Let's break down and solve each part of the problem provided.

Part (a): Finding the General Solution

Given the system of differential equations: $\mathbf{x}' = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix} \mathbf{x}$ we are tasked to find the general solution for $\mathbf{x}(t)$.

1. Find the eigenvalues of the matrix: $A = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}$ The characteristic polynomial is given by $\det(A - \lambda I) = 0$: $\det \begin{pmatrix} 3 - \lambda & -2 \\ 2 & -2 - \lambda \end{pmatrix} = (3 - \lambda)(-2 - \lambda) - (-2)(2) = 0$ Simplifying, we get the characteristic polynomial: $\lambda^2 - \lambda - 6 = 0$ Solving for $\lambda$, we get the eigenvalues $\lambda_1 = 2$ and $\lambda