Problem 23 - Calculus for Biology and Medicine

Question

Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{cc} 4 & -7 \\ 2 & -5 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=13 \text { and } x_{2}(0)=3 \text { . } $$

Step-by-Step Solution

Answer

The particular solution is \( x_1(t) = 14e^{2t} - e^{-3t} \) and \( x_2(t) = 4e^{2t} - e^{-3t} \).

Step 1: Write the System of Differential Equations

The given matrix equation can be expanded into a system of linear differential equations: \( \frac{d x_1}{d t} = 4x_1 - 7x_2 \) and \( \frac{d x_2}{d t} = 2x_1 - 5x_2 \).

Step 2: Find the Eigenvalues of the Coefficient Matrix

The coefficient matrix \( A \) is \( \begin{pmatrix} 4 & -7 \ 2 & -5 \end{pmatrix} \). Find the eigenvalues by solving \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Compute: \[ \begin{vmatrix} 4-\lambda & -7 \ 2 & -5-\lambda \end{vmatrix} = (4-\lambda)(-5-\lambda) - (-7)(2) \]. This simplifies to \( \lambda^2 + \lambda - 6 = 0 \).

Step 3: Solve the Characteristic Equation

The characteristic equation \( \lambda^2 + \lambda - 6 = 0 \) can be factored as \( (\lambda - 2)(\lambda + 3) = 0 \). Thus, the eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \).

Step 4: Find the Eigenvectors

For \( \lambda_1 = 2 \), solve \( (A - 2I)\mathbf{v_1} = \mathbf{0} \). This yields the augmented matrix: \( \begin{pmatrix} 2 & -7 \ 2 & -7 \end{pmatrix} \rightarrow \mathbf{v_1} = \begin{pmatrix} 7 \ 2 \end{pmatrix} \). For \( \lambda_2 = -3 \), solve \( (A + 3I)\mathbf{v_2} = \mathbf{0} \), giving: \( \begin{pmatrix} 7 & -7 \ 2 & -2 \end{pmatrix} \rightarrow \mathbf{v_2} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 5: Construct the General Solution

The general solution of the system is a linear combination of the eigenvectors with exponential terms: \( \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} = c_1 \begin{pmatrix} 7 \ 2 \end{pmatrix}e^{2t} + c_2 \begin{pmatrix} 1 \ 1 \end{pmatrix}e^{-3t} \).

Step 6: Apply Initial Conditions

Substitute \( t = 0 \), \( x_1(0) = 13 \), and \( x_2(0) = 3 \) into the general solution to find \( c_1 \) and \( c_2 \). Solve: \[ 7c_1 + c_2 = 13 \] \[ 2c_1 + c_2 = 3 \]. Subtract the second equation from the first to get \( 5c_1 = 10 \), so \( c_1 = 2 \). Substitute \( c_1 = 2 \) back into \( 2c_1 + c_2 = 3 \) to find \( c_2 = -1 \).

Step 7: Write the Particular Solution

Substitute \( c_1 = 2 \) and \( c_2 = -1 \) into the general solution: \( \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} = 2 \begin{pmatrix} 7 \ 2 \end{pmatrix} e^{2t} - \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{-3t} \).

Key Concepts

Linear Differential Equations

Linear differential equations form a crucial part of calculus and mathematical analysis. They involve derivatives of a function and are called 'linear' because the function and its derivatives appear in a linear form. In an initial-value problem, you not only solve these equations but also find a specific solution that satisfies given starting conditions. For instance, in our problem, we deal with a system of linear differential equations:

• \( \frac{dx_1}{dt} = 4x_1 - 7x_2 \)
• \( \frac{dx_2}{dt} = 2x_1 - 5x_2 \)

This setup means that the rate of change of each function is linearly dependent on both functions. To find the unique solution, we have initial conditions. These are values of the variables at \( t = 0 \), such as \( x_1(0) = 13 \) and \( x_2(0) = 3 \), which allow us to determine specific constants in the solution.

Eigenvalues and Eigenvectors

In the context of linear differential equations, eigenvalues and eigenvectors offer powerful tools to simplify the problem. They are related to matrices, particularly when we encounter a system of equations that can be represented in matrix form. Here, the matrix is given by:

• \( A = \begin{pmatrix} 4 & -7 \ 2 & -5 \end{pmatrix} \)

Eigenvalues are special numbers associated with a matrix, which, when related to an eigenvector, indicate that the matrix stretches or shrinks this vector without changing its direction. We find eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \). In our example, this results in the eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \).
Eigenvectors are then found by substituting each eigenvalue back into the matrix equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). For \( \lambda_1 = 2 \), the eigenvector is \( \begin{pmatrix} 7 \ 2 \end{pmatrix} \), and for \( \lambda_2 = -3 \), it is \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
These eigenvectors form the basis of the solution to the differential equation, as a linear combination of these vectors determines the behavior of the system over time.

Matrix Algebra

Matrix algebra provides a structured way to handle linear equations, especially when dealing with multiple variables and their relationships. In our exercise, the system of differential equations can be compactly written in matrix form, which allows us to apply matrix operations and concepts like eigenvalues and eigenvectors more efficiently.
The matrix \( A = \begin{pmatrix} 4 & -7 \ 2 & -5 \end{pmatrix} \) captures the coefficients of the variables in the system and directly relates to how the system evolves over time. By representing the system in matrix form, you utilize the power of linear transformations, where the matrix operates on a vector (in this case, the state of the system \( \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} \)) to produce its derivative.

• This approach streamlines the solving process, helping to leverage linear algebraic methods like finding determinants and eigenvector decomposition to solve the system.
• In practice, it allows for easier computation and clearer insight into the system's behavior, particularly when handling complex systems with many variables.

Thus, matrix algebra not only offers a concise representation but also connects the theoretical with the computational aspects of solving linear differential equations.