Problem 11 - Advanced Engineering Mathematics

Question

In Problems 11-16, verify that the vector \(\mathbf{X}\) is a solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=3 x-4 y \\ &\frac{d y}{d t}=4 x-7 y ; \quad \mathbf{X}=\left(\begin{array}{l} 1 \\ 2 \end{array}\right) e^{-5 t} \end{aligned} $$

Step-by-Step Solution

Answer

\( \mathbf{X}(t) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{-5t} \) is a solution to the system.

Step 1: Verify the Form of X

Identify the form of the given solution vector \( \mathbf{X} = \begin{bmatrix} 1 \ 2 \end{bmatrix} e^{-5t} \). Thus, \, \mathbf{X}(t) = \begin{bmatrix} x(t) \ y(t) \end{bmatrix} = \begin{bmatrix} 1e^{-5t} \ 2e^{-5t} \end{bmatrix}."

Step 2: Differentiate x(t) and y(t)

Calculate the derivatives: 1. \( \frac{dx}{dt} = -5e^{-5t} \) since \( x(t) = e^{-5t} \).2. \( \frac{dy}{dt} = -10e^{-5t} \) since \( y(t) = 2e^{-5t} \).

Step 3: Substitute into the System

Substitute \( x(t) = e^{-5t} \) and \( y(t) = 2e^{-5t} \) into the original system equations and verify the derivatives:\[\frac{dx}{dt} = 3x - 4y = 3(e^{-5t}) - 4(2e^{-5t}) = 3e^{-5t} - 8e^{-5t} = -5e^{-5t}\]\[\frac{dy}{dt} = 4x - 7y = 4(e^{-5t}) - 7(2e^{-5t}) = 4e^{-5t} - 14e^{-5t} = -10e^{-5t}\]

Step 4: Confirmation of Solution

Since both derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) from Step 2 are equal to the right side of the system in Step 3, \( \mathbf{X} \) is confirmed to be a solution of the system.

Key Concepts

Solution Verification

Solution verification is basically the process of checking whether a proposed solution to a mathematical problem or system is correct. When you verify a solution, you plug the solution back into the original problem and see if it satisfies all given conditions. This is a crucial step to ensure that the process you followed is indeed correct.

For example, consider a system of differential equations like the one in the exercise. We need to ensure that the solution vector \( \mathbf{X} = \begin{bmatrix} 1 \ 2 \end{bmatrix} e^{-5t} \) indeed satisfies the given differential equations. This involves calculating the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) from this vector and substituting these values into the system of equations.

Once the derivatives are calculated and substituted, both sides of each equation should match, just as they did in the step-by-step solution. If they do, then the vector is a valid solution. Think of it like a mathematical proof to validate your answers.

Vector Differential Equations

Vector differential equations are systems of equations where the unknowns are vectors. These equations involve taking derivatives of vector components with respect to a variable, usually time (\( t \)).

In the provided exercise, the system is represented by two differential equations for \( x(t) \) and \( y(t) \), which together form the vector \( \mathbf{X}(t) \). Each component of the vector is governed by its own differential equation. Thus, the system can describe more complex relationships between the components, as opposed to scalar equations that involve only one variable.

Solving vector differential equations usually requires:

• Determining the form of the solution, which often involves exponential functions.
• Calculating derivatives of these components, which can then be substituted into the original system.
• Confirming the solution by verification.

Understanding this kind of system is essential in fields like engineering and physics, where multiple interdependent processes need to be modeled simultaneously.

Linear Differential Systems

Linear differential systems are composed of linear differential equations, where each term is either a constant or a product of a constant and a single variable. This linearity simplifies the process of finding solutions since solutions can often be represented as the sum of basis solutions.

In our exercise, both equations are linear: \( \frac{dx}{dt} = 3x - 4y \) and \( \frac{dy}{dt} = 4x - 7y \). The terms are either constants or constants multiplied by variables \( x \) and \( y \).

Characteristics of linear systems include:

• Superposition principle: The sum of any two solutions is also a solution.
• They often yield solutions in terms of exponential functions, which translate into real-world processes like decay or growth.
• Greater simplification in terms of stability analysis and modeling compared to nonlinear systems.

Linear systems are highly prevalent in real-life applications, so mastering them provides a solid foundation for understanding more complex systems.