Problem 24 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solution is \( x_1(t) = -e^{-5t} \) and \( x_2(t) = \frac{1}{2} e^{-5t} \).

Step 1: Write the System as a Matrix Equation

The given problem is a system of linear differential equations. It can be represented in matrix form as \[ \mathbf{x}'(t) = A\mathbf{x}(t) \], where \( \mathbf{x}(t) = \left[ \begin{array}{c} x_1(t) \ x_2(t) \end{array} \right] \) and \( A = \left[ \begin{array}{cc} -3 & 4 \ -1 & 2 \end{array} \right] \).

Step 2: Solve for Eigenvalues of the Matrix

To solve the system, we first need the eigenvalues of matrix \( A \). The eigenvalues \( \lambda \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Calculating the determinant, we have \[ \det\left( \begin{array}{cc} -3-\lambda & 4 \ -1 & 2-\lambda \end{array} \right) = (\lambda + 3)(\lambda - 2) + 4 = \lambda^2 + \lambda - 10 = 0. \] Solving this quadratic equation gives \( \lambda = 2 \) and \( \lambda = -5 \).

Step 3: Find the Eigenvectors

For each eigenvalue, we find the corresponding eigenvector by solving \( (A - \lambda I)\mathbf{v} = 0 \). For \( \lambda = 2 \), solving \[ \begin{array}{cc} -5 & 4 \ -1 & 0 \end{array} \] gives the eigenvector \( \mathbf{v}_1 = \left[ \begin{array}{c} 4 \ 1 \end{array} \right] \). For \( \lambda = -5 \), solving \[ \begin{array}{cc} 2 & 4 \ -1 & 7 \end{array} \] gives the eigenvector \( \mathbf{v}_2 = \left[ \begin{array}{c} 2 \ -1 \end{array} \right] \).

Step 4: Form the General Solution

With eigenvalues \( \, \lambda_1 = 2 \, \) and \( \, \lambda_2 = -5 \, \), and corresponding eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), the general solution to the system is \[ \mathbf{x}(t) = c_1 \begin{bmatrix} 4 \ 1 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 2 \ -1 \end{bmatrix} e^{-5t}. \]

Step 5: Apply Initial Conditions

Given \( x_1(0) = 1 \) and \( x_2(0) = 2 \), substitute into the solution \( \mathbf{x}(0) = c_1 \begin{bmatrix} 4 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} 2 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \). Solve the system \( 4c_1 + 2c_2 = 1 \) and \( c_1 - c_2 = 2 \). Solving, \( c_1 = 0 \) and \( c_2 = -1/2 \).

Step 6: Substitute Values to Find Particular Solution

Insert \( c_1 = 0 \) and \( c_2 = -1/2 \) into the general solution, yielding \[ \mathbf{x}(t) = -\frac{1}{2} \begin{bmatrix} 2 \ -1 \end{bmatrix} e^{-5t} = \begin{bmatrix} -e^{-5t} \ \frac{1}{2}e^{-5t} \end{bmatrix}. \] Thus, \( x_1(t) = -e^{-5t} \) and \( x_2(t) = \frac{1}{2} e^{-5t} \).

Key Concepts

Eigenvalues

Eigenvalues are crucial in understanding matrix equations, especially when solving differential equations. They are scalars that give insight into the behavior of a system described by a matrix. To find the eigenvalues of a matrix, you need to solve the characteristic equation. This involves subtracting a variable \( \lambda \) times the identity matrix from the original matrix, and then setting the determinant of the resulting matrix to zero.

In step 2 of our problem, we computed the eigenvalues \( \lambda \) for the matrix \( A \). The characteristic equation is derived from the determinant: - First, subtract \( \lambda \) from the diagonal elements of the matrix. - Then, compute the determinant, which involves multiplying the diagonal elements and subtracting the product of the off-diagonal elements.

For the matrix \( A \) given, this results in the quadratic equation \( \lambda^2 + \lambda - 10 = 0 \). Solving this equation gave us the eigenvalues \( \lambda = 2 \) and \( \lambda = -5 \), each of which describes a different dynamic behavior of the system.

Eigenvectors

Eigenvectors are vectors associated with eigenvalues, providing a direction along which a linear transformation acts by stretching or compressing. Once you have the eigenvalues, finding their corresponding eigenvectors involves solving the equation \( (A - \lambda I)\mathbf{v} = 0 \).

This equation helps identify the vectors that remain in the same span (line through the origin) under the transformation described by \( A \). Practically, this means setting up a system of linear equations using the found eigenvalue and solving for vector \( \mathbf{v} \).

For our exercise, once \( \lambda \) values were identified:

• We found the first eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 4 \ 1 \end{bmatrix} \) for \( \lambda = 2 \)
• And the second eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 2 \ -1 \end{bmatrix} \) for \( \lambda = -5 \)

These eigenvectors help construct the general solution of the differential equation by determining the direction and nature of the system's trajectories.

Matrix Equations

Matrix equations are used to express complex systems of linear equations succinctly. They are particularly effective at encapsulating differential equations into a form that can leverage powerful linear algebra techniques.

The system of differential equations we began with is represented in matrix form as \( \mathbf{x}'(t) = A\mathbf{x}(t) \). Here, \( \mathbf{x}(t) \) is a vector of functions, and \( A \) is a square matrix. This setup translates the problem of finding solutions to the differential equation into a problem of linear algebra.

The matrix equation setup helps:

• Identify eigenvalues and eigenvectors which provide the solution's structure
• Simplify the process of handling simultaneous differential equations

In our exercise, using matrix representation facilitated an organized approach to handle the initial-value problem, moving systematically from understanding system behavior to finding specific solutions that match given initial conditions.