Problem 14 - 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. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 2 & 1 \\ 4 & -1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Step-by-Step Solution

Answer

The general solution is \( x(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} -1 \\ 4 \end{bmatrix} e^{-2t} \).

Step 1: Rewrite the System of Differential Equations

The given system of differential equations is: \( \dot{x} = A x \), where \( x = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) and \( A = \begin{bmatrix} 2 & 1 \ 4 & -1 \end{bmatrix} \). Our task is to find the general solution of this system.

Step 2: Find the Eigenvalues of Matrix A

The eigenvalues of matrix \( A \) are determined by solving the characteristic equation \( \det(A - \lambda I) = 0 \). This leads to:\[\det \begin{bmatrix} 2 - \lambda & 1 \ 4 & -1 - \lambda \end{bmatrix} = (2 - \lambda)(-1 - \lambda) - 4 \cdot 1 = 0\]Expanding and simplifying, we get:\[\lambda^2 - \lambda - 6 = 0\]Solving for \( \lambda \) using the quadratic formula gives \( \lambda = 3 \) and \( \lambda = -2 \).

Step 3: Find the Eigenvectors for Each Eigenvalue

1. **For \( \lambda = 3 \):** Solve \( (A - 3I)v = 0 \) with \( A - 3I = \begin{bmatrix} -1 & 1 \ 4 & -4 \end{bmatrix} \). Solving this gives \( v_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).2. **For \( \lambda = -2 \):** Solve \( (A + 2I)v = 0 \) with \( A + 2I = \begin{bmatrix} 4 & 1 \ 4 & 1 \end{bmatrix} \). Solving this gives \( v_2 = \begin{bmatrix} -1 \ 4 \end{bmatrix} \).

Step 4: Write the General Solution

The general solution is a linear combination of the solutions along each eigenvector:\[ x(t) = c_1 \begin{bmatrix} 1 \ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} -1 \ 4 \end{bmatrix} e^{-2t} \].Where \( c_1 \) and \( c_2 \) are arbitrary constants determined by initial conditions.

Step 5: Sketch the Lines in the Direction of Eigenvectors

On a coordinate plane, sketch the line \( x_1 = x_2 \) corresponding to \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( x_2 = -4x_1 \) corresponding to \( \begin{bmatrix} -1 \ 4 \end{bmatrix} \). Arrows on the lines indicate the direction of movement. On \( x_1 = x_2 \), direction is away from origin (growing exponentially with \( e^{3t} \)); on \( x_2 = -4x_1 \), direction points toward origin (decaying with \( e^{-2t} \)).

Key Concepts

Eigenvalues

In dealing with systems of differential equations, finding eigenvalues is an essential step towards understanding how the system behaves. The eigenvalues of a matrix offer insight into whether the solutions to the system grow, shrink, or oscillate over time. These values are derived from the characteristic equation of a matrix. In our case, for the matrix \( A = \begin{bmatrix} 2 & 1 \ 4 & -1 \end{bmatrix} \), we form the characteristic equation by setting the determinant of \( (A - \lambda I) \) equal to zero: \[(2 - \lambda)(-1 - \lambda) - 4 \cdot 1 = 0\]Solving the above equation leads us to a quadratic equation:\[\lambda^2 - \lambda - 6 = 0\]Utilizing the quadratic formula, we find the roots, or eigenvalues, as \( \lambda = 3 \) and \( \lambda = -2 \). These calculations reveal significant aspects of the system's character:

• Positive eigenvalue \( \lambda = 3 \) suggests solutions that grow exponentially.
• Negative eigenvalue \( \lambda = -2 \) suggests solutions that decay exponentially.

Eigenvectors

Once we have the eigenvalues of a matrix, the next step is to find the corresponding eigenvectors. Eigenvectors are vectors that, when multiplied by the matrix, only stretch or compress and do not change direction. They indicate the directions along which the transformation by the matrix is just a scaling transformation.

Finding Eigenvectors

To find eigenvectors for an eigenvalue \( \lambda \), we solve the equation \( (A - \lambda I)v = 0 \):

• **For \( \lambda = 3 \)**: The matrix \( A - 3I = \begin{bmatrix} -1 & 1 \ 4 & -4 \end{bmatrix} \) leads to solving for vector \( v_1 = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \). Solving gives \( v_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
• **For \( \lambda = -2 \)**: The matrix \( A + 2I = \begin{bmatrix} 4 & 1 \ 4 & 1 \end{bmatrix} \) results in solving for vector \( v_2 = \begin{bmatrix} -1 \ 4 \end{bmatrix} \).

These vectors guide the direction of solutions in our system of differential equations. They provide a map of how solutions evolve over time related to the system's structure.

System of Differential Equations

A system of differential equations is a set of interrelated equations involving derivatives of unknown functions. These systems are fundamental in modeling dynamics where multiple variables change regarding time or another metric. In the context of our particular exercise, our system is given by:\[\frac{d}{dt}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \ 4 & -1 \end{bmatrix} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\]This setup implies the derivatives \( \frac{dx_1}{dt} \) and \( \frac{dx_2}{dt} \) are linarly dependent upon the current values of \( x_1 \) and \( x_2 \). To solve such systems, we employ the method of finding eigenvalues and eigenvectors of the coefficient matrix.The solution to the system expresses in terms of the eigenvectors and related exponential functions:\[x(t) = c_1 \begin{bmatrix} 1 \ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} -1 \ 4 \end{bmatrix} e^{-2t}\]Where \( c_1 \) and \( c_2 \) are constants determined by initial conditions, illustrating a superposition principle inherent in linear systems. This expression effectively captures all possible behaviors of the solution paths through time, driven by the initial setup of the differential equations.