Problem 18 - Calculus for Biology and Medicine

Question

Find the general solution of each system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line of eigenvectors the direction in which the solution would move if it starts on that line. (Figure 11.37) $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{cc} 7 & 4 \\ 2 & 5 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Step-by-Step Solution

Answer

The general solution is \(\mathbf{x}(t) = c_1 \begin{bmatrix}2 \\ 1\end{bmatrix} e^{9t} + c_2 \begin{bmatrix}1 \\ -1\end{bmatrix} e^{3t}\), with solutions moving away from the origin along eigenvector lines.

Step 1: Write the System in Matrix Form

The given system of differential equations is already provided in matrix form as \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\), with \(A = \begin{bmatrix} 7 & 4 \ 2 & 5 \end{bmatrix}\) and \(\mathbf{x} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\). Our task is to find the general solution for this system.

Step 2: Find the Eigenvalues of Matrix A

To find the eigenvalues, solve the characteristic equation \(\det(A - \lambda I) = 0\). Here,\[\det\begin{bmatrix}7 - \lambda & 4 \ 2 & 5 - \lambda \end{bmatrix} = (7 - \lambda)(5 - \lambda) - (4)(2) = 0.\]Simplifying this results in the characteristic polynomial:\[(7 - \lambda)(5 - \lambda) - 8 = \lambda^2 - 12\lambda + 27 = 0.\]Solve for \(\lambda\) using the quadratic formula:

Step 3: Solve for Eigenvalues

The quadratic formula is \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-12\), and \(c=27\). Substitute the values:\[\lambda = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 27}}{2} = \frac{12 \pm \sqrt{36}}{2}.\]This simplifies to:\[\lambda = \frac{12 \pm 6}{2}.\]The roots are \(\lambda_1 = 9\) and \(\lambda_2 = 3\).

Step 4: Find the Eigenvectors Corresponding to Each Eigenvalue

First, consider \(\lambda_1 = 9\). To find the eigenvector \(\mathbf{v_1}\), solve \((A - 9I)\mathbf{v_1} = \mathbf{0}\):\[\begin{bmatrix}7 - 9 & 4 \ 2 & 5 - 9 \end{bmatrix}\begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}0 \ 0\end{bmatrix}.\]Simplifies to:\[\begin{bmatrix}-2 & 4 \ 2 & -4\end{bmatrix}\begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}0 \ 0\end{bmatrix}.\]This gives equations \(-2x + 4y = 0\) and \(2x - 4y = 0\). Solving:\[\mathbf{v_1} = \begin{bmatrix}2 \ 1\end{bmatrix}.\]

Step 5: Find the Second Eigenvector

Now consider \(\lambda_2 = 3\). Solve \((A - 3I)\mathbf{v_2} = \mathbf{0}\):\[\begin{bmatrix}7 - 3 & 4 \ 2 & 5 - 3 \end{bmatrix}\begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}0 \ 0\end{bmatrix}\]Simplifies to:\[\begin{bmatrix}4 & 4 \ 2 & 2\end{bmatrix}\begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}0 \ 0\end{bmatrix}.\]This gives equations \(2x + 2y = 0\) and \(x = -y\). Solving:\[\mathbf{v_2} = \begin{bmatrix}1 \ -1\end{bmatrix}.\]

Step 6: Write the General Solution

The general solution for the system can be written as a linear combination of the solutions associated with each eigenpair:\[\mathbf{x}(t) = c_1 \begin{bmatrix}2 \ 1\end{bmatrix} e^{9t} + c_2 \begin{bmatrix}1 \ -1\end{bmatrix} e^{3t},\]where \(c_1\) and \(c_2\) are constants determined by initial conditions.

Step 7: Sketch Direction Lines of Eigenvectors

The direction lines of the eigenvectors are \(\mathbf{v_1} = \begin{bmatrix}2 \ 1\end{bmatrix}\) and \(\mathbf{v_2} = \begin{bmatrix}1 \ -1\end{bmatrix}\). For \(\mathbf{v_1}\), draw a line in the direction \(\begin{bmatrix}2 \ 1\end{bmatrix}\) with solutions moving away from the origin as \(t\) increases because \(e^{9t}\) grows rapidly. For \(\mathbf{v_2}\), draw a line in the direction \(\begin{bmatrix}1 \ -1\end{bmatrix}\). Solutions also move away from the origin for increasing \(t\) due to \(e^{3t}\), but more slowly.

