Question

\(\frac{d P}{d t}=0.0002 P(1200-P) \text { and } P=20 \text { when } t=0\)

Short answer

The final solution will give a model for the change in population over time.

Step-by-step solution

Rewrite the Equation

Firstly, rewrite the differential equation to separate the variables: \(\frac{dP}{P(1200-P)}=0.0002 dt\).

Integrate Both Sides

Now, integrate both sides of the equation. The left side requires partial fraction decomposition before integrating: \[\int \frac{1}{P} + \frac{1}{1200-P} dP = \int 0.0002 dt \] This integral can be performed as: \[ln|P| - ln|1200-P| = 0.0002t + C \] or in a simplified form: \[ln| \frac{P}{1200-P} | = 0.0002t + C\].

Solve for P(t)

Apply the exponent to remove the natural logarithm and solve for P(t): \[\frac{P}{1200-P} = e^{0.0002t+C} = Ae^{0.0002t} \] where \(A = e^C\). Therefore, \[P(t)= \frac{1200Ae^{0.0002t}}{1+Ae^{0.0002t}}\].

Use the Initial Conditions to Solve for A

We're given that when \(t=0\), \(P=20\). Substituting these values into the equation, we get: \[20 = \frac{1200A}{1+A}\]. Solving this equation should give us the value of A.

Final Solution

Substitute the obtained value of A into the equation for P(t) to get the final solution.

Key concepts

Separation of Variables

Separation of Variables is a widely-used method for solving ordinary differential equations. When dealing with equations that describe physical phenomena, such as population growth or decay, the method involves rearranging the equation to isolate the variables on different sides of the equation.

For example, consider the differential equation \(\frac{dP}{dt} = 0.0002 P(1200-P)\) from our exercise regarding population growth. The first step in using separation of variables is to rewrite the equation so that all terms involving the variable P are on one side, and all terms involving t are on the other side: \(\frac{dP}{P(1200-P)} = 0.0002 dt\).

After this, you can integrate both sides of the equation separately, which brings us closer to the solution. It’s essential that the equation is separable for this method to work, and luckily, many equations in physics and engineering are.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used in calculus to break down complex rational expressions into simpler ones that are more manageable to integrate. In our exercise, after separating variables, we face an integral on the P side that is not straightforward. The expression \(\frac{1}{P(1200-P)}\) can be decomposed into simpler parts: \(\int\frac{1}{P} + \frac{1}{1200-P}dP\).

This is achieved by expressing the original fraction as a sum of fractions with simpler denominators. Each of these simpler fractions corresponds to a partial fraction that we can then integrate. By performing the integration, we move towards finding the general solution to the original differential equation.

Exponential Growth and Decay

Understanding Exponential Processes

Exponential growth and decay reflect processes that change at rates proportional to their current values, typical in populations, radioactive substances, and finance. The equation \(\frac{dP}{dt} = 0.0002 P(1200-P)\) provided is a model for a population where the rate of change is proportional to both the current population and the remaining population capacity of the environment.

The solution to these equations frequently involves the exponential function \(e^x\). After integrating and applying the initial conditions, we obtain an equation of the form \(P(t) = \frac{1200Ae^{0.0002t}}{1+Ae^{0.0002t}}\), which exhibits exponential growth behavior regulated by the constant 0.0002 and an initial size determined by A.

Initial Value Problems

Initial value problems are a subset of differential equations where you are given the value of the function or its derivatives at a particular point. This information is crucial to finding a unique solution for a differential equation. In our example, we know that at time \(t=0\), the population \(P\) is 20: \(P(0) = 20\).

By substituting the initial value into our general solution, we can solve for the constant A. This constant is what makes our solution specific to the initial conditions provided. Without an initial value, there would be infinitely many solutions to the differential equation since A could be anything. The initial value anchors our solution, providing a complete picture of how the system behaves over time.