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

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

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

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

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

Need to separate the variables.

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

Integrating both sides.

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

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

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

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

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

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

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

Thanks in advance

Craig