Given the following initial value problem, find the time t for the when the population P is half of the limiting value:
$$\frac{dP}{dt}=P(10^{-1}-10^{-7}P), P(0)=5,000$$
I transformed the equation to the following to find the limiting value (capacity of the population k = 1,000,000).
$$\frac{dP}{dt}=\frac{1}{10}P(1-\frac{P}{1,000,000})$$
The general solution will then be:
$$\ln(P)-\ln(1-\frac{P}{1,000,000})=\frac{1}{10}t+C$$
Letting P(0)=5,000 and finding the constanc C:
$$C=\ln(\frac{5\times10^{9}}{995,000})$$
Thus, the particular solution is:
$$t(P)=10(\ln(P)-\ln(1-\frac{P}{1,000,000})-\ln(\frac{5\times10^{9}}{995,000})$$
Thus the answer I get from t(500,000) becomes 52.9 which is 10 times greater than the expected answer 5.29. Did I miss any step while finding the solution?
Check feasibility: You start with $P=5000$ at $t=0$ and $P$ stays below $500000$ until the time $t_0$ we are looking for. That makes $\frac{dP}{dt}<50000$ for $0\le t\le t_0$, hence $P(t)\le 5000+50000t$ for $0\le t\le t_0$. From this extremely simple bound, $$t_0\ge\frac{500000-5000}{50000}\approx 10 $$ so that $t_0=5.29$ is impossible. There might be some error in the problem statement.