Problem 15 - 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. $$ \left[\begin{array}{l} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{rr} -3 & 3 \\ 6 & 4 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{X}(t) = c_1 e^{6t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{-5t} \begin{pmatrix} -3 \\ 2 \end{pmatrix} \).

Step 1: Write the System as a Matrix Equation

The given system of differential equations can be represented in matrix form as \( \frac{dX}{dt} = AX \), where \( X = \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} \) and \( A = \begin{pmatrix} -3 & 3 \ 6 & 4 \end{pmatrix} \).

Step 2: Find the Eigenvalues of the Matrix A

To find the eigenvalues of matrix \( A \), solve the characteristic equation \( \det(A - \lambda I) = 0 \). This gives \[ \det \begin{pmatrix} -3 - \lambda & 3 \ 6 & 4 - \lambda \end{pmatrix} = 0. \] Simplifying, the characteristic polynomial is \( (\lambda + 3)(\lambda - 4) - 18 = \lambda^2 - \lambda - 30 = 0 \). Solving for \( \lambda \), we factor to find \( (\lambda - 6)(\lambda + 5) = 0 \). Thus, the eigenvalues are \( \lambda_1 = 6 \) and \( \lambda_2 = -5 \).

Step 3: Find the Eigenvectors for Each Eigenvalue

For \( \lambda_1 = 6 \), solve \( (A - 6I)v = 0 \):\[ \begin{pmatrix} -9 & 3 \ 6 & -2 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0. \]This results in the equation \( v_2 = 3v_1 \), leading to the eigenvector \( v_1 = \begin{pmatrix} 1 \ 3 \end{pmatrix} \).For \( \lambda_2 = -5 \), solve \( (A + 5I)v = 0 \):\[ \begin{pmatrix} 2 & 3 \ 6 & 9 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0. \]This results in the equation \( v_1 = -\frac{3}{2}v_2 \), leading to the eigenvector \( v_2 = \begin{pmatrix} -3 \ 2 \end{pmatrix} \).

Step 4: Construct the General Solution

Using the eigenvalues and eigenvectors found, the general solution of the system can be expressed as\[ X(t) = c_1 e^{6t} \begin{pmatrix} 1 \ 3 \end{pmatrix} + c_2 e^{-5t} \begin{pmatrix} -3 \ 2 \end{pmatrix}, \]where \( c_1 \) and \( c_2 \) are arbitrary constants determined by initial conditions.

Step 5: Sketch Eigenvector Lines and Indicate Solution Direction

Draw the lines through the origin in the directions of the eigenvectors. For the eigenvector \( \begin{pmatrix} 1 \ 3 \end{pmatrix} \), the line is in the direction of \( x_2 = 3x_1 \). Solutions along this line move away from the origin as \( t \) increases, due to positive \( \lambda_1 = 6 \). For the eigenvector \( \begin{pmatrix} -3 \ 2 \end{pmatrix} \), the line direction is \( x_2 = -\frac{2}{3}x_1 \). Solutions along this line move towards the origin as \( t \) increases, due to negative \( \lambda_2 = -5 \).

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial for solving systems of differential equations. Eigenvalues are special scalars that determine whether solutions of the system exhibit exponential growth or decay over time. To find these, you solve the characteristic equation, which is formed from the matrix associated with the system. This involves subtracting \( \lambda I \) (where \( I \) is the identity matrix) from the original matrix, then setting the determinant equal to zero. This comes from solving \( \text{det}(A - \lambda I) = 0 \), where \( A \) is the coefficient matrix. In our exercise, we've found the eigenvalues \( \lambda_1 = 6 \) and \( \lambda_2 = -5 \).
Eigenvectors, on the other hand, help describe the directions of these exponential changes and are critical when forming the general solution. Once you have your eigenvalues, you can find the corresponding eigenvectors by solving the equation \((A - \lambda I)v = 0\). Each eigenvector corresponds to a unique eigenvalue and gives a direction in which the system's solutions project.
In this task, the eigenvector corresponding to \( \lambda_1 = 6 \) is \( \begin{pmatrix} 1 \ 3 \end{pmatrix} \), and for \( \lambda_2 = -5 \) is \( \begin{pmatrix} -3 \ 2 \end{pmatrix} \). These vectors not only help form the general solution but are also used to sketch direction fields, showing potential trajectories of the system.

Matrix Equation

A matrix equation is a concise way of expressing a system of differential equations. This allows for more intuitive handling of complicated systems. In the matrix equation \( \frac{dX}{dt} = AX \), we encapsulate all variables and their derivatives.
The matrix \( A \) contains all the coefficients from the system of equations. Each element of \( A \) directly affects the behavior and direction of the system’s solutions. Writing the system as a matrix equation simplifies finding solutions, especially when involving eigenvalues and eigenvectors.
Consider \[ A = \begin{pmatrix} -3 & 3 \ 6 & 4 \end{pmatrix} \] from our exercise. It takes all differential coefficients and represents them in a compact form. Solving \( \frac{dX}{dt} = AX \) after finding \( A \) is a conduit to using linear algebra methods like diagonalization or finding eigenpairs, key elements in our journey from differential equations to general solutions.

General Solution of Differential Equations

The ultimate goal when dealing with systems of differential equations is often to find the general solution. This solution isn’t justone path through the phase space of our system, but all possible paths depending on initial conditions. The general solution is normallyformed using both the eigenvalues and eigenvectors found earlier. In our case, it combines both exponential functions (based oneigenvalues) and the direction of eigenvectors.
For example, with eigenvalues \( \lambda_1 = 6 \) and \( \lambda_2 = -5 \), and their corresponding eigenvectors, the generalsolution becomes \[ X(t) = c_1 e^{6t} \begin{pmatrix} 1 \ 3 \end{pmatrix} + c_2 e^{-5t} \begin{pmatrix} -3 \ 2 \end{pmatrix} \].
Arbitrary constants \( c_1 \) and \( c_2 \) appear in the solution, representing that the specific path depends on the system'sinitial state. This formula shows us every possible way our system could evolve over time. Understanding this representation is critical, as it permits insight into the behavior and stability of dynamic systems. Solutions emphasize how certain dynamicsspeed up or slow down with factors like \( e^{6t} \) driving them exponentionally fast, in contrast to slower decays shownby \( e^{-5t} \).