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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?

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    Your calculation is correct. – 2017-02-02
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    First check the problem statement and the point is to check if the time is in _days_, _years_ or anything else that makes your answer 10 times grater. – 2017-02-02
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    I knew it, the book was then wrong - probably a typo. I also tried solving the problem by definite integrals to skip the part where I still have to find the constant. It results to the same answer 52.9 months. I don't think my units are wrong. – 2017-02-02

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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.