Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find the solutions to a. \(w_{0}=3 \quad w_{1}=1\) b. \(w_{0}=0 \quad w_{1}=0\) c. \(w_{0}=2 \quad w_{1}=1\) d. \(w_{0}=1 \quad w_{1}=1\) $$ \begin{aligned} w_{t+2}-5 w_{t+1}+6 w_{t} &=0 \\ w_{t+2}+8 w_{t+1}+12 w_{t} &=0 \\ w_{t+2}-6 w_{t+1}+8 w_{t} &=0 \\ w_{t+2}-5 w_{t+1}+4 w_{t} &=0 \end{aligned} $$

Step-by-Step Solution

Answer

Solve each characteristic equation using initial conditions; verify results by substitution.

Step 1: Identify the Characteristic Equation

For each given difference equation, identify the characteristic equation. The general form for a second-order linear homogeneous difference equation is: \[w_{t+2} + b w_{t+1} + c w_t = 0\] The characteristic equation is found by replacing \(w_t\) with \(r^t\), leading to: \[r^2 + br + c = 0\].

Step 2: Solve the Characteristic Equation for Each Case

Solve the quadratic characteristic equations derived from each difference equation from Step 1 to find their general solutions:a. \(r^2 - 5r + 6 = 0\)b. \(r^2 + 8r + 12 = 0\)c. \(r^2 - 6r + 8 = 0\)d. \(r^2 - 5r + 4 = 0\).

Step 3: Apply Initial Conditions to Find Specific Solutions

Utilize the initial conditions \(w_0\) and \(w_1\) to determine the specific constants for each solution:a. Use \(w_0 = 3\) and \(w_1 = 1\) for the equation solved under case a.b. Use \(w_0 = 0\) and \(w_1 = 0\) for the equation solved under case b.c. Use \(w_0 = 2\) and \(w_1 = 1\) for the equation solved under case c.d. Use \(w_0 = 1\) and \(w_1 = 1\) for the equation solved under case d.

Step 4: Compute General Solutions with Initial Conditions

For each specific case, compute the sequence that satisfies both the general form and the initial conditions. This involves substituting the values for \(C_1\) and \(C_2\) back into the general solution obtained from the roots.

Step 5: Verify the Solutions

Verify that the solutions obtained satisfy both the original difference equation and the initial conditions specified. This can involve substituting values back into the original equations to ensure correctness.

Key Concepts

Characteristic Equations in Difference Equations

To solve any second-order linear homogeneous difference equation, like the ones given in the exercise, we begin by identifying the characteristic equation. It's a powerful tool that transforms the difference equation into a more workable form. Whenever you're working with a difference equation of the form: \[ w_{t+2} + b w_{t+1} + c w_t = 0 \], the corresponding characteristic equation can be formulated by assuming solutions of the form \(w_t = r^t\). This assumption simplifies the equation to a classic quadratic form: \[ r^2 + br + c = 0 \]. Once you have formulated the characteristic equation, solving it will help in determining the potential forms of solutions \(w_t = C_1 r_1^t + C_2 r_2^t\), where \(r_1\) and \(r_2\) are roots of the characteristic equation. Depending on whether these roots are real and distinct, repeated, or complex, the structure of your solution will change. This is why mastering characteristic equations is so crucial in solving difference equations.

The Role of Initial Conditions

Initial conditions in difference equations provide us the necessary information to find a specific solution from the general solution. In essence, they pin down which solution (out of the infinite possibilities given by the general solution) applies to our particular problem. Each difference equation was provided with initial conditions like:

• \( w_0 \), the known initial value when \( t = 0 \),
• \( w_1 \), the known value at the next step when \( t = 1 \).

These values are substituted into the general solution formula to solve for the constants \( C_1 \) and \( C_2 \). For example, if your general solution is \( w_t = C_1 r_1^t + C_2 r_2^t \), you would substitute \( t = 0 \) and \( t = 1 \) to get two equations:

• \( C_1 + C_2 = w_0 \)
• \( C_1 r_1 + C_2 r_2 = w_1 \)

Solving this system will give the specific constants needed, making the solution uniquely applicable with your initial conditions.

Understanding General Solutions

The general solution to a difference equation represents a family of possible solutions for that equation. Once the characteristic equation is solved, we generally arrive at a formula involving arbitrary constants, which represent this family. For instance, if you solve a characteristic equation and find the roots \( r_1 \) and \( r_2 \), your general solution will typically be expressed as: \[ w_t = C_1 r_1^t + C_2 r_2^t \] where \( C_1 \) and \( C_2 \) are constants. These constants will adjust based on the initial conditions given, making the general solution flexible. This flexibility allows you to match the solution to specific situations, described by the initial conditions. It's critical to notice if the roots are real, repeated, or complex, each of which brings its own form of general solutions:

• Distinct Real Roots: As shown above.
• Repeated Roots: Form \( w_t = (C_1 + C_2 t) r^t \).
• Complex Roots: Involve trigonometric functions and are expressed as \( w_t = r^t (C_1 \cos(\theta t) + C_2 \sin(\theta t)) \) where \( r e^{i\theta} \) is the complex root.

Understanding these nuances will deepen your grasp of both difference equations and their applications.