Problem 26 - Calculus for Biology and Medicine

Question

Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 2 & 6 \\ 1 & 3 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=-3 \text { and } x_{2}(0)=1 \text { . } $$

Step-by-Step Solution

Answer

The particular solution is \( x_1(t) = -\frac{3}{5}e^{6t} - \frac{12}{5}e^{-t} \) and \( x_2(t) = -\frac{1}{5}e^{6t} + \frac{6}{5}e^{-t} \).

Step 1: Write the system in matrix form

The system of differential equations can be written in matrix form as: \[ \frac{d\mathbf{x}}{dt} = A\mathbf{x} \]where \( \mathbf{x} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) and \( A = \begin{bmatrix} 2 & 6 \ 1 & 3 \end{bmatrix} \).

Step 2: Determine the eigenvalues of matrix A

To find the eigenvalues of matrix \( A \), solve for \( \lambda \) in the equation: \[ \text{det}(A - \lambda I) = 0 \]where \( I \) is the identity matrix. Calculate:\[ A - \lambda I = \begin{bmatrix} 2-\lambda & 6 \ 1 & 3-\lambda \end{bmatrix} \]Find the determinant:\[ (2-\lambda)(3-\lambda) - 6(1) = \lambda^2 - 5\lambda - 6 = 0 \].Solve for \( \lambda \): The roots of \( \lambda^2 - 5\lambda - 6 = 0 \) are \( \lambda = 6 \) and \( \lambda = -1 \).

Step 3: Find eigenvectors for each eigenvalue

For \( \lambda = 6 \): solve \((A - 6I)\mathbf{v} = 0\). Calculate:\[ A - 6I = \begin{bmatrix} -4 & 6 \ 1 & -3 \end{bmatrix} \]Find a non-zero vector \( \mathbf{v_1} \): The eigenvector can be chosen as \( \mathbf{v_1} = \begin{bmatrix} 3 \ 1 \end{bmatrix} \).For \( \lambda = -1 \): solve \((A + I)\mathbf{v} = 0\). Calculate:\[ A + I = \begin{bmatrix} 3 & 6 \ 1 & 4 \end{bmatrix} \]Find a non-zero vector \( \mathbf{v_2} \): The eigenvector can be chosen as \( \mathbf{v_2} = \begin{bmatrix} -2 \ 1 \end{bmatrix} \).

Step 4: Write the general solution

The general solution of the system is:\[ \mathbf{x}(t) = c_1 e^{6t} \begin{bmatrix} 3 \ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2 \ 1 \end{bmatrix} \] where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

Step 5: Apply initial conditions

Substitute \( t = 0 \) into the general solution and use the initial conditions \( x_1(0) = -3 \) and \( x_2(0) = 1 \):\[ \begin{bmatrix} -3 \ 1 \end{bmatrix} = c_1 \begin{bmatrix} 3 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} -2 \ 1 \end{bmatrix} \]. This gives the system of equations:1. \( 3c_1 - 2c_2 = -3 \)2. \( c_1 + c_2 = 1 \).Solve these equations for \( c_1 \) and \( c_2 \).

Step 6: Solve system of equations for c1 and c2

From \( c_1 + c_2 = 1 \), we have \( c_2 = 1 - c_1 \). Substitute \( c_2 = 1 - c_1 \) into the first equation:\[ 3c_1 - 2(1 - c_1) = -3 \]which simplifies to:\[ 3c_1 - 2 + 2c_1 = -3 \] \[ 5c_1 = -1 \].Therefore, \( c_1 = -\frac{1}{5} \).Substitute back to find \( c_2 \):\[ c_2 = 1 + \frac{1}{5} = \frac{6}{5} \].

Step 7: Write the particular solution

Substitute \( c_1 = -\frac{1}{5} \) and \( c_2 = \frac{6}{5} \) into the general solution to get:\[ \mathbf{x}(t) = -\frac{1}{5} e^{6t} \begin{bmatrix} 3 \ 1 \end{bmatrix} + \frac{6}{5} e^{-t} \begin{bmatrix} -2 \ 1 \end{bmatrix} \].This can also be expressed as:\[ \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} -\frac{3}{5}e^{6t} - \frac{12}{5}e^{-t} \ -\frac{1}{5}e^{6t} + \frac{6}{5}e^{-t} \end{bmatrix} \].

Key Concepts

Eigenvalues and Eigenvectors

In the context of linear algebra, eigenvalues and eigenvectors are fundamental concepts that can help solve system of differential equations. When dealing with a matrix, eigenvalues are special scalars, while eigenvectors are non-zero vectors associated with these scalars. If you think of a matrix as a function that transforms vectors, then finding an eigenvector means identifying the vector that doesn't change direction under this transformation, only its magnitude, which is scaled by its eigenvalue.

Mathematically, this relationship is expressed as \( A\mathbf{v} = \lambda \mathbf{v} \), where \( A \) is a matrix, \( \mathbf{v} \) is an eigenvector, and \( \lambda \) is its corresponding eigenvalue. To find the eigenvalues of a matrix, one typically solves the characteristic equation \( \det(A - \lambda I) = 0 \). This involves subtracting \( \lambda \) along the diagonal of the matrix and equating its determinant to zero.

In our exercise, we calculated the eigenvalues \( \lambda = 6 \) and \( \lambda = -1 \) for the matrix \( A = \begin{bmatrix} 2 & 6 \ 1 & 3 \end{bmatrix} \). For each eigenvalue, we find a corresponding eigenvector by solving \( (A - \lambda I)\mathbf{v} = 0 \). This gives us the direction in which the system evolves. These steps help us form the general solution to our system of differential equations.

System of Differential Equations

A system of differential equations is a set of equations that involve derivatives of multiple interdependent variables. These systems often model real-world phenomena like population dynamics, electrical circuits, or mechanical systems, where multiple variables influence each other over time. To solve such systems mathematically, we often represent them in matrix form, which succinctly describes the relationships between variables.

Consider the problem at hand, modeled as \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), where \( \mathbf{x} \) is a vector of functions. Here, \( A \) is the matrix that determines how each function interacts with others in the system. Such a representation allows us to apply techniques from linear algebra, like finding eigenvalues and eigenvectors, thus simplifying the task of solving the system.

The matrix approach consolidates the differential equations into one concise equation. It paves the way for utilizing the powerful tools of linear algebra, allowing us to explore solutions that describe how the system evolves over time. The general solution comprises combinations of exponentials with eigenvalues and eigenvectors, representing the system's behavior under various initial conditions.

Initial Value Problems

Initial value problems (IVPs) are a class of differential equations where the solution is conditioned on specific initial values for the variables. These problems are common in applications where you are interested in how a system evolves from a particular starting state, for example, how a population grows from an initial size or how a capacitor charges from an initial voltage level.

The solution process begins by solving the general differential equation, often resulting in a solution with constants. The given initial conditions then provide specific values for these constants. In our exercise, we have the initial conditions \( x_1(0) = -3 \) and \( x_2(0) = 1 \). These initial conditions guide us in calculating the constants in the solution's expression.

By substituting these values into the general solution, we solve for the specific constants \( c_1 \) and \( c_2 \). This transforms the general solution into a particular solution, which fulfills the initial conditions and precisely describes the system's state over time. Solving IVPs gives a concrete outcome to abstract systems of equations, revealing the distinct path a system follows from its initial state.