Problem 2 - Biocalculus Calculus for the Life Sciences

Question

Show that \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) . \(\begin{array}{l}{A=\left[ \begin{array}{rr}{-2} & {1} \\ {1} & {-2}\end{array}\right]} \\ {x_{1}(t)=\frac{1}{2}\left(e^{-3 t}+e^{-1}\right),} & {x_{2}(t)=\frac{1}{2}\left(-e^{-3 t}+e^{-t}\right)}\end{array}\)

Step-by-Step Solution

Answer

Both functions satisfy the differential equation system.

Step 1: Understand the System of Equations

The system is defined by \(\frac{d \mathbf{x}}{d t} = A \mathbf{x}\) where \(A\) is a matrix. Specifically, \(A = \left[ \begin{array}{rr} -2 & 1 \ 1 & -2 \end{array} \right]\). We need to show that both \(x_1(t)\) and \(x_2(t)\) are solutions to this differential equation system.

Step 2: Calculate the Derivatives of \(x_1(t)\) and \(x_2(t)\)

For \(x_1(t) = \frac{1}{2}(e^{-3t} + e^{-t})\), calculate \(\frac{d}{dt}x_1(t) = \frac{1}{2}(-3e^{-3t} - e^{-t})\). For \(x_2(t) = \frac{1}{2}(-e^{-3t} + e^{-t})\), calculate \(\frac{d}{dt}x_2(t) = \frac{1}{2}(3e^{-3t} - e^{-t})\).

Step 3: Express x in Vector Form

Express the solution as a vector: \(\mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\). This results in \(\mathbf{x}(t) = \begin{bmatrix} \frac{1}{2}(e^{-3t} + e^{-t}) \ \frac{1}{2}(-e^{-3t} + e^{-t}) \end{bmatrix}\).

Step 4: Perform Matrix Multiplication

Perform \(A \mathbf{x}(t)\):\[A \mathbf{x}(t) = \begin{bmatrix} -2 & 1 \ 1 & -2 \end{bmatrix} \begin{bmatrix} \frac{1}{2}(e^{-3t} + e^{-t}) \ \frac{1}{2}(-e^{-3t} + e^{-t}) \end{bmatrix}\]Calculate each component for the result vector:- For the first component: \(-2 \cdot \frac{1}{2}(e^{-3t} + e^{-t}) + 1 \cdot \frac{1}{2}(-e^{-3t} + e^{-t}) = \frac{1}{2}(-3e^{-3t} - e^{-t})\).- For the second component: \(1 \cdot \frac{1}{2}(e^{-3t} + e^{-t}) - 2 \cdot \frac{1}{2}(-e^{-3t} + e^{-t}) = \frac{1}{2}(3e^{-3t} - e^{-t})\).

Step 5: Verify the Solution

Compare the calculated derivatives in Step 2 with the result of the matrix multiplication in Step 4. Notice both derivatives from Step 2 match the corresponding components resulting from \(A \mathbf{x}(t)\), confirming \(\frac{d \mathbf{x}}{d t} = A \mathbf{x}\). Thus, both \(x_1(t)\) and \(x_2(t)\) satisfy the system of differential equations.

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, essential for solving systems of linear equations. It involves the combination of two matrices to produce a new matrix. To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. The resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.
For example, consider multiplying matrix \( A \) by a vector \( \mathbf{x}(t) \):

• Matrix \( A \) is given by \( \begin{bmatrix} -2 & 1 \ 1 & -2 \end{bmatrix} \).
• The vector \( \mathbf{x}(t) \) is \( \begin{bmatrix} \frac{1}{2}(e^{-3t} + e^{-t}) \ \frac{1}{2}(-e^{-3t} + e^{-t}) \end{bmatrix} \).

To calculate \( A \mathbf{x}(t) \), you take the dot product of each row of \( A \) with the column vector \( \mathbf{x}(t) \). This involves multiplying corresponding entries and summing the products:

• First component: \(-2 \times \frac{1}{2}(e^{-3t} + e^{-t}) + 1 \times \frac{1}{2}(-e^{-3t} + e^{-t}) = \frac{1}{2}(-3e^{-3t} - e^{-t}) \).
• Second component: \(1 \times \frac{1}{2}(e^{-3t} + e^{-t}) - 2 \times \frac{1}{2}(-e^{-3t} + e^{-t}) = \frac{1}{2}(3e^{-3t} - e^{-t}) \).

Thus, matrix multiplication allows us to transform vectors and understand complex transformations in vector spaces.

System of Equations

A system of equations is a collection of two or more equations with a set of variables. Solutions to these equations must satisfy each equation in the system simultaneously. When a system is represented using matrices, it can be written in the form \( \frac{d \mathbf{x}}{d t} = A \mathbf{x} \), where \( A \) is a matrix, and \( \mathbf{x} \) is a vector consisting of functions of time \( t \).
In this scenario, the system of differential equations given is:\[\frac{d}{dt}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} -2 & 1 \ 1 & -2 \end{bmatrix}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\]
This indicates a relationship between derivatives \( \frac{d}{dt}x_1(t) \) and \( \frac{d}{dt}x_2(t) \) and the functions \( x_1(t) \) and \( x_2(t) \).
To solve this system, you find functions \( x_1(t) \) and \( x_2(t) \) that satisfy both equations simultaneously with their respective derivatives matching the transformation imposed by matrix \( A \). This is crucial in applications such as physics and engineering, where systems are modeled to predict behaviors and responses.

Vector Calculus

Vector calculus is an extension of calculus involving functions with multiple variables. It is instrumental in analyzing and understanding the behavior of physical systems in multi-dimensional space. In the context of differential equations, vector calculus allows us to treat derivatives of vectors efficiently.
For vector-valued functions like \( \mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \), the derivative \( \frac{d \mathbf{x}}{dt} \) represents the rate of change of both components \( x_1(t) \) and \( x_2(t) \) with respect to time \( t \).

• The derivative \( \frac{d}{dt} x_1(t) \) shows how the function \( x_1(t) \) changes over time.
• Similarly, \( \frac{d}{dt} x_2(t) \) provides insight into the temporal evolution of \( x_2(t) \).

By utilizing vector calculus, particularly in conjunction with a matrix such as \( A \), we can efficiently predict how systems evolve:

• The notation \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) encapsulates the idea that the rate of change of the vector \( \mathbf{x} \) is influenced by both its current state and the dynamics described by \( A \).

Thus, vector calculus not only simplifies complex multi-variable problems but also offers insights into the interconnectedness of the components of dynamical systems.