# What does c represent in this differential equation?

If I have this equation:

$p'(t)-p(t)\alpha =0$

I can say that $p$ is a function that represents the size of a population at time t. The rate at which the population grows is constant. The solution will show that the size of the population is proportional to the initial size.

If I have this equation:

$p'(t)-p(t)f(t) =0$

I can say the rate at which the population grows is determined by $f$. The size of the population is still proportional to the initial size.

But if I have:

$p'(t)-p(t)\alpha = h(t)$

It's difficult to determine from the solution what role the initial size of the population, $p(0)$, has. The solution is:

$p(t)=\bigl(\int e^{-\alpha t}h(t)\ dt + c\bigr)\ e^{\alpha t}$

So my question is this: if I'm interpreting these differential equations as growth functions, what does $c$ represent in the last equation? In the previous equations, $c=p(0)$.

• $p(0)=c$ still means that $c$ is the initial size of the population if $p(t)$ represents the size of the population at time $t$. – Matt Sep 12 '12 at 18:20
• For constant $h(t)$ (breeding plus immigration), there is a fairly natural interpretation as a "virtual" population, since $p(t)$ is a constant plus an exponential term. – André Nicolas Sep 12 '12 at 19:18

The equation $$p'(t)=f(t)\ p(t)$$ says: The rate $p'$ by which the population $P$ grows is proportional to the current size $p$ of $P$; but the proportionality factor $f$ valid at time $t$ depends on time. If, e.g., the growth rate depends on the seasons, the function $t\mapsto f(t)$ would be a certain periodic function with period 365 days. At any rate, the size of $P$ at some later time $t$ will be proportional to the initial size $p(0)$.
The equation $$p'(t)=\alpha p(t)+h(t)$$ with constant $\alpha\in{\mathbb R}$ says: The population $P$ would rise (or decline) at the constant rate $\alpha$, and therefore would be given by $p(t)=p(0)\ e^{\alpha t}$, if it were not for an additional extraneous influx (or decrease) $h(t)$ dependent only on time $t$, but not on the current size $p(t)$ of the population. In this case it may very well be that for large $t>0$ the initial value $p(0)$ has almost no effect on the actual value $p(t)$. This is the case, e.g., if $\alpha<0$, and $h$ is some periodic function with period $T$. Then for large $t$ there will be a certain "stable" periodic behavior. This "limiting" behavior depends only on $h$, but not on the initial value $p(0)$.
Concerning your last question: When you write the solution in the "more correct" form $$p(t)=\int_0^t e^{\alpha(t-\tau)}\ h(\tau)\ d\tau + p(0)e^{\alpha t}$$ then you can verify by inspection that $p(0)$ plays no rôle for large $t$ when $\alpha<0$.
• I've sloppily figured out that, if $h$ were a constant instead, then $c=p(0)-\frac{h(t)}{\alpha}$, so as $\alpha \to \infty, c \to p(0)$. But h is not a constant, so this is moot (and probably wrong.) – Korgan Rivera Sep 12 '12 at 23:46
In your first equation, the solution is $p(t)=ce^{at}$. As you say, you choose $c$ to be $p(0)$ if that is the initial condition you are given. If you are given that $p(2)=10000$, that there were $10000$ individuals at year $2$, you would find $c=10000e^{-2a}$ Similarly, $c$ in the last equation is a constant of integration which you set to match some initial condition you are given.
• In the first and second equation, I don't choose $c$ to be $p(t)$. $p(0)$ is in fact $c$. If I'm given $p(2)= 10000$, $p(0)$ is still $c$. So it's easy to read that the population is proportional to the initial size: the c term in the solutions for the first and second equations I know is p(0). But that's not so for the 3rd equation. – Korgan Rivera Sep 12 '12 at 18:40
• @KorganRivera:If you start the integral in the third equation at 0, then $c=p(0)$ again-the integral is zero and the denominator is 1. If you start the integral at a different time, $c$ will be different. This is like any indefinite integral. If you write $\int x dx = x^2+c$, the $c$ will adjust according to the lower limit you use. – Ross Millikan Sep 12 '12 at 18:44