\) is.

Instead, you have to split the differentials (the \(d~\Box\)):
\[\begin{align*}\frac{dP}{dt}&=0.25P(1-0.0004P)\\\\
\frac{dP}{0.25P(1-0.0004P)}&=dt
\end{align*}\]
Now you can integrate both sides w.r.t. their appropriate variables:
\[\int\frac{dP}{0.25P(1-0.0004P)}=\int dt\]