The problem in question:
The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 800,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 80,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time. (Note that the variable t represents days.)
From this information I get:
$$\frac{dP}{dt}= \frac{ln(2)}{7}P-80,000$$
Since this equation is of the form: $\frac{dP}{dt}=rp-k$ , I know $P(t)=Ce^{rt}+\frac{k}{r}$
Plug in the values of $r$ and $k$:
$$P(t)=Ce^{\frac{ln(2)}{7}t}+\frac{560,000}{ln(2)}$$
$P(0)=800,000 $, therefore:
$C=800,000-\frac{560,000}{ln(2)}$
Plug in C for the final equation:
$$P(t)=(800,000-\frac{560,000}{ln(2)})e^{\frac{ln(2)}{7}t}+\frac{560,000}{ln(2)}$$
Where did I go wrong here?
Best Answer
Just plug back in to the differential equation to check your work.
$$ P(t) = (800,000 - \frac{560,000}{\ln 2} ) e^{\frac{\ln 2}{7} t} + \frac{560000}{\ln 2}\\ \frac{dP}{dt} = (800,000 - \frac{560,000}{\ln 2} ) e^{\frac{\ln 2}{7} t} \frac{\ln 2}{7}\\ = (800,000 - \frac{560,000}{\ln 2} ) e^{\frac{\ln 2}{7} t} \frac{\ln 2}{7} + \frac{560000}{\ln 2} \frac{\ln 2}{7} - \frac{560000}{\ln 2} \frac{\ln 2}{7}\\ = \frac{\ln 2}{7} \big( (800,000 - \frac{560,000}{\ln 2} ) e^{\frac{\ln 2}{7} t} + \frac{560000}{\ln 2} \big) - \frac{560000}{7}\\ = \frac{\ln 2}{7} P(t) - 80000 $$
and
$$ P(0)=(800,000 - \frac{560,000}{\ln 2} ) e^{\frac{\ln 2}{7} *0} + \frac{560000}{\ln 2}\\ = (800,000 - \frac{560,000}{\ln 2} ) + \frac{560000}{\ln 2}\\ = 800,000 $$
So your reasoning is fine.