Question

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

Short 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-by-step solution

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. \(\frac{d P} {P(5000-P)} = 10^{-5} dt\)

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.

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: \(P(t) = \frac{5000e^{10^{-5}t + C}}{1 + e^{10^{-5}t + C}}\).

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.