I have the DE, $\displaystyle \frac{dP}{dt} = B_0e^{-at}P$, which I've solved to give $\displaystyle P = P_0 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}$.

I'm also given that $\displaystyle B(t) = B_0e^{-at}$ and that $\displaystyle P(0) = P_0$.

Q) Suppose that at time $\displaystyle t = 0$ there are $\displaystyle P_0 = 10^6$ cells and that P(t) is increasing at the rate of $\displaystyle 3 \times 10^5$ cells per month. After 6 months the tumor has doubled (size and number of cells). Find a.

A)

We can say that $\displaystyle P = 10^6 e^{\frac{B_0}{a}\left(1 - e^{-at}\right)}$.

If the tumor has doubled in 6 months, then does this mean that $\displaystyle P(6) = 2P(0)$

$\displaystyle 10^6 e^{\frac{B_0}{a}\left(1 - e^{-a6}\right)} = 2(10^6 e^{\frac{B_0}{a}\left(1 - e^{0}\right)})$, therefore $\displaystyle e^{\frac{B_0}{a}\left(1 - e^{-6a}\right)} = 10^6(1)$

Also if $\displaystyle P(t)$ is increasing at a rate of $\displaystyle 3 \times 10^5$, I presume that $\displaystyle \frac{dP}{dt}=3 \times 10^6$, or is it $\displaystyle B(t)= 3 \times 10^6$?

If someone could give me a gentle push in the right direction it would be much appreciated

Thanks a lot

Craig