Solving Population and Growth Problems in Differential Equations

To successfully solve population and growth problems using differential equations, a student must first learn to translate word problems into mathematical rates. The core premise of exponential growth is that the rate of change of a population is directly proportional to the current population size. Mathematically, this dynamic relationship is modeled by the first-order differential equation `\frac{dP}{dt} = kP`, where `P` represents the population size, `t` represents time, and `k` is the constant of proportionality or growth rate.

The first practical step in solving these problems is finding the general solution to the differential equation, which is almost always done via the separation of variables method. By rearranging the core equation to group all population terms on one side and time terms on the other, you obtain `\frac{1}{P} dP = k dt`. Integrating both sides yields `\ln|P| = kt + C`, which can be rewritten exponentially as `P(t) = C e^{kt}` where `C` represents the initial population at `t = 0` (often denoted as `P_0`).

The next phase involves extracting key data from the problem text to solve for the missing constants, `P_0` and `k`. Usually, a problem will provide two distinct data points, such as the initial population at the starting time and the population size at a later milestone. Substituting the initial condition gives you `P_0` immediately, while plugging the second data point into the equation allows you to isolate and solve for the growth constant `k` using natural logarithms.

Once both constants are explicitly determined, you can construct the specific particular solution model for that population. This complete equation acts as a predictive tool, allowing you to answer the final prompt questions. Whether the problem asks you to calculate the future population size at a specific time or find the exact duration required for a population to double, you can easily find the answer through standard algebraic manipulation.

Let us look at a practical example: a colony of bacteria starts with an initial population of 100 cells, and after 2 hours, the population grows to 400 cells. We want to find the exact particular equation that models this growth and determine the population after 5 hours.

We begin with our exponential model `P(t) = P_0 e^{kt}`. Since the initial population is 100, we substitute `P_0 = 100` to get `P(t) = 100 e^{kt}`. Next, we use our second data point, `P(2) = 400`, which gives the algebraic equation `400 = 100 e^{k(2)}`. Dividing by 100 yields `4 = e^{2k}`, and taking the natural logarithm of both sides gives `\ln(4) = 2k`, which simplifies to `k = \frac{\ln(4)}{2} \approx 0.6931`.

With both constants found, our specific model is `P(t) = 100 e^{0.6931 t}` or simply `P(t) = 100 \cdot 4^{t/2}`. To find the population after 5 hours, we substitute `t = 5` into our model, which gives `P(5) = 100 \cdot 4^{5/2} = 100 \cdot 32 = 3200` cells. This systematic approach guarantees a precise and correct solution every time.