Problem 13 - Calculus for Biology and Medicine

Question

In Problems 13-18, 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 11.32) $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 1 & 3 \\ 5 & 3 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Step-by-Step Solution

Answer

The general solution is \(\mathbf{x}(t) = c_1 e^{6t} \begin{bmatrix}1 \\ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix}-1/3 \\ 1 \end{bmatrix}\). Solutions move away on \(\mathbf{v}_1\) and toward origin on \(\mathbf{v}_2\).

Step 1: Write the System of Equations

Given the system of differential equations can be written as matrix multiplication: \(\begin{bmatrix}\frac{dx_1}{dt} \\frac{dx_2}{dt} \end{bmatrix} = \begin{bmatrix}1 & 3 \5 & 3\end{bmatrix}\begin{bmatrix}x_1(t) \x_2(t) \end{bmatrix}\)We need to solve for the general solution of this system.

Step 2: Find Eigenvalues

To find the eigenvalues of matrix \(A\), use the characteristic equation \(\det(A - \lambda I) = 0\). The matrix \(A\) is:\[\begin{bmatrix}1 & 3 \5 & 3\end{bmatrix}\]The characteristic equation is:\[\begin{vmatrix}1-\lambda & 3 \5 & 3-\lambda\end{vmatrix} = (1-\lambda)(3-\lambda) - 15 = 0\]Solve the resulting quadratic equation for \(\lambda\). This gives \(\lambda^2 - 4\lambda - 12 = 0\).

Step 3: Solve for Eigenvalues

Solve the quadratic equation \(\lambda^2 - 4\lambda - 12 = 0\) using the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For \(a = 1, b = -4, c = -12\), the eigenvalues are:\[\lambda_1 = 6, \quad \lambda_2 = -2\]

Step 4: Find Eigenvectors

For each eigenvalue \(\lambda\), solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\) to find the corresponding eigenvectors. For \(\lambda_1 = 6\):\[\begin{bmatrix}1-6 & 3 \5 & 3-6\end{bmatrix} \begin{bmatrix}v_1 \v_2 \end{bmatrix} = \begin{bmatrix}0 \0\end{bmatrix}\]Solve to find \(\mathbf{v}_1\).Similarly for \(\lambda_2 = -2\) solve the equation to find \(\mathbf{v}_2\).

Step 5: Solve for Eigenvectors

For \(\lambda_1 = 6\), solving gives the eigenvector \(\mathbf{v}_1 = \begin{bmatrix}1 \1\end{bmatrix}\).For \(\lambda_2 = -2\), solving gives the eigenvector \(\mathbf{v}_2 = \begin{bmatrix}-1/3 \1\end{bmatrix}\).

Step 6: Form General Solution

The general solution of the system is given by:\[\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\]Substituting the values:\[\mathbf{x}(t) = c_1 e^{6t} \begin{bmatrix}1 \1\end{bmatrix} + c_2 e^{-2t} \begin{bmatrix}-1/3 \1\end{bmatrix}\]

Step 7: Sketch and Analyze Direction

Sketch the eigenvectors as lines on the plane. For \(\mathbf{v}_1 = \begin{bmatrix}1 \1\end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix}-1/3 \1\end{bmatrix}\), draw lines along these directions.The solutions move **away** from the origin along \(\mathbf{v}_1\) (since \(e^{6t}\) grows) and move **towards** the origin along \(\mathbf{v}_2\) (since \(e^{-2t}\) decays).

Key Concepts

Eigenvalues

In the world of linear algebra and differential equations, eigenvalues play a pivotal role. They are values that help us understand the behavior of linear transformations. For a given matrix \( A \), obtaining eigenvalues requires solving the characteristic equation \( \det(A - \lambda I) = 0 \). This involves subtracting \( \lambda \) times the identity matrix from \( A \) and then calculating the determinant. In our problem, solving the characteristic equation gives eigenvalues \( \lambda_1 = 6 \) and \( \lambda_2 = -2 \). Each of these values provides vital information about how a system behaves over time. The eigenvalues indicate growth or decay:

• Positive eigenvalues like \( \lambda_1 = 6 \) suggest exponential growth.
• Negative eigenvalues like \( \lambda_2 = -2 \) suggest exponential decay.

Understanding these values can tell you how the solutions of a differential equation system will evolve.

Eigenvectors

Associated with each eigenvalue is at least one corresponding eigenvector. Eigenvectors are non-zero vectors that change at most by a scalar factor when the corresponding linear transformation is applied. To find eigenvectors, solve the equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). In this exercise, for eigenvalue \( \lambda_1 = 6 \), the eigenvector is \( \mathbf{v}_1 = \begin{bmatrix}1 \ 1\end{bmatrix} \). For \( \lambda_2 = -2 \), the eigenvector is \( \mathbf{v}_2 = \begin{bmatrix}-1/3 \ 1\end{bmatrix} \). These vectors are crucial because:

• They define directions along which the system expands or contracts.
• They help form the general solution by providing a basis for the solution space.

The directions indicated by eigenvectors help us understand the movement along these lines in the solution space.

Matrix Multiplication

Matrix multiplication is an essential operation in linear algebra used to solve systems of differential equations. Consider the system expressed in matrix form as:\[\begin{bmatrix}\frac{dx_1}{dt} \ \frac{dx_2}{dt} \end{bmatrix} = \begin{bmatrix}1 & 3 \ 5 & 3\end{bmatrix}\begin{bmatrix}x_1(t) \ x_2(t) \end{bmatrix}\]When multiplying matrices, each element of the resulting matrix is calculated as the dot product of the corresponding row of the first matrix and the column of the second matrix:

• Multiply elements in the row by corresponding column elements.
• Sum up these products.

In the context of differential equations, this operation links derivatives of variables with coefficients, enabling the study of systems in a compact form. Proper understanding of matrix multiplication helps in setting up and solving such systems efficiently.

General Solution

The general solution of a system of linear differential equations provides a complete description of the system's behavior over time. It is constructed using the eigenvalues and eigenvectors. The solution is expressed as:\[\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\]where \( c_1 \) and \( c_2 \) are constants determined by initial conditions. In the given problem:

• \( \lambda_1 = 6 \), \( \mathbf{v}_1 = \begin{bmatrix}1 \ 1\end{bmatrix} \) leads to exponential growth away from the origin.
• \( \lambda_2 = -2 \), \( \mathbf{v}_2 = \begin{bmatrix}-1/3 \ 1\end{bmatrix} \) indicates decay towards the origin.

Understanding the general solution allows us to capture the complete dynamics of the system, predicting long-term behavior based on initial conditions.