Problem 21 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$\begin{aligned} &\frac{d x}{d t}=-5 x-y\\\ &\frac{d y}{d t}=4 x-y\\\ &x(1)=0, y(1)=1 \end{aligned}$$

Step-by-Step Solution

Answer

Find the general solution using eigenvalues, apply initial conditions to solve for constants, then determine \( x(t) \) and \( y(t) \).

Step 1: Understand the Problem

We are given a system of first-order linear differential equations with initial conditions. We need to solve for the functions \( x(t) \) and \( y(t) \) using the given initial conditions \( x(1) = 0 \) and \( y(1) = 1 \).

Step 2: Set Up the Equations for Eigenvalues

To solve this system using a matrix approach, we first write it in matrix form. The system is: \[\frac{d}{dt} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -5 & -1 \ 4 & -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}\]Find the eigenvalues of the matrix \( A = \begin{pmatrix} -5 & -1 \ 4 & -1 \end{pmatrix} \).

Step 3: Compute Eigenvalues

The eigenvalues, \( \lambda \), can be found by solving the characteristic equation:\[\text{det}(A - \lambda I) = 0.\]For matrix \( A - \lambda I \), we have:\[\begin{vmatrix} -5-\lambda & -1 \ 4 & -1-\lambda \end{vmatrix} = 0.\]Calculate the determinant to get:\((\lambda + 5)(\lambda + 1) - (-4) = 0\). Solve this for \( \lambda \).

Step 4: Solve the Characteristic Equation

The determinant simplifies to:\((\lambda + 5)(\lambda + 1) - 4 = \lambda^2 + 6 \lambda + 5 - 4 = \lambda^2 + 6 \lambda + 1 = 0\).Solve \( \lambda^2 + 6 \lambda + 1 = 0 \) using the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]where \( a = 1, b = 6, c = 1 \).

Step 5: Find the Eigenvalues

Using the quadratic formula:\[\lambda = \frac{-6 \pm \sqrt{36 - 4}}{2} = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2} = -3 \pm 2\sqrt{2}\]So, the eigenvalues are \( \lambda_1 = -3 + 2\sqrt{2} \) and \( \lambda_2 = -3 - 2\sqrt{2} \).

Step 6: Solve for Eigenvectors

For each eigenvalue, solve the equation \((A - \lambda I)\mathbf{v} = \mathbf{0}\). Begin with \( \lambda_1 = -3 + 2\sqrt{2} \) and then for \( \lambda_2 = -3 - 2\sqrt{2} \). This gives us a corresponding eigenvector for each eigenvalue.

Step 7: Construct the General Solution

The general solution for the system is a linear combination of its eigenvectors and exponential terms:\[\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t}\]where \(c_1\) and \(c_2\) are constants determined by initial conditions.

Step 8: Apply Initial Conditions

Using \( x(1) = 0 \) and \( y(1) = 1 \), substitute these values into the general solution and solve the system of equations for \( c_1 \) and \( c_2 \).

Step 9: Determine the Specific Solution

After determining \( c_1 \) and \( c_2 \), write out the specific forms of \( x(t) \) and \( y(t) \) in terms of the constants and eigenvectors. These are your solutions to the initial-value problem.

Key Concepts

Eigenvalues and Eigenvectors

A key part of solving systems of differential equations using the matrix approach involves finding eigenvalues and eigenvectors. An eigenvalue, \( \lambda \), is a number that characterizes the dynamics of the system. To find them, you solve the characteristic equation obtained from the matrix representation of the system. The characteristic equation is derived from \( \text{det}(A - \lambda I) = 0 \), where \( A \) is the coefficient matrix and \( I \) is the identity matrix.

Once eigenvalues are found, eigenvectors can be determined. They are vectors that, when multiplied by the matrix, only get scaled by the eigenvalue. In our case, for each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) to find the corresponding eigenvector \( \mathbf{v} \). The pair of eigenvalues and eigenvectors together forms the basis for expressing solutions to the system of differential equations.

• Eigenvalues provide information on the stability and type of behavior of the system.
• Eigenvectors help in constructing the specific solution by forming a basis of solutions.

Initial-Value Problem

An initial-value problem involves finding a solution to a differential equation that satisfies certain specified values at a given initial point. Typically, this includes initial conditions, which are specific values for the function or its derivatives at a certain point.

In the given problem, we have initial conditions that are necessary to find the constants in the general solution. These are \( x(1)=0 \) and \( y(1)=1 \). By substituting the initial conditions into the general solution, we can find the specific constants (namely \( c_1 \) and \( c_2 \)) that satisfy the equations at \( t = 1 \).

• Initial conditions turn the general solution into a specific solution.
• They are essential because they allow us to understand how the solution behaves at the start.

Without initial conditions, infinitely many solutions are possible, so they are a crucial part of solving an initial-value problem.

Matrix Approach in Differential Equations

Solving differential equations using matrices involves expressing the system in the form of matrix equations for efficient computation and analysis.

Here is how it works: A system of equations can be rewritten as a matrix equation \( \frac{d}{dt} \mathbf{x} = A \mathbf{x} \), where \( \mathbf{x} \) is a vector containing the functions we are solving for, and \( A \) is a matrix containing the coefficients of these functions.

• This approach simplifies complex systems by using linear algebra tools.
• Allows gaining insights into the system's behavior through eigen-analysis.
• Makes computation more straightforward for both analytical and numerical solutions.

By writing the system in a matrix form, we use eigenvalues and eigenvectors to form the fundamental structure of the solution, which is a combination of exponential functions modified by these eigen-elements. The matrix approach effectively makes complicated problems more manageable.