Problem 39 - Advanced Engineering Mathematics

Question

(a) \(\operatorname{From} d P / d t=\left(k_{1}-k_{2}\right) P\) we obtain \(P=P_{0} e^{\left(k_{1}-k_{2}\right) t}\) where \(P_{0}=P(0)\). (b) If \(k_{1} >k_{2}\) then \(P \rightarrow \infty\) as \(t \rightarrow \infty\). If \(k_{1}=k_{2}\) then \(P=P_{0}\) for every \(t .\) If \(k_{1}< k_{2}\) then \(P \rightarrow 0\) as \(t \rightarrow \infty\).

Step-by-Step Solution

Answer

(a) \( P = P_0 e^{(k_1 - k_2)t} \). (b) If \( k_1 > k_2 \), \( P \rightarrow \infty \); if \( k_1 = k_2 \), \( P = P_0 \); if \( k_1 < k_2 \), \( P \rightarrow 0 \).

Step 1: Interpret the Differential Equation

We are given the differential equation \( \frac{dP}{dt} = (k_1 - k_2)P \). The equation describes the rate of change of \( P \) concerning time \( t \), where \( k_1 \) and \( k_2 \) are constants.

Step 2: Solve the Differential Equation

To solve \( \frac{dP}{dt} = (k_1 - k_2)P \), we recognize it as a separable differential equation. Rearrange terms: \( \frac{1}{P} \frac{dP}{dt} = k_1 - k_2 \). Integrate both sides with respect to their variables: \( \int \frac{1}{P} \, dP = \int (k_1 - k_2) \, dt \). This gives \( \ln|P| = (k_1 - k_2)t + C \), where \( C \) is the integration constant.

Step 3: Solve for P

Exponentiate both sides to solve for \( P \): \( e^{\ln|P|} = e^{(k_1 - k_2)t + C} \). Hence, \( |P| = e^C \cdot e^{(k_1 - k_2)t} \). Let \( e^C = P_0 \), the initial condition when \( t = 0 \), which simplifies to \( P = P_0 e^{(k_1 - k_2)t} \).

Step 4: Analyze Different Cases

1. **Case \( k_1 > k_2 \):** \( (k_1 - k_2) > 0 \) implies the exponent in \( P = P_0 e^{(k_1 - k_2)t} \) is positive. As \( t \rightarrow \infty \), \( P \rightarrow \infty \).2. **Case \( k_1 = k_2 \):** \( (k_1 - k_2) = 0 \) makes the exponent zero: \( P = P_0 e^0 = P_0 \). Thus, \( P \) stays constant at \( P_0 \) for all \( t \).3. **Case \( k_1 < k_2 \):** \( (k_1 - k_2) < 0 \) indicates that the exponent becomes negative. As \( t \rightarrow \infty \), \( P \rightarrow 0 \).

Key Concepts

Exponential Growth and Decay

Exponential growth and decay describe scenarios where a quantity grows or shrinks at a rate proportional to its current value. The differential equation given in the problem is a prime example of exponential growth and decay. It shows how the rate of change of a population, or any quantity, depends on its current state.

- When you have exponential growth, such as when an organism reproduces without restraint, the equation has a positive exponent, leading to increasing values over time. In our case, this happens when \( k_1 > k_2 \), meaning the growth rate surpasses the decay rate, causing the population \( P \) to grow without bounds as time \( t \) becomes very large.
- Exponential decay, on the other hand, describes situations where a quantity decreases over time at a rate proportional to its size. Here, if \( k_1 < k_2 \), the population decreases over time, leading to \( P \) approaching zero as \( t \) increases.
- If \( k_1 = k_2 \), there is no net change, thus making the population constant over time. This balance point means the growth and decay are in perfect equilibrium.

Separable Differential Equations

Separable differential equations allow us to manipulate and solve them using integration techniques. Such equations can be expressed as products of functions of different variables.

In the given problem, the differential equation \( \frac{dP}{dt} = (k_1 - k_2)P \) is separable. We can rearrange it to isolate relationships involving only \( P \) on one side and \( t \) on the other. This process is done through:

• Moving all terms involving \( P \) to one side: \( \frac{1}{P} \frac{dP}{dt} \) on the left.
• Leaving terms in \( t \) on the right: \( k_1 - k_2 \).

By integrating both sides, we translate the equation from its differential form into a more comprehensible solution. On integrating each side, you get:
- The left side gives \( \ln|P| \),
- The right side results in \( (k_1 - k_2)t + C \) where \( C \) is the constant of integration.
This process helps find the general solution for the differential equation, which can then be used to solve specific cases with initial values.

Initial Value Problems

Initial Value Problems (IVPs) involve differential equations together with a specific starting condition, allowing us to find particular solutions.

In our case, knowing \( P_0 \), the initial population or initial state when \( t=0 \), is critical for determining the constant \( C \). When we had integrated the separable differential equation, we introduced the constant \( C \), which is crucial for finding specific solutions that satisfy the given initial conditions.

By substituting the given initial values into the general solution, one can solve for \( C \) or directly express it through \( P_0 \). Specifically, setting \( t = 0 \) translates to:\

• \( P = P_0 e^{(k_1 - k_2)0} = P_0 \cdot 1 = P_0 \)

This condition allows us to express the solution uniquely based on the problem's context, leading to \( P = P_0 e^{(k_1 - k_2)t} \). Understanding IVPs is essential for transitioning from general equations to specific solutions that realistically model systems.