Problem 25 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solution is \( x_1(t) = \frac{7}{2} e^{5t} - \frac{1}{2} e^{-3t} \) and \( x_2(t) = \frac{1}{2} e^{5t} + \frac{1}{2} e^{-3t} \).

Step 1: Set Up the System of Equations

The given differential equation can be expressed as a vector equation. We write it as \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), where \( A = \begin{bmatrix} 4 & 7 \ 1 & -2 \end{bmatrix} \) and \( \mathbf{x} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \). The initial condition is \( \mathbf{x}(0) = \begin{bmatrix} -3 \ 1 \end{bmatrix} \).

Step 2: Find the Eigenvalues of the Matrix

Determine the eigenvalues of matrix \( A \). Calculate the characteristic equation: \( \det(A - \lambda I) = 0 \). Here, \( A - \lambda I = \begin{bmatrix} 4-\lambda & 7 \ 1 & -2-\lambda \end{bmatrix} \). Compute the determinant: \( (4-\lambda)(-2-\lambda) - 7 = \lambda^2 - 2\lambda - 15 \). Solve \( \lambda^2 - 2\lambda - 15 = 0 \) to find \( \lambda = 5 \) and \( \lambda = -3 \).

Step 3: Find the Eigenvectors

For each eigenvalue, find its corresponding eigenvector by solving \( (A - \lambda I)\mathbf{v} = 0 \). For \( \lambda = 5 \): Solve \( \begin{bmatrix} -1 & 7 \ 1 & -7 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = 0 \), resulting in \( \mathbf{v} = \begin{bmatrix} 7 \ 1 \end{bmatrix} \). For \( \lambda = -3 \): Solve \( \begin{bmatrix} 7 & 7 \ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = 0 \), giving \( \mathbf{v} = \begin{bmatrix} -1 \ 1 \end{bmatrix} \).

Step 4: Construct the General Solution

The general solution to the system \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \) is of the form \( \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \), where \( \lambda_1 = 5 \), \( \lambda_2 = -3 \), \( \mathbf{v}_1 = \begin{bmatrix} 7 \ 1 \end{bmatrix} \), and \( \mathbf{v}_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \). Thus, \( \mathbf{x}(t) = c_1 e^{5t} \begin{bmatrix} 7 \ 1 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} -1 \ 1 \end{bmatrix} \).

Step 5: Apply Initial Conditions

Use the initial condition \( \mathbf{x}(0) = \begin{bmatrix} -3 \ 1 \end{bmatrix} \) to solve for constants \( c_1 \) and \( c_2 \). Substitute \( t = 0 \) into the general solution: \( c_1 \begin{bmatrix} 7 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \ 1 \end{bmatrix} = \begin{bmatrix} -3 \ 1 \end{bmatrix} \). Solving, \( 7c_1 - c_2 = -3 \) and \( c_1 + c_2 = 1 \). Solving these, \( c_1 = 1/2 \) and \( c_2 = 1/2 \).

Step 6: Final Solution

Substitute \( c_1 = 1/2 \) and \( c_2 = 1/2 \) back into the general solution to get the particular solution: \( \mathbf{x}(t) = \frac{1}{2} e^{5t} \begin{bmatrix} 7 \ 1 \end{bmatrix} + \frac{1}{2} e^{-3t} \begin{bmatrix} -1 \ 1 \end{bmatrix} \). The individual components are \( x_1(t) = \frac{7}{2} e^{5t} - \frac{1}{2} e^{-3t} \) and \( x_2(t) = \frac{1}{2} e^{5t} + \frac{1}{2} e^{-3t} \).

Key Concepts

Eigenvalues and Eigenvectors

If you are studying systems of linear differential equations, understanding eigenvalues and eigenvectors is crucial. Eigenvalues provide us insight into the behavior of the system and whether the solutions will grow, decay, or oscillate over time. Eigenvectors give direction to these behaviors.

To find the eigenvalues of a matrix, you have to solve the characteristic equation. This equation is derived from the determinant of \(A - \lambda I\), where \(A\) is your matrix and \(\lambda\) represents eigenvalues. In simpler terms, you replace \(\lambda\) with variable values until you find which values make the determinant zero.

Once you have the eigenvalues, eigenvectors are found by solving \( (A - \lambda I)\mathbf{v} = 0\). Here's a straightforward way to think about it:

• Plug in each eigenvalue to find the corresponding eigenvector.
• For each \(\lambda\), subtract it from the diagonal of matrix \(A\).
• Solve the resulting system for vector \(\mathbf{v}\).

The solution vector \(\mathbf{v}\) that results is your eigenvector.

Initial Value Problem

An initial value problem involves solving a differential equation with given conditions at a specific point, often \(t = 0\). In this exercise, it was about determining the particular solution to the given system by incorporating initial conditions.

Initial conditions provide the necessary information to pinpoint exact solutions among many potential ones offered by the general solutions. In simpler terms, when you find a general solution through solving differential equations, it will contain arbitrary constants. Initial conditions help you solve for these constants.

For example:

• You find the general solution involving constants \(c_1\) and \(c_2\).
• Plug in the initial conditions such as \(\mathbf{x}(0) = \begin{bmatrix} -3 \ 1 \end{bmatrix}\) to create system equations.
• Solve these equations to find the values of \(c_1\) and \(c_2\).

The particular solution is the one that matches both the differential equation and the initial conditions.

Systems of Linear Differential Equations

Systems of linear differential equations can be viewed as a collection of equations working together to describe a complex dynamic system. In such systems, the rate of change of multiple variables depends not only on those variables themselves but also on others.

These types of equations often arise in modeling real-world phenomena across numerous fields such as physics, engineering, and finance. Understanding how to convert these systems into vector equations is key.


The vector form \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\) is efficient because it unifies various differential equations into a single expression. Here:

• \(\mathbf{x}\) is a vector that contains the system's variables, for instance, \(\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\).
• \(A\) is a matrix that describes how each variable in the system is influenced by the others.

By understanding how to manipulate these forms, you can analyze the dynamics and predict the behavior of the system over time.