A city has a growing population at a rate proportional to the current population, that is:
$$\frac{dP}{dx}=kP.$$
- Verify that $P(t)=P_0e^{kt}$, $t>0$ is a solution of the equation.
- If the population on 1st January 2006 which is $t=1$ was 147,200 and on 1st January 2007 when $t=2$ was 154,800, find the initial population and the value of $k$. Round your answer down to the 3dpl.
- Find the population on 1st January 2012.
- Find the time it takes for the population to double.
This question is really confusing me, I have done i. but I'm not sure how to do the rest. Detailed help would be very appreciated as I have many more questions similar to this. Thanks.
Part (ii) has given you two boundary conditions $P(1) = 147200$ and $P(2) = 154800$. Since your general solution has two unknown parameters $k$ and $P_0$, you can use the information to find two equations involving them, namely $147200 = P_0 e^{k}$ and $154800 = P_0 e^{2k}$, which you can then solve. (Hint: divide). This gives a particular solution to the ODE.
Once you have those, you can evaluate $P$ at the new time given to solve (iii). For (iv), observe that $P_0$ is the initial population at time $t = 0$, so the population will have doubled at time $\tau$ satisfying $P(\tau) = 2P_0$. You can solve this by taking $\log$ of both sides of your particular solution from part (ii).