Problem 8 - Advanced Engineering Mathematics

Question

In Problems 5-14, solve the given linear system by the methods of this chapter. $$ \mathbf{X}^{\prime}=\left(\begin{array}{ll} -2 & 5 \\ -2 & 4 \end{array}\right) \mathbf{X} $$

Step-by-Step Solution

Answer

Solve the system: \( \mathbf{X}(t) = c_1 e^{(1+\sqrt{3})t} \begin{pmatrix} 5 \\ 2+\sqrt{3} \end{pmatrix} + c_2 e^{(1-\sqrt{3})t} \begin{pmatrix} 5 \\ 2-\sqrt{3} \end{pmatrix} \).

Step 1: Write the System of Differential Equations

The given matrix equation can be written as a system of two differential equations. Let \( \mathbf{X} = \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} \). Then the system is: \( x_1' = -2x_1 + 5x_2 \) and \( x_2' = -2x_1 + 4x_2 \).

Step 2: Find the Eigenvalues of the Coefficient Matrix

The coefficient matrix is \( A = \begin{pmatrix} -2 & 5 \ -2 & 4 \end{pmatrix} \). Find the eigenvalues by solving the characteristic equation \( |A - \lambda I| = 0 \). The characteristic equation is: \((-2 - \lambda)(4 - \lambda) - (5)(-2) = 0\). Simplify to find the polynomial \( \lambda^2 - 2\lambda - 2 = 0 \).

Step 3: Solve the Characteristic Equation

Solve \( \lambda^2 - 2\lambda - 2 = 0 \) using the quadratic formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2, c = -2 \). Plug in these values to find: \( \lambda = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} \). Simplify to get \( \lambda_1 = 1 + \sqrt{3} \) and \( \lambda_2 = 1 - \sqrt{3} \).

Step 4: Find the Eigenvectors for Each Eigenvalue

For \( \lambda_1 = 1 + \sqrt{3} \), substitute back into \((A - \lambda I) \mathbf{v} = 0\) and solve to get eigenvectors. Do the same for \( \lambda_2 = 1 - \sqrt{3} \). When \( A - \lambda_1 I \) is solved, one possible eigenvector \( \mathbf{v}_1 \) is \( \begin{pmatrix} 5 \ 2 + \sqrt{3} \end{pmatrix} \). For \( \lambda_2 \), solve to get another eigenvector \( \mathbf{v}_2 \) as \( \begin{pmatrix} 5 \ 2 - \sqrt{3} \end{pmatrix} \).

Step 5: Write the General Solution

The general solution is a linear combination of the solutions generated by the eigenvalues and their corresponding eigenvectors: \( \mathbf{X}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \), where \( \lambda_1 = 1 + \sqrt{3} \) and \( \lambda_2 = 1 - \sqrt{3} \). Therefore, \( \mathbf{X}(t) = c_1 e^{(1+\sqrt{3})t} \begin{pmatrix} 5 \ 2+\sqrt{3} \end{pmatrix} + c_2 e^{(1-\sqrt{3})t} \begin{pmatrix} 5 \ 2-\sqrt{3} \end{pmatrix} \).

Step 6: Determine Constants Based on Initial Conditions (if provided)

If specific initial conditions are given, substitute them into the general solution to find the values of \( c_1 \) and \( c_2 \). Otherwise, the general solution stands as is, representing the family of solutions to the system.

Key Concepts

Eigenvalues and Eigenvectors

Understanding the role of **eigenvalues and eigenvectors** is crucial when dealing with linear differential equations, especially in the realm of solving systems. These two components provide insight into the behavior of a system over time.

• **Eigenvalues** (\(\lambda\)) can be thought of as factors that scale eigenvectors when a matrix is applied;
• they tell us about the stability and nature of equilibrium points in a system.

More precisely, they show how fast or slow solutions grow or decay.

On the other hand, **eigenvectors** are non-zero vectors that only change by a scalar factor (specifically, an eigenvalue) during a linear transformation.

• In terms of differential equations, they help describe the direction of these changes.

Together, eigenvalues and their corresponding eigenvectors provide a more comprehensive picture of how a matrix affects space.

In the original problem, after finding the eigenvalues (\(\lambda_1 = 1 + \sqrt{3}\), \(\lambda_2 = 1 - \sqrt{3}\)), we proceed to find their eigenvectors (\(\mathbf{v}_1\) and \(\mathbf{v}_2\)). Each pair allows us to construct part of the solution for the system.

Characteristic Equation

The **characteristic equation** is a polynomial equation derived from a square matrix, key in determining eigenvalues.

• It is typically written as \(|A - \lambda I| = 0\), where \(A\) is the matrix, \(\lambda\) represents eigenvalues, and \(I\) is the identity matrix.
• This results in a polynomial of degree equal to the order of the matrix (quadratic for a 2x2 system).

Solving this polynomial gives the eigenvalues of the matrix.

In the original problem, substituting into the characteristic formula yields:
\((-2 - \lambda)(4 - \lambda) - (5)(-2) = 0\), resulting in the quadratic equation:
\(\lambda^2 - 2\lambda - 2 = 0\).

Using the quadratic formula, we found two solutions which serve as eigenvalues for the system matrices. Solving the characteristic equation is therefore a pivotal step in analyzing and solving linear differential equations.

General Solution of Differential Equations

The **general solution** of a linear differential equation captures the complete set of possible solutions for the system. It's expressed as a combination of the solutions associated with each eigenvalue and its corresponding eigenvector.

• Typically represented as \(\mathbf{X}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\).
• This general form allows for applying initial conditions, if present, to find specific constants \(c_1\) and \(c_2\), describing specific solutions.

This general solution is formed after discovering the eigenvalues and eigenvectors.

In practical terms, the general solution explains the possible behaviors of a system over time, incorporating all variations due to the initial state. So, for our original problem, having arrived at the eigenvalues and eigenvectors, we can construct this general solution to summarize all possible behaviors described by the differential equation system. It becomes a tool, not only for prediction but also for further analysis, particularly in systems modeling.