Problem 14 - Advanced Engineering Mathematics

Question

In Problems 11-16, verify that the vector \(\mathbf{X}\) is a solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 2 & 1 \\ -1 & 0 \end{array}\right) \mathbf{X} ; \quad \mathbf{X}=\left(\begin{array}{l} 1 \\ 3 \end{array}\right) e^{t}+\left(\begin{array}{r} 4 \\ -4 \end{array}\right) t e^{t} $$

Step-by-Step Solution

Answer

\( \mathbf{X} \) is a solution because \( \mathbf{X}' \) equals the right-hand side of the given system.

Step 1: Compute the Derivative of \( \mathbf{X} \)

First, find the derivative \( \mathbf{X}' \) of the vector \( \mathbf{X} \). Given:\[ \mathbf{X} = \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \begin{pmatrix} 4 \ -4 \end{pmatrix} t e^t \]We use the product rule for the term including \( t e^t \):\[ \mathbf{X}' = \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \left[ \begin{pmatrix} 4 \ -4 \end{pmatrix} (t \cdot e^t)' \right] \]The derivative \( (t \cdot e^t)' \) is \( e^t + t e^t \) by the product rule. So:\[ \mathbf{X}' = \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \begin{pmatrix} 4 \ -4 \end{pmatrix} \left( e^t + t e^t \right) \]Multiply and combine the terms to simplify.

Step 2: Simplify \( \mathbf{X}' \)

Expand the expression for the derivative from the previous step and simplify:\[ \mathbf{X}' = \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \begin{pmatrix} 4 e^t + 4 t e^t \ -4 e^t - 4 t e^t \end{pmatrix} \]Combine like terms:\[ \mathbf{X}' = \begin{pmatrix} 1 + 4 \ 3 - 4 \end{pmatrix} e^t + \begin{pmatrix} 4t \ -4t \end{pmatrix} e^t \]\[ \mathbf{X}' = \begin{pmatrix} 5 \ -1 \end{pmatrix} e^t + \begin{pmatrix} 4t \ -4t \end{pmatrix} e^t \]

Step 3: Compute the Right-Hand Side of the Equation

Now, compute the matrix-vector multiplication on the right-hand side of the given equation:\[ \begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix} \mathbf{X} \]Plug in \( \mathbf{X} \):\[ \begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix} \left( \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \begin{pmatrix} 4 \ -4 \end{pmatrix} t e^t \right) \]Distribute the matrix:\[ \begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix} \begin{pmatrix} 4 \ -4 \end{pmatrix} t e^t \]

Step 4: Simplify the Right-Hand Side Result

Calculate each matrix-vector product:For \( \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t \):\[ \begin{pmatrix} 2 \times 1 + 1 \times 3 \ -1 \times 1 + 0 \times 3 \end{pmatrix} = \begin{pmatrix} 5 \ -1 \end{pmatrix} \]Multiply each by \( e^t \):\[ \begin{pmatrix} 5 \ -1 \end{pmatrix} e^t \]For \( \begin{pmatrix} 4 \ -4 \end{pmatrix} t e^t \):\[ \begin{pmatrix} 2 \times 4 + 1 \times (-4) \ -1 \times 4 + 0 \times (-4) \end{pmatrix} = \begin{pmatrix} 4 \ -4 \end{pmatrix} \]Multiply each by \( t e^t \):\[ \begin{pmatrix} 4 \times t \ -4 \times t \end{pmatrix} e^t \]

Step 5: Compare Both Sides

From Steps 2 and 4, we have:Left-hand side (\( \mathbf{X}' \)):\[ \begin{pmatrix} 5 \ -1 \end{pmatrix} e^t + \begin{pmatrix} 4t \ -4t \end{pmatrix} e^t \]Right-hand side:\[ \begin{pmatrix} 5 \ -1 \end{pmatrix} e^t + \begin{pmatrix} 4t \ -4t \end{pmatrix} e^t \]Both sides are equal, confirming \( \mathbf{X} \) is indeed a solution.

Key Concepts

Vector Solution Verification

Vector solution verification refers to the process of checking whether a given vector indeed satisfies a system of differential equations. This involves finding the derivative of the vector and ensuring that both sides of the original equation match.

To verify a vector solution, we first compute the derivative of the given vector, which represents changes over time or space. By comparing this derivative to the function defined by the system, we can confirm the vector satisfies the conditions laid out by the equations.

In our case, the solution vector was \[ \mathbf{X} = \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t + \begin{pmatrix} 4 \ -4 \end{pmatrix} t e^t \]and after calculating its derivative and equating it to the function described by the system, we could confidently declare that\[ \begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix} \] is indeed satisfied by this vector. This meticulous check assures us that the vector solution is precise and valid.

Matrix Multiplication

Matrix multiplication is a key operation in solving systems of linear equations, including differential equations. This mathematical operation involves multiplying rows of a matrix by columns of another, yielding a new matrix of potential solutions.

To multiply matrices correctly, you match each element of a row from the first matrix with each element of a column of the second matrix, summing the products. It is important to ensure the first matrix's number of columns matches the second matrix's number of rows for multiplication to be possible.

For example, in the given problem, the matrix \( \begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix} \)multiplies with the vector\( \begin{pmatrix} 1 \ 3 \end{pmatrix} e^t \)to yield a new vector as shown:

• First element: \((2 \times 1) + (1 \times 3) = 5\)
• Second element: \((-1 \times 1) +(0 \times 3) = -1\)

Post multiplication, each element is then multiplied by \( e^t \) as required by our equation.

Product Rule

The product rule is a calculus principle that explains derivation of products of two functions. It is especially useful in differential equations where expressions often involve terms multiplied together involving both variables and exponentials like \( e^t \).

The product rule states that if you have a product of two functions, \( u(t) \) and \( v(t) \), their derivative is given by:\[(uv)' = u'v + uv'\]

In this exercise, \( \begin{pmatrix} 4 \ -4 \end{pmatrix} t e^t \)is treated using the product rule, with \( u = t \) and \( v = e^t \). Here,

• \( t' = 1 \),\( e^t' = e^t \)

Thus the derivative becomes \( e^t + t e^t \), allowing us to systematically handle even complex expressions involving product functions.

Eigenvectors and Eigenvalues

Eigenvectors and eigenvalues are crucial concepts for systems of differential equations. They are used to analyze linear transformations and simplify solutions.

An eigenvector of a matrix is a non-zero vector that changes at most by a scalar factor when that matrix is applied. The corresponding scalar factor is called the eigenvalue. These concepts help in understanding systems' behaviors because they decipher stable states or directions of change within the system.

In differential equations, particularly matrix equations, these eigenvectors are often looked at to predict long-term behavior. By determining eigenvalues, we can ascertain if solutions grow, decay, or maintain their state over time, offering a greater comprehension of dynamic systems.