Problem 21 - Advanced Engineering Mathematics

Question

In Problems 21 and 22 , 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

The solution is \(x(t) = -e^{-3t}\), \(y(t) = e^{-3t}(2t + 2)\).

Step 1: Identify the Problem Type

The given problem is a set of first-order linear differential equations with initial conditions specified at a particular point. It is a system of differential equations.

Step 2: Solve the System Using Eigenvalues and Eigenvectors

We can represent the system as the matrix equation \( \mathbf{X}' = A \mathbf{X} \), where \(A = \begin{bmatrix} -5 & -1 \ 4 & -1 \end{bmatrix}\) and \(\mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix}\). To solve, find the eigenvalues \(\lambda\) by solving \( \text{det}(A - \lambda I) = 0 \). The characteristic equation is \( (\lambda + 5)(\lambda + 1) + 4 = 0 \), simplifying to \( \lambda^2 + 6\lambda + 9 = 0 \). Solve this quadratic equation to find eigenvalues.

Step 3: Calculate Eigenvalues

The characteristic equation is \( \lambda^2 + 6\lambda + 9 = 0 \), which factors into \((\lambda + 3)^2 = 0\). So, \(\lambda = -3\) is a repeated root.

Step 4: Find Eigenvectors

For \(\lambda = -3\), solve \( (A + 3I)\mathbf{v} = 0 \). Here, \( A + 3I = \begin{bmatrix} -2 & -1 \ 4 & 2 \end{bmatrix} \). Solving \( \begin{bmatrix} -2 & -1 \ 4 & 2 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) gives \( v_1 = -\frac{1}{2}v_2 \). Choose \(v_2 = 2 \), then \( v_1 = -1\), so an eigenvector is \( \mathbf{v_1} = \begin{bmatrix} -1 \ 2 \end{bmatrix} \).

Step 5: Find General Solution

Since \(\lambda = -3\) is a repeated eigenvalue, find a generalized eigenvector \(\mathbf{v_2}\). Solve \((A + 3I)\mathbf{v_2} = \mathbf{v_1}\). This yields the generalized eigenvector \(\mathbf{v_2} = \begin{bmatrix} 0 \ 1 \end{bmatrix}\). The general solution is \(\mathbf{X}(t) = c_1 e^{-3t} \begin{bmatrix} -1 \ 2 \end{bmatrix} + c_2 e^{-3t} (\begin{bmatrix} -1 \ 2 \end{bmatrix} t + \begin{bmatrix} 0 \ 1 \end{bmatrix})\).

Step 6: Apply Initial Conditions

Use \(x(1) = 0\) and \(y(1) = 1\) to solve for constants \(c_1\) and \(c_2\). Set \(\begin{bmatrix} x(1) \ y(1) \end{bmatrix} = c_1 e^{-3} \begin{bmatrix} -1 \ 2 \end{bmatrix} + c_2 e^{-3} (\begin{bmatrix} -1 \ 2 \end{bmatrix} + \begin{bmatrix} 0 \ 1 \end{bmatrix})\) equal to \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\). Solve this system to find \(c_1 = 1\) and \(c_2 = 1\).

Step 7: Write the Particular Solution

Substitute the values of \(c_1\) and \(c_2\) into the general solution: \(\mathbf{X}(t) = e^{-3t} \begin{bmatrix} -1 \ 2t+2 \end{bmatrix} \). Thus, \(x(t) = -e^{-3t}\) and \(y(t) = e^{-3t}(2t + 2)\).

Key Concepts

First-Order Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. A **first-order differential equation** specifically involves only the first derivative of the function. In simpler terms, it shows how a function is changing over time or space.

In this particular exercise, we deal with a system comprising two first-order differential equations. Here's what they look like:

• \( \frac{dx}{dt} = -5x - y \)
• \( \frac{dy}{dt} = 4x - y \)

These equations suggest a relationship between the functions \(x(t)\) and \(y(t)\) and their rates of change \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Our goal is to discover explicit expressions for \(x(t)\) and \(y(t)\), given the initial conditions \(x(1) = 0\) and \(y(1) = 1\).

Learning how to solve such equations helps in predicting future behavior of dynamic systems and is fundamental in fields like physics and engineering.

Eigenvalues and Eigenvectors

When solving systems of differential equations, the concepts of **eigenvalues** and **eigenvectors** become crucial. These are properties of matrices that help simplify complex systems into an easier form. For a given matrix \(A\), eigenvectors are non-zero vectors \(\mathbf{v}\) that, when multiplied by \(A\), yield the same vector scaled by a scalar \(\lambda\), which is the eigenvalue:

\[ A \mathbf{v} = \lambda \mathbf{v} \]

To find eigenvalues, we solve the **characteristic equation**, which is derived from the matrix \(A - \lambda I\). For this problem, the characteristic equation was:

\[ \lambda^2 + 6\lambda + 9 = 0 \]

Solving gives a repeated eigenvalue \(\lambda = -3\). This indicates that the system might require additional steps, such as finding a generalized eigenvector, for a complete solution. Identifying eigenvectors helps construct solutions that satisfy the original differential equations effectively.

Systems of Differential Equations

A **system of differential equations** consists of multiple differential equations that describe a set of relationships between several functions and their derivatives. Understanding how these functions influence each other provides us insight into the behavior of complex systems, such as economic models or ecological interactions.

The solution involves finding a pair of functions, here \(x(t)\) and \(y(t)\), which satisfy both individual equations simultaneously. We treat them as a vector \(\mathbf{X}(t) = \begin{bmatrix} x(t) \ y(t) \end{bmatrix}\) and express the entire system as a matrix equation:

\[ \mathbf{X}'(t) = A \mathbf{X}(t) \]

In this exercise, the system is linear and homogeneous, characterized by the matrix \(A = \begin{bmatrix} -5 & -1 \ 4 & -1 \end{bmatrix}\). The aim is to determine how \(x(t)\) and \(y(t)\) interact and evolve, revealing more about the multi-dimensional nature of the system.

Matrix Methods in Differential Equations

Matrix methods are powerful tools in solving systems of differential equations. By representing the system of equations as a matrix equation \(\mathbf{X}' = A \mathbf{X} \), we can leverage linear algebra techniques to find solutions.

Here, the matrix \(A = \begin{bmatrix} -5 & -1 \ 4 & -1 \end{bmatrix}\) comes into play. The solution technique involves several linear algebra steps:

• Finding eigenvalues and eigenvectors of matrix \(A\).
• Constructing solutions using these eigenvectors and eigenvalues.
• Using initial conditions to determine specific constants in the general solution.

After solving for eigenvalues, we needed a generalized eigenvector due to the repeated root, which further addressed the system's complexity. By applying these matrix methodologies, we transform a potentially overwhelming system of equations into a manageable solution process, illustrating the utility and importance of these techniques in problem-solving across various scientific fields.