Problem 23 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solution is \( x(t) = 2e^{2t}\begin{bmatrix} 7 \\ 2 \end{bmatrix} - e^{-3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} \)."

Step 1: Write the System of Equations

The system of differential equations is given as \(\frac{d x_1}{dt} = 4x_1 - 7x_2\) and \(\frac{d x_2}{dt} = 2x_1 - 5x_2\). This matrix form involves the vector \(\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\) and the matrix \(\begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix}\).

Step 2: Find the Eigenvalues of the Coefficient Matrix

Calculate the eigenvalues of the matrix \(\begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix}\). The eigenvalues can be found by solving the characteristic equation: \(\det(\begin{bmatrix} 4-\lambda & -7 \ 2 & -5-\lambda \end{bmatrix}) = 0\).

Step 3: Solve the Characteristic Equation

The characteristic equation is \((4 - \lambda)(-5 - \lambda) - (-7)(2) = 0\). Simplifying yields \(\lambda^2 + \lambda - 6 = 0\). Factor the quadratic to get \( (\lambda - 2)(\lambda + 3) = 0 \), giving the eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = -3\).

Step 4: Find the Eigenvectors

For each eigenvalue, solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\). For \(\lambda_1 = 2\), solve \(\begin{bmatrix} 2 & -7 \ 2 & -7 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), resulting in eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 7 \ 2 \end{bmatrix}\). For \(\lambda_2 = -3\), solve \(\begin{bmatrix} 7 & -7 \ 2 & -2 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), resulting in eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\).

Step 5: Form the General Solution

The general solution is \( \mathbf{x}(t) = c_1 e^{2t} \begin{bmatrix} 7 \ 2 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \). This combines the eigenvectors with exponential functions of eigenvalues.

Step 6: Apply Initial Conditions

Use initial conditions \(x_1(0)=13\) and \(x_2(0)=3\) to solve for constants \(c_1\) and \(c_2\). At \(t = 0\), the system becomes \(13 = 7c_1 + c_2\) and \(3 = 2c_1 + c_2\). Solve the system to get \(c_1 = 2\) and \(c_2 = -1\).

Step 7: Write the Particular Solution

Substitute the constants \(c_1 = 2\) and \(c_2 = -1\) back into the general solution to get the specific solution: \( \mathbf{x}(t) = 2e^{2t} \begin{bmatrix} 7 \ 2 \end{bmatrix} - e^{-3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \).

Key Concepts

System of Differential Equations

In the context of mathematics, a system of differential equations is a collection of two or more interrelated differential equations. These equations describe how multiple functions change over time, interacting with each other. Consider our example where we have:

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

Here, each function \( x_1(t) \) and \( x_2(t) \) is influenced by both current values of itself and the other function. This interdependence is often expressed using matrices, transforming the system into a form that is easier to analyze and solve.

The beauty of using matrices lies in their power to summarize complex systems succinctly. The initial value problem, with its condition specifying values at \( t = 0 \), further guides us to specific solutions from the general solutions possible.

Eigenvalues and Eigenvectors

To solve a system of differential equations, finding eigenvalues and eigenvectors is a crucial step. Eigenvalues indicate the specific scaling factors of transformations, whereas eigenvectors determine the direction of these transformations.

In our example, we consider the matrix:

• \( \begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix} \)

We find the eigenvalues by solving for \( \lambda \'s \) that make the matrix \( A - \lambda I \) singular. Upon solving, we discover:

• Eigenvalues: \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \)

Each eigenvalue leads to a specific eigenvector, which we determine by finding nontrivial solutions to \( (A - \lambda I)\mathbf{v} = \mathbf{0} \).
For these eigenvalues, we identify the eigenvectors:

• For \( \lambda_1 = 2 \), the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 7 \ 2 \end{bmatrix} \)
• For \( \lambda_2 = -3 \), the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \)

These elements are foundational to forming the general solution of the system.

Characteristic Equation

The characteristic equation is pivotal in finding eigenvalues of a matrix, and hence, essential in solving systems of differential equations. It is derived from the determinant of the matrix \( A - \lambda I \), where \( I \) denotes the identity matrix.

For our system, the matrix:

• \( \begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix} \)

has the characteristic equation expressed as:\[\det(\begin{bmatrix} 4-\lambda & -7 \ 2 & -5-\lambda \end{bmatrix}) = 0\]Solving this equation gives us the quadratic equation:\[\lambda^2 + \lambda - 6 = 0\]By factoring, or using the quadratic formula, we find solutions for \( \lambda \) - our eigenvalues. The solutions \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \) reflect how the system evolves, with each solution pairing with respective eigenvectors to influence the behavior of the solution over time.

Thus, the characteristic equation not only leads us to the eigenvalues but ultimately unveils important dynamics of the system's behavior.