Problem 23 - Advanced Engineering Mathematics

Question

Verify that the vector \(\mathbf{X}_{p}\) is a particular solution of the given system. $$ \begin{aligned} &\mathbf{X}^{\prime}=\left(\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right) \mathbf{X}-\left(\begin{array}{l} 1 \\ 7 \end{array}\right) e^{t} ; \\ &\mathbf{X}_{p}=\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{t}+\left(\begin{array}{r} 1 \\ -1 \end{array}\right) t e^{t} \end{aligned} $$

Step-by-Step Solution

Answer

\( \mathbf{X}_p \) is verified as a particular solution of the system.

Step 1: Differentiate the guess solution

Differentiate \( \mathbf{X}_{p} = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{t} + \begin{pmatrix} 1 \ -1 \end{pmatrix} t e^{t} \) term by term. Use the product rule on \( te^t \). This gives: \[ \mathbf{X}_{p}' = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{t} + \begin{pmatrix} (1 \cdot t e^t)' \ (-1 \cdot t e^t)' \end{pmatrix} = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{t} + \begin{pmatrix} (e^t + te^t) \ -(e^t + te^t) \end{pmatrix}. \] Simplifying, \( \mathbf{X}_{p}' = \begin{pmatrix} 1 + e^t + te^t \ 1 - e^t - te^t \end{pmatrix}. \)

Step 2: Compute the right-hand side of the system

Substitute \( \mathbf{X}_p = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \) into the system: \( \mathbf{X}' = A\mathbf{X} - \begin{pmatrix} 1 \ 7 \end{pmatrix} e^t \), where \( A = \begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix}. \) Compute \( A \mathbf{X}_p \): \[ A \mathbf{X}_p = \begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \left( \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} t e^t \right) \]This results in:\[ \begin{pmatrix} 3 \ 7 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} t e^t. \] Subtract \( \begin{pmatrix} 1 \ 7 \end{pmatrix} e^t \) to obtain the full right-hand side:\[ \begin{pmatrix} 3 \ 7 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} t e^t - \begin{pmatrix} 1 \ 7 \end{pmatrix} e^t = \begin{pmatrix} 2 \ 0 \end{pmatrix} e^t + \begin{pmatrix} t e^t \ -t e^t \end{pmatrix}. \]

Step 3: Compare both sides

Our derivative from Step 1 is \( \begin{pmatrix} 1 + e^t + te^t \ 1 - e^t - te^t \end{pmatrix} \) and our transformed right-hand side from Step 2 is \( \begin{pmatrix} 2 \ 0 \end{pmatrix} e^t + \begin{pmatrix} t e^t \ -t e^t \end{pmatrix} \). Substitute \( 1 + e^t \) from Step 1 and \( 2 \) from Step 2: The expressions equate:\[ \begin{pmatrix} 2e^t + te^t \ 0 \ \end{pmatrix} = \begin{pmatrix} 2e^t + te^t \ 0 \end{pmatrix}. \] This confirms equality of the derivative with the system transformation, showing \( \mathbf{X}_p \) satisfies the system as a particular solution.

Key Concepts

Matrix Operations

Matrix operations are fundamental tools used in solving systems of differential equations. They include operations like addition, multiplication, and finding inverses. These operations are crucial for handling vectors and matrices, which represent systems of equations in a compact form. Understanding how to perform these operations helps in analyzing linear transformations and solving linear algebraic equations.

• **Matrix Addition**: This involves adding corresponding elements of two matrices of the same dimension to produce a new matrix. For example, given matrices \( A \) and \( B \), their sum \( A + B \) is obtained by adding each entry \( a_{ij} \) with \( b_{ij} \).
• **Matrix Multiplication**: A bit more complex, this involves rows and columns. To multiply matrix \( A \) by matrix \( B \), calculate the dot product of rows of \( A \) with columns of \( B \). The element in the result matrix \( C \) at position \( c_{ij} \) is the sum of the products of corresponding entries.

These operations are pivotal when transforming the equation \( \mathbf{X}' = A\mathbf{X} - \mathbf{C} \) in our differential equation, where \( A \) is a matrix and \( \mathbf{X} \) is a vector.

Matrix operations are essential for combining terms and simplifying expressions to accurately solve and verify solutions for systems of equations.

Product Rule

The Product Rule is a key principle in calculus used to differentiate products of two functions. It states that if you have a function \( f(x) = u(x) \, v(x) \), the derivative \( f'(x) \) is given by \( u'(x) \, v(x) + u(x) \, v'(x) \).

In vector calculus, where functions are often expressed as vectors, applying the product rule can be slightly more complex especially when both functions are compounded with exponential terms, as seen in the expression \( te^t \). For example:

• If \( v(x) = e^x \) and \( u(x) = x \), then \( v'(x) = e^x \) and \( u'(x) = 1 \).
• Applying the product rule, the derivative \( (xe^x)' = 1 \cdot e^x + x \cdot e^x = e^x + xe^x \).

This rule is used in our exercise to compute the derivative of the part containing \( t e^t \), allowing us to differentiate each component of the vector individually.

It's essential to apply the product rule precisely, as it ensures the accuracy of the resultant differential expressions.

Vector Calculus

Vector calculus extends calculus to vector fields and is crucial for understanding physical phenomena like fluid flow and electromagnetic fields. It incorporates operations like differentiation and integration of vector functions.

In our context, vectors \( \mathbf{X}_p \) are combinations of basic vector functions. Differentiation of such vectors involves applying calculus rules, including the product rule, to each component independently.

• **Differentiating Vectors**: When differentiating vector functions, treat each component separately, apply the derivative to each, and reassemble them into the vector form.
• **Exponential Components**: With expressions like \( \begin{pmatrix} 1 \cr 1 \end{pmatrix} e^t \), exponential terms are differentiated as normal, often requiring product and chain rules.

Vector calculus is not just about computation but also about visualization and understanding how changes in one part of a vector field affect another. This knowledge is integral for verifying that our computed solution satisfies the differential equation given by the original problem.