Problem 5 - Advanced Engineering Mathematics

Question

Write the linear system in matrix form. $$ \begin{aligned} &\frac{d x}{d t}=x-y+z+t-1 \\ &\frac{d y}{d t}=2 x+y-z-3 t^{2} \\ &\frac{d z}{d t}=x+y+z+t^{2}-t+2 \end{aligned} $$

Step-by-Step Solution

Answer

The matrix form is: \( \begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \\ \frac{dz}{dt} \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} + \begin{pmatrix} t - 1 \\ -3t^2 \\ t^2 - t + 2 \end{pmatrix} \)

Step 1: Identify Functions and Their Derivatives

In the given system, the three differential equations represent the derivatives \( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \). These derivatives can be expressed as linear combinations of the variables \(x\), \(y\), \(z\), and \(t\).

Step 2: Organize Coefficients

For each equation, list the coefficients of \(x\), \(y\), \(z\), and any constant terms:1. \(\frac{dx}{dt} = x - y + z + t - 1\) gives us the coefficients: 1, -1, 1, and -1 (for the constant term).2. \(\frac{dy}{dt} = 2x + y - z - 3t^2\) gives us: 2, 1, -1, and 0 (no constant term).3. \(\frac{dz}{dt} = x + y + z + t^2 - t + 2\) gives us: 1, 1, 1, 2 (the constant term is accounted for by summing t-related terms).

Step 3: Write in Matrix Form

Using the coefficients, express the system in matrix form. The system can be represented as:\[\begin{pmatrix}\frac{dx}{dt} \\frac{dy}{dt} \\frac{dz}{dt}\end{pmatrix} =\begin{pmatrix}1 & -1 & 1 \2 & 1 & -1 \1 & 1 & 1\end{pmatrix} \begin{pmatrix}x \y \z\end{pmatrix} +\begin{pmatrix}t - 1 \-3t^2 \t^2 - t + 2\end{pmatrix}\]

Step 4: Simplify and Validate

Double-check the matrix elements and constants before finalizing. Ensure each term corresponds to its respective variable and equation in the original system. The additional vector accounts for the non-variable terms such as constants and other expressions involving \(t\).

Key Concepts

Differential Equations

Differential equations are vital in modeling a wide range of real-world phenomena. They tell us how a function, often representing physical quantities, changes over time or space. In the provided exercise, we have a system of differential equations that involves the derivatives of functions with respect to time:

• \(\frac{d x}{d t}\) gives the rate of change of \(x\).
• \(\frac{d y}{d t}\) gives the rate of change of \(y\).
• \(\frac{d z}{d t}\) gives the rate of change of \(z\).

These equations are linear because each derivative is expressed as a linear combination of the variables \(x\), \(y\), \(z\), and a linear function of \(t\), without products of these terms or nonlinear functions like exponents or trigonometric functions. The goal is to capture the relationships and dynamics between these variables and time.

Matrix Representation

Representing a system of differential equations in matrix form can simplify the process of solving and analyzing complex systems. A matrix representation expresses the system as an equation involving matrices and vectors:Let's break down the components:

• The vector \(\begin{pmatrix}\frac{dx}{dt} \ \frac{dy}{dt} \ \frac{dz}{dt}\end{pmatrix}\) holds the derivatives.
• The matrix \(\begin{pmatrix}1 & -1 & 1 \ 2 & 1 & -1 \ 1 & 1 & 1\end{pmatrix}\) contains the coefficients from the differential equations.
• The vector \(\begin{pmatrix}x \ y \ z\end{pmatrix}\) represents the variables whose rates are being described.
• Finally, the vector \(\begin{pmatrix}t - 1 \ -3t^2 \ t^2 - t + 2\end{pmatrix}\) accounts for terms independent of the primary variables, involving just \(t\) and constants.

Using this structure provides clarity by organizing all variable-related terms in one matrix equation, making it easier to manipulate and solve.

Coefficients Identification

Identifying coefficients in differential equations is crucial before writing them in matrix form. It involves extracting numerical factors applied to each variable, understanding constants, and noting variable-independent terms.For each equation:

• In \(\frac{dx}{dt} = x - y + z + t - 1\), identify coefficients: 1 for \(x\), -1 for \(y\), 1 for \(z\), and a constant term of -1 (adjusted for the presence of \(t\)).
• In \(\frac{dy}{dt} = 2x + y - z - 3t^2\), extract: 2 for \(x\), 1 for \(y\), -1 for \(z\), and no constant, since all is accounted by \(t\).
• In \(\frac{dz}{dt} = x + y + z + t^2 - t + 2\), the coefficients are 1 for each \(x\), \(y\), \(z\), with the constant being a combination of the terms depending on \(t\).

Labeling these coefficients is essential for correctly building the system in matrix form, ensuring all parts of the original equations are well-represented.