Problem 12 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

For each equation find a number \(E\) such that \(P_{t}=E\) is a solution. a. \(P_{t+1}-0.9 P_{t}=2\) b. \(\quad P_{t+1}+0.9 P_{t}=2\) c. \(P_{t+1}-0.5 P_{t}=3\) d. \(P_{t+2}-0.2 P_{t+1}-0.6 P_{t}=2\) e. \(P_{t+1}+0.9 P_{t}=-5\) f. \(P_{t+2}+0.2 P_{t+1}-0.4 P_{t}=2\)

Step-by-Step Solution

Answer

E = 20, 1.053, 6, 10, -2.632, 2.5 for equations a, b, c, d, e, f respectively.

Step 1: Understanding Equations with Solution Approach

To find a solution where \( P_t = E \), substitute \( P_{t+1} = E \) and \( P_t = E \) into each equation, simplifying the problem to solve for \( E \).

Step 2a: Solve Equation for Part (a)

Substitute \( P_{t+1} = E \) and \( P_t = E \) in the equation \( P_{t+1} - 0.9 P_t = 2 \). This results in:\[ E - 0.9E = 2 \]Simplify to:\[ 0.1E = 2 \]Solve for \( E \):\[ E = \frac{2}{0.1} = 20 \]

Step 2b: Solve Equation for Part (b)

Substitute \( P_{t+1} = E \) and \( P_t = E \) in the equation \( P_{t+1} + 0.9 P_t = 2 \). This results in:\[ E + 0.9E = 2 \]Simplify to:\[ 1.9E = 2 \]Solve for \( E \):\[ E = \frac{2}{1.9} \approx 1.053 \]

Step 2c: Solve Equation for Part (c)

Substitute \( P_{t+1} = E \) and \( P_t = E \) in the equation \( P_{t+1} - 0.5 P_t = 3 \). This results in:\[ E - 0.5E = 3 \]Simplify to:\[ 0.5E = 3 \]Solve for \( E \):\[ E = \frac{3}{0.5} = 6 \]

Step 2d: Solve Equation for Part (d)

Substitute \( P_{t+2} = E \), \( P_{t+1} = E \), and \( P_t = E \) in the equation \( P_{t+2} - 0.2 P_{t+1} - 0.6 P_t = 2 \). This results in:\[ E - 0.2E - 0.6E = 2 \]Simplify to:\[ 0.2E = 2 \]Solve for \( E \):\[ E = \frac{2}{0.2} = 10 \]

Step 2e: Solve Equation for Part (e)

Substitute \( P_{t+1} = E \) and \( P_t = E \) in the equation \( P_{t+1} + 0.9 P_t = -5 \). This results in:\[ E + 0.9E = -5 \]Simplify to:\[ 1.9E = -5 \]Solve for \( E \):\[ E = \frac{-5}{1.9} \approx -2.632 \]

Step 2f: Solve Equation for Part (f)

Substitute \( P_{t+2} = E \), \( P_{t+1} = E \), and \( P_t = E \) in the equation \( P_{t+2} + 0.2 P_{t+1} - 0.4 P_t = 2 \). This results in:\[ E + 0.2E - 0.4E = 2 \]Simplify to:\[ 0.8E = 2 \]Solve for \( E \):\[ E = \frac{2}{0.8} = 2.5 \]

Key Concepts

Constant Solutions

In mathematics, particularly in the study of differential and difference equations, a constant solution refers to a solution to an equation where the value remains the same throughout, regardless of independent variables like time. This concept is essential when addressing problems like the ones mentioned, because constant solutions provide a straightforward answer, simplifying complex systems.

Consider the equation given in the exercise:

• For instance, let’s take part (a) which is: \( P_{t+1} - 0.9P_t = 2 \).
• To find a constant solution, we assume \( P_{t+1} = E \) and \( P_t = E \) where \( E \) is constant.

Substituting these into the equation results in: \[ E - 0.9E = 2 \]
Simplifying gives us \( 0.1E = 2 \) and solving that, \( E = 20 \). Therefore, 20 is a constant solution.

This method shows that, for such linear equations, constant solutions can be obtained by equating the terms where all \( P \)'s merge into a single constant, showing stability or equilibrium over time.

Linear Recurrence Relations

Linear recurrence relations represent sequences where each term is defined as a linear function of preceding terms. They involve constant coefficients and are pivotal in analyzing discrete-time processes, like the changes in investments, populations, or resources over time. This kind of equation is evident in various parts of our exercise.

For example, part (a) of the exercise:

• \( P_{t+1} - 0.9P_t = 2 \)

This is a first-order linear recurrence relation, as each term \( P_{t+1} \) is defined based on the immediately previous term \( P_t \) with the addition of some constant.

Key features of these relations include:

• Stability – How solutions change as we move from one term to the next.
• Predictability – Helps in forecasting future values.

Linear recurrence relations are ubiquitous in various fields, particularly in computer science for algorithms, financial forecasting, and modeling natural phenomena.

Equilibrium Solutions

Equilibrium solutions in the context of differential and difference equations refer to situations where the system stabilizes, meaning the state does not change over time. These solutions are critical for understanding the long-term behavior of a system.

In our exercise, equilibrium solutions are akin to constant solutions, because they too hold the state steady across terms. For example:

• Looking at exercise part (b), \( P_{t+1} + 0.9P_t = 2 \), we can observe equilibrium where the state \( P_t = E \) such that the additions balance out to keep the system stable.

Substitute \( P_{t+1} = E \) and \( P_t = E \), and simplify:\[ E + 0.9E = 2 \]From here, solving gives us \( E = \frac{2}{1.9} \approx 1.053 \).
Thus, the system, if it starts at 1.053, will stay at that level indefinitely if not perturbed - hence achieving equilibrium.

Understanding equilibrium is crucial for engineers and scientists when designing systems that need to be self-regulating or sustainable over time.