Key Concepts

Eigenvalues

Eigenvalues play a crucial role in solving systems of differential equations. They are the scalars \(\lambda\) in the equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \(A\) is a square matrix and \(\mathbf{v}\) is a nonzero vector, called the eigenvector corresponding to \(\lambda\). To find the eigenvalues of a matrix, we use the **characteristic equation**, which we derive from the determinant formula:

• Subtract \(\lambda\) times the identity matrix, \(I\), from \(A\).
• Calculate the determinant of \(A - \lambda I\).
• Set the determinant to zero: \(\det(A - \lambda I) = 0\).

Solving this equation provides the eigenvalues. In our problem, the characteristic polynomial is \(\lambda^2 - 12\lambda + 27 = 0\), and the eigenvalues found are \(\lambda_1 = 9\) and \(\lambda_2 = 3\). These values help determine the behavior and stability of the system of differential equations.

Eigenvectors

Once we have the eigenvalues, the next step is finding the corresponding eigenvectors. Eigenvectors are vectors that satisfy the equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \(\lambda\) is a known eigenvalue. The process is as follows:

• Plug an eigenvalue \(\lambda\) into \(A - \lambda I\) to form a new matrix.
• Solve the system \( (A - \lambda I)\mathbf{v} = \mathbf{0} \).

For instance, with \(\lambda_1 = 9\), the system becomes \(\begin{bmatrix}-2 & 4 \ 2 & -4\end{bmatrix}\mathbf{v_1} = \mathbf{0}\), resulting in the eigenvector \(\mathbf{v_1} = \begin{bmatrix}2 \ 1\end{bmatrix}\). Similarly, for \(\lambda_2 = 3\), the eigenvector is found as \(\mathbf{v_2} = \begin{bmatrix}1 \ -1\end{bmatrix}\). These vectors indicate the directions of stretching or compressing in the system.

Characteristic Equation

The characteristic equation is a polynomial equation obtained directly from an \(n \times n\) matrix. It involves the determinant of the matrix formed by subtracting \(\lambda I\) from \(A\). Its roots are the eigenvalues.Here's how you form it:

• Consider \(A - \lambda I\), where \(I\) is the identity matrix of the same size as \(A\).
• Calculate the determinant \(\det(A - \lambda I)\).
• For our matrix, this results in the equation \(\lambda^2 - 12\lambda + 27 = 0\).

Solving this equation gives us the eigenvalues required to solve the system of differential equations. Understanding this equation is critical, as it ties together the matrix algebra with systems of differential equations, aiding in discovering the intrinsic properties of the system.

General Solution

To find the general solution of a system of differential equations, we combine the eigenvalues and eigenvectors. The general solution is a combination of terms, each associated with an eigenpair (eigenvalue-eigenvector pair):\[\mathbf{x}(t) = c_1 \mathbf{v_1} e^{\lambda_1 t} + c_2 \mathbf{v_2} e^{\lambda_2 t},\]where \(c_1\) and \(c_2\) are constants determined by initial conditions, \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are eigenvectors, and \(\lambda_1\) and \(\lambda_2\) are eigenvalues.

• Each term \(e^{\lambda t}\) describes the growth or decay behavior relative to time \(t\).
• The constants \(c_1\) and \(c_2\) adjust the weight of each solution based on starting conditions.

In our example, we have solutions linked to eigenpairs, leading to the complete expression describing the system's dynamic behavior.

Matrix Form

Matrix form is a way of representing systems of differential equations using matrices, which makes complex systems easier to handle. For a system of linear differential equations \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\), the matrix form involves two key components:

• **Matrix \(A\):** Contains coefficients that determine how each variable affects the rate of change of others.
• **Vector \(\mathbf{x}(t)\):** Represents the variables in the system.

Writing systems of equations in matrix form simplifies the process of finding eigenvalues, eigenvectors, and eventually the general solution. This compact representation allows for a broader analysis of the system's properties, such as stability and response to initial conditions. In our scenario, \(A = \begin{bmatrix} 7 & 4 \ 2 & 5 \end{bmatrix}\) shows interactions in the system of equations effectively, showcasing the utility of the matrix approach in differential equations.