Problem 18 - Calculus for Biology and Medicine

Question

Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the direction in which the solution would move if it starts on that line. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 5 & 2 \\ 1 & 6 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{x}(t) = c_1 e^{7t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{4t} \begin{bmatrix} -2 \\ 1 \end{bmatrix} \).

Step 1: Write the System in Matrix Form

The system of differential equations is given by: \[ \frac{d}{dt} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} 5 & 2 \ 1 & 6 \end{bmatrix} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \] This can be expressed as \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), where \( A = \begin{bmatrix} 5 & 2 \ 1 & 6 \end{bmatrix} \).

Step 2: Find Eigenvalues of Matrix

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). For matrix \( A \), the characteristic equation is:\[ \det \left(\begin{bmatrix} 5 - \lambda & 2 \ 1 & 6 - \lambda \end{bmatrix}\right) = (5 - \lambda)(6 - \lambda) - 2 \]\[ = \lambda^2 - 11\lambda + 28 \]Setting \( \lambda^2 - 11\lambda + 28 = 0 \), we solve for \( \lambda \).

Step 3: Calculate Eigenvalues

The characteristic equation can be factored as \[ (\lambda - 7)(\lambda - 4) = 0 \]The eigenvalues are \( \lambda_1 = 7 \) and \( \lambda_2 = 4 \).

Step 4: Find Eigenvectors for Each Eigenvalue

Substitute \( \lambda_1 = 7 \) into \( (A - \lambda I)\mathbf{v} = 0 \), giving:\[ \begin{bmatrix} -2 & 2 \ 1 & -1 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]From this, \( v_1 = v_2 \), so an eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).Do the same for \( \lambda_2 = 4 \):\[ \begin{bmatrix} 1 & 2 \ 1 & 2 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \] Leading to \( v_1 = -2v_2 \), resulting in eigenvector \( \mathbf{v}_2 = \begin{bmatrix} -2 \ 1 \end{bmatrix} \).

Step 5: Write the General Solution

The general solution to the system is given by:\[ \mathbf{x}(t) = c_1 e^{7t} \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 e^{4t} \begin{bmatrix} -2 \ 1 \end{bmatrix} \]where \( c_1 \) and \( c_2 \) are arbitrary constants.

Step 6: Sketch the Eigenvectors and Describe the Solution Trajectories

On a coordinate plane, sketch the lines corresponding to the eigenvectors \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} -2 \ 1 \end{bmatrix} \). Solutions move outward along the line due to positive eigenvalues, indicating trajectories moving away from the origin as time \( t \) increases. For \( \lambda_1 = 7 \), the growth is faster along the direction of \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) compared to \( \lambda_2 = 4 \).

Key Concepts

Eigenvalues

Eigenvalues are crucial in understanding the behavior of differential equations, especially in systems of equations. They are scalars associated with a square matrix, in our case matrix \( A \), which help us determine the nature of the solutions. To find eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \). This results in a polynomial whose roots are the eigenvalues.

In our problem, we find two eigenvalues: \( \lambda_1 = 7 \) and \( \lambda_2 = 4 \). These eigenvalues tell us about the growth rate of solutions over time.

• When the eigenvalues are positive, like \( 7 \) and \( 4 \), the solutions move outward, indicating a system that moves away from equilibrium or the origin.
• The size of the eigenvalue defines the rate of divergence from the origin. For example, the eigenvalue \( 7 \) indicates a faster growth compared to \( 4 \).

Understanding eigenvalues is key to predicting the long-term behavior of differential systems.

Eigenvectors

Eigenvectors are vectors that remain along the same direction after transformation by the matrix, even though they might be scaled by the eigenvalue. Once we have our eigenvalues, the next step is to find the corresponding eigenvectors. Each solution vector in the differential equation system is aligned along these eigenvectors.

In our given system, for the eigenvalue \( \lambda_1 = 7 \), we derive the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 1 \end{bmatrix} \). Similarly, for \( \lambda_2 = 4 \), the eigenvector is \( \mathbf{v}_2 = \begin{bmatrix} -2 1 \end{bmatrix} \).

• These vectors give us the principal directions or lines along which the behavior of the system can be analyzed.
• Each eigenvector corresponds to a specific equation, forming lines in the phase space.

These vectors, when graphed, show us the direction of movement as time progresses.

System of Equations

A system of differential equations involves a set of equations, usually more than one, including derivatives of unknown functions. It models various processes and phenomena in fields like physics, engineering, economics, etc., where multiple variables change with respect to each other and time.

In our exercise, we deal with a system expressed in matrix form:

• The matrix \( A = \begin{bmatrix} 5 2 1 6 \end{bmatrix} \) acts on the vector \( \mathbf{x} = \begin{bmatrix} x_1(t) x_2(t) \end{bmatrix} \) representing the variables of the system.
• This compact notation helps streamline complex operations and makes solving with linear algebra techniques feasible.

With systems like these, the interplay of variables can be explored through matrices, eigenvalues, and eigenvectors, leading to general solutions that describe how the system evolves over time.