1. ## Population growth DE

$\frac{dP}{dt} = B_0e^{-at}P$

We know that $P(0) = P_0$, and we know that $B(t) =B_0e^{-at}$ where $a$ and $B_0$ are positive and real constants.

I need to solve this for $P(t)$ and find the limiting size.

Here's how far I got, could someone please take a look and see if this seems ok.

$\frac{dP}{dt} = B_0e^{-at}P$

Need to separate the variables.

$\frac{1}{P} dP = B_0e^{-at} dt$

Integrating both sides.

$\ln{P} = \frac{-B_0e^{-at}}{a} + C$

If we now take both sides to the power $e$

$P = e^{\frac{-B_0e^{-at}}{a} + C} = Ae^{\frac{-B_0e^{-at}}{a}}$ - **Sorry I'm not sure how to enlarge the latex code**

Now we know the initial condition, $P(0) = P_0$, this then gives us:

$P_0 = Ae^{\frac{-B_0}{a}}$, so $A = P_0e^{\frac{B_0}{a}}$

Giving us $P = P_0 \cdot e^{\frac{B_0}{a}} \cdot e^{\frac{-B_0e^{-at}}{a}}$.

The reason I'm asking is that it doesn't look the tidiest of equations.

Craig

2. Looks fine to me. You could make it a bit prettier.
$P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}
$

3. Thanks

4. Just one more little thing I've just found

I also need to find out the limiting size of the population.

$P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}$

As $t \to \infty$, then $(1 - e^{-at}) \to 1$.

So am I correct in saying that as $t \to \infty$, then $P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)} \to P_0 e^{\frac{B_0}{a}}$

Thanks again

5. No,
$
\lim_{t \to \infty} (1 - e^{-at}) = 1
$
,
but
$
P_{\infty} = \lim_{t \to \infty} P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)} = P_0 e^{\frac{B_0}{a}}
$

6. Originally Posted by lvleph
No,
$
\lim_{t \to \infty} (1 - e^{-at}) = 1
$
,
but
$
P_{\infty} = \lim_{t \to \infty} P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)} = P_0 e^{\frac{B_0}{a}}
$
Sorry there was a typo in my work above, was meant to be $\lim_{t \to \infty} (1 - e^{-at}) = 1$.