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

Question

Compute and graph the solutions to $$ y^{\prime \prime}+p y^{\prime}+y=0 \quad y(0)=1 \quad y^{\prime}(0)=0 $$ for \(p=4, p=2, p=1, p=0,\) and \(p=-2\)

Step-by-Step Solution

Answer

Solve the characteristic equation for roots based on the given p-values, apply initial conditions to find constant values, and graph each solution.

Step 1: Set up the Differential Equation

The given differential equation is \( y'' + p y' + y = 0 \). This is a second-order linear homogeneous differential equation.

Step 2: Find the Characteristic Equation

Replace \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1, giving us the characteristic equation \( r^2 + pr + 1 = 0 \).

Step 3: Solve the Characteristic Equation for Each p

The solutions of the characteristic equation \( r^2 + pr + 1 = 0 \) are given by the quadratic formula \( r = \frac{-p \pm \sqrt{p^2 - 4}}{2} \). Calculate this for each value of \( p \).

Step 4: Analyze Roots for p=4

For \( p = 4 \), the characteristic equation becomes \( r^2 + 4r + 1 = 0 \). Solving this gives roots, \( r = -2 \pm \sqrt{3} \). These are real and distinct, giving the general solution \( y = C_1 e^{(-2 + \sqrt{3})x} + C_2 e^{(-2 - \sqrt{3})x} \).

Step 5: Analyze Roots for p=2

For \( p = 2 \), the characteristic equation is \( r^2 + 2r + 1 = 0 \). Solving, we find a double root: \( r = -1 \). The solution takes the form \( y = C_1 e^{-x} + C_2 x e^{-x} \).

Step 6: Analyze Roots for p=1

For \( p = 1 \), the equation becomes \( r^2 + r + 1 = 0 \). The roots are \( r = \frac{-1 \pm \sqrt{-3}}{2} \), which simplifies to \( r = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \). The solution is \( y = e^{-\frac{x}{2}}(C_1 \cos(\frac{\sqrt{3}}{2} x) + C_2 \sin(\frac{\sqrt{3}}{2} x)) \).

Step 7: Analyze Roots for p=0

For \( p = 0 \), the equation is \( r^2 + 1 = 0 \), giving roots \( r = \pm i \), which corresponds to \( y = C_1 \cos(x) + C_2 \sin(x) \).

Step 8: Analyze Roots for p=-2

For \( p = -2 \), the characteristic equation \( r^2 - 2r + 1 = 0 \) gives \( r = 1 \) as a double root. The solution is \( y = C_1 e^{x} + C_2 x e^{x} \).

Step 9: Apply Initial Conditions for Each Case

Use \( y(0) = 1 \) and \( y'(0) = 0 \) to find \( C_1 \) and \( C_2 \) for each case. Substitute initial conditions into the respective solution equations.

Step 10: Graphing the Solutions

For each value of \( p \), graph \( y \) using the derived expressions with constants determined from initial conditions. Analyze stability and convergence or oscillations.

Key Concepts

Characteristic Equation

When dealing with second-order differential equations like the one given in the exercise, a common method for finding solutions is to use the characteristic equation. For the differential equation \( y'' + py' + y = 0 \), the characteristic equation is obtained by replacing the derivatives with powers of a variable \( r \).

• Replace \( y'' \) with \( r^2 \)
• Replace \( y' \) with \( r \)
• Replace \( y \) with \( 1 \)

This transformation gives the quadratic characteristic equation \( r^2 + pr + 1 = 0 \). Solving this equation for \( r \) provides the roots that help determine the type of solution to the differential equation.
The solutions to the characteristic equation can be found using the quadratic formula: \[ r = \frac{-p \pm \sqrt{p^2 - 4}}{2} \] Depending on the value of \( p \), the nature of the roots (real, complex, or repeated) will affect the form of the solution.

Initial Conditions

Initial conditions are necessary to find specific solutions and determine any constants in general solutions of differential equations. In the exercise, the given initial conditions are \( y(0) = 1 \) and \( y'(0) = 0 \).
These conditions provide specific values of the function and its derivative at a particular point, often at \( x = 0 \). By substituting these values into the general solutions, you can solve for the constants \( C_1 \) and \( C_2 \).
For example, if our general solution has the form \( y = C_1 e^{ax} + C_2 e^{bx} \), plug in the initial conditions:

• \( C_1 + C_2 = 1 \)
• The derivative, \( y'(0) = 0 \), should also be calculated and solved for \( C_1 \) and \( C_2 \).

These constants provide the particular solution that satisfies the given initial conditions. Each set of initial conditions can lead to different behaviors of solutions, such as oscillations or exponential growth/decay.

Homogeneous Differential Equations

A homogeneous differential equation is one in which every term is dependent on the unknown function and its derivatives. The equation \( y'' + py' + y = 0 \) is homogeneous because all terms involve either \( y \) itself or its derivatives.
These types of equations signify that the solution is purely determined by the characteristics of the system itself rather than by any external inputs or forces. Solving homogeneous equations involves understanding their characteristic roots and leads to solutions that often include exponential functions or trigonometric components, as influenced by the nature of the roots:

• Real and distinct roots lead to solutions with exponential terms.
• Complex roots introduce trigonometric functions.
• Repeated roots yield solutions that involve polynomial and exponential factors.

The absence of non-homogeneous terms (external inputs) simplifies the process of finding general solutions.

Solutions and Graphing

Once you have the general solution for a differential equation, applying initial conditions gives you a specific solution that can be graphically represented. The behavior of the graph is highly dependent on the nature of the characteristic roots.

• Real distinct roots lead to exponential decay or growth, often showing convergence or divergence on the graph.
• Complex roots introduce oscillatory behavior, resembling sine and cosine waveforms.
• Repeated roots can show polynomial growth or decay.

Graphing these solutions involves plotting the function \( y \) against \( x \). It helps visualize how solutions behave over time, demonstrating stability (where solutions settle or return to equilibrium), periodic motion, or divergence.
By analyzing the graphs, one can gain insights into the dynamic nature of the system modeled by the differential equation. Key takeaways from the graph might include determining the solution's stability, detecting oscillations, and identifying long-term behavior.