Problem 30 - Calculus: Graphical, Numerical, Algebraic

Question

\(\frac{d P}{d t}=10^{-5} P(5000-P) \text { and } P=50 \text { when } t=0\)

Step-by-Step Solution

Answer

The particular solution to the differential equation is \(P(t) = \frac{5000e^{10^{-5}t + \frac{50}{4950}}}{1 + e^{10^{-5}t + \frac{50}{4950}}}\).

Step 1: Separate variables

Rewrite the differential equation in the form where all terms involving \(P\) are on one side and all terms involving \(t\) are on the other side. \(\frac{d P} {P(5000-P)} = 10^{-5} dt\)

Step 2: Integrate both sides

When we integrate both sides, we get the integral equation. Be careful when integrating the left-hand side; it is a logarithmic integration. Here is what you get when you perform the integration: \(\int \frac{1} {P(5000-P)} dP = 10^{-5}\int dt\). The left side is the integral of the sum of two logarithms, so we can split it into two integrals: \(\int \frac{1} {P} dP + \int \frac{1} {5000 - P} dP = 10^{-5}t + C\), where \(C\) is the constant of integration.

Step 3: Solve for P

Use logarithm rules to combine the two logarithms: \(\ln|P| - \ln|5000 - P| = 10^{-5}t + C\). This can be further simplified to \(\ln\left|\frac{P}{5000 - P}\right| = 10^{-5}t + C\). Apply the exponential function to both sides to solve for \(P\): \(\frac{P}{5000 - P} = e^{10^{-5}t + C}\). Finally, isolate \(P\) to get the general solution: \(P(t) = \frac{5000e^{10^{-5}t + C}}{1 + e^{10^{-5}t + C}}\).

Step 4: Apply initial condition and solve for C

We know that when \(t=0\), \(P=50\). Substituting those values in to solve for \(C\), we get \(\ln|C| = \ln\left|\frac{50}{4950}\right|\). Solving for \(C\) gives \(C = \frac{50}{4950}\). Substitute \(C\) back into \(P(t)\) to get the particular solution.

Key Concepts

Separation of Variables

In differential equations, separation of variables is a powerful technique used to solve equations by isolating different variables onto different sides of an equation. This allows each variable to be integrated independently. In our exercise, the differential equation given is \( \frac{d P}{d t}=10^{-5} P(5000-P) \). To separate the variables:

• Move all terms involving \( P \) to one side of the equation.
• Place all terms involving \( t \) on the other side.

This rearrangement results in \( \frac{d P} {P(5000-P)} = 10^{-5} dt \). By achieving this separation, we prepare the equation for integration, laying the groundwork for finding the solution to the differential equation.

Logarithmic Integration

Logarithmic integration arises in cases where a fraction of functions is involved, particularly with polynomials in the denominator. When integrating,

• Write the equation as the sum of two partial fractions.
• Each fraction can then be integrated separately.

For instance, the integral \( \int \frac{1} {P(5000-P)} dP \) can be split into two terms: \( \int \frac{1} {P} dP + \int \frac{1} {5000 - P} dP \). The result of integrating each part is a logarithmic expression, which transforms into \( \ln|P| - \ln|5000 - P| \). Using properties of logarithms, this is simplified to \( \ln\left|\frac{P}{5000 - P}\right| \). By converting to exponential form, one can advance towards solving the equation completely.

Initial Conditions

Initial conditions are crucial for determining a unique solution to a differential equation. They specify the value of the function at a particular point. In our exercise, we know that at \( t=0 \), \( P=50 \). This information allows us to find the constant of integration \( C \).

• Substitute the initial conditions into your general solution.
• Solve for \( C \) based on these values.

Here, substituting \( t=0 \) and \( P=50 \) into \( \ln\left|\frac{P}{5000-P}\right| = 10^{-5}t + C \) gives us \( C = \ln\left|\frac{50}{4950}\right| \). With this value of \( C \), you substitute back into your solution for \( P(t) \) to yield the particular solution, accurately describing the behavior of \( P \) over time given the initial conditions.