Question

Verify that the given vector satisfies the given differential equation. \(\mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\ {0} & {-1} & {1}\end{array}\right) \mathbf{x}, \quad \mathbf{x}=\left(\begin{array}{r}{6} \\ {-8} \\ {-4}\end{array}\right) e^{-t}+2\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) e^{2 t}\)

Short answer

Yes, the given vector \(\mathbf{x}\) satisfies the differential equation \(\mathbf{x'} = A\mathbf{x}\) as we have shown that \(\mathbf{x'} = A\mathbf{x}\).

Step-by-step solution

1. Calculate the derivative of \(\mathbf{x}\)

Using the chain rule, the derivative of \(\mathbf{x}\) is given by: \(\mathbf{x'} = \frac{d}{dt}(\begin{pmatrix} 6 \\ -8 \\ -4 \end{pmatrix}e^{-t} + 2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}e^{2t}) = \begin{pmatrix} 6 \\ -8 \\ -4 \end{pmatrix}\frac{d}{dt}(e^{-t}) + 2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}\frac{d}{dt}(e^{2t})\). Now we compute the derivatives of the exponentials: \(\frac{d}{dt}(e^{-t}) = -e^{-t}\), and \(\frac{d}{dt}(e^{2t}) = 2e^{2t}\). Substitute these back into the expression for \(\mathbf{x'}\): \(\mathbf{x'} = \begin{pmatrix} 6 \\ -8 \\ -4 \end{pmatrix}(-e^{-t}) + 2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}(2e^{2t}) = \begin{pmatrix} -6e^{-t} \\ 8e^{-t} \\ 4e^{-t} \end{pmatrix} + \begin{pmatrix} 0 \\ 4e^{2t} \\ -4e^{2t} \end{pmatrix}\).

2. Multiply A with the given vector \(\mathbf{x}\)

Now, we'll perform the matrix-vector multiplication: \(A\mathbf{x} = A\left(\begin{pmatrix} 6 \\ -8 \\ -4 \end{pmatrix}e^{-t} + 2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}e^{2t}\right) = A\begin{pmatrix} 6e^{-t} \\ -8e^{-t} \\ -4e^{-t} \end{pmatrix} + A\begin{pmatrix} 0 \\ 2e^{2t} \\ -2e^{2t} \end{pmatrix}\). Now, we compute the products: \(A\begin{pmatrix} 6e^{-t} \\ -8e^{-t} \\ -4e^{-t} \end{pmatrix} = \begin{pmatrix} -6e^{-t} \\ 8e^{-t} \\ 4e^{-t} \end{pmatrix}\), and \(A\begin{pmatrix} 0 \\ 2e^{2t} \\ -2e^{2t} \end{pmatrix} = \begin{pmatrix} 0 \\ 4e^{2t} \\ -4e^{2t} \end{pmatrix}\). Thus, \(A\mathbf{x} = \begin{pmatrix} -6e^{-t} \\ 8e^{-t} \\ 4e^{-t} \end{pmatrix} + \begin{pmatrix} 0 \\ 4e^{2t} \\ -4e^{2t} \end{pmatrix}\).

3. Check if the results are equal

Now, comparing the expressions for \(\mathbf{x'}\) and \(A\mathbf{x}\), we can see that: \(\mathbf{x'} = \begin{pmatrix} -6e^{-t} \\ 8e^{-t} \\ 4e^{-t} \end{pmatrix} + \begin{pmatrix} 0 \\ 4e^{2t} \\ -4e^{2t} \end{pmatrix}\), and \(A\mathbf{x} = \begin{pmatrix} -6e^{-t} \\ 8e^{-t} \\ 4e^{-t} \end{pmatrix} + \begin{pmatrix} 0 \\ 4e^{2t} \\ -4e^{2t} \end{pmatrix}\), which implies \(\mathbf{x'} = A\mathbf{x}\). Hence, the given vector \(\mathbf{x}\) satisfies the given differential equation.

Key concepts

Matrix-Vector Multiplication

Matrix-vector multiplication is a fundamental operation in linear algebra where a matrix is multiplied by a vector. This process involves taking a combination of the rows of the matrix with the corresponding elements of the vector. In the context of differential equations, particularly when dealing with systems of equations, matrix multiplication helps to transform a vector representing the current state of the system into another vector representing the state's rate of change.

To perform matrix-vector multiplication, you multiply each element of a row in the matrix by the corresponding element of the vector, then sum the products to form a single number. This process is repeated for each row in the matrix, resulting in a new vector. For instance, if you multiply a matrix A by a vector x, each element of the resulting vector is obtained by the dot product of the corresponding row in A with the vector x.

In the provided exercise, the matrix A is multiplied by a vector-valued function of time x(t). The computation is done by separately considering the scalar multiples of the matrix A, which involves the exponential terms e^{-t} and e^{2t}. Afterward, these results are added together to form the product Ax. Doing this accurately confirms if the vector-valued function x(t) satisfies the differential equation by comparing Ax with x', the time derivative of x(t).

Chain Rule for Derivatives

The chain rule is a principle in calculus for finding the derivative of the composition of two or more functions. This rule is essential when dealing with differential equations involving functions multiplied by other functions, especially when the functions are variable dependent, like exponential functions involving time t.

The chain rule states that the derivative of a composite function f(g(x)) with respect to x is the derivative of f with respect to g(x) multiplied by the derivative of g with respect to x. Mathematically, it is expressed as \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \).

In the given exercise, the chain rule is used to differentiate the vector-valued function x(t) that includes scalar multiplication of an exponential function. It is applied to find the derivative of e^{-t} and e^{2t}, which are composed inside the vector elements. This step is crucial for subsequently verifying that x(t) satisfies the differential equation by comparing its derivative x' with the product resulting from matrix-vector multiplication Ax.

Exponential Functions

Exponential functions are an integral part of many mathematical areas, including the study of differential equations. They are defined as functions of the form \( f(t) = e^{kt} \), where \( e \) is the base of natural logarithms, approximately equal to 2.71828, \( t \) is the variable, and \( k \) is a constant which dictates the rate of growth or decay of the function.

Exponential functions are particularly important in solving differential equations because they have unique properties that model growth and decay processes, such as in populations, finance, or radioactive decay. One of the notable properties is that the derivative of an exponential function is proportional to the function itself, \( \frac{d}{dt}e^{kt} = ke^{kt} \).

This property simplifies the process of finding the derivative when applying the chain rule, as demonstrated in the solution of the exercise. When \( k \) is negative, the exponential function represents a decay process; when \( k \) is positive, it represents growth. In our exercise, both processes occur as the vector-valued function x(t) includes terms with e^{-t} and e^{2t}.

System of Differential Equations

A system of differential equations consists of multiple differential equations that relate to several unknown functions and their derivatives. These systems often appear in real-world scenarios where multiple related phenomena are modeled simultaneously.

Systems can be linear or nonlinear, and can vary in terms of how many differential equations they contain and how these equations interact with each other. In linear systems, the unknown functions and their derivatives are directly proportional to each other, and the equations can often be written in matrix form, allowing for solutions that involve matrix-vector multiplication.

In the exercise provided, we're dealing with a linear system represented in matrix form, where a matrix A captures the relationship between the vector of functions x(t) and their derivatives x'(t). This type of system can often be solved by finding a set of functions that transform according to specific rules, such as exponential functions, which when differentiated, yield the original function scaled by a constant. The solution to the system is confirmed by verifying that the derivative of the proposed solution vector x(t), found through the chain rule and matrix-vector multiplication, matches the form described by the system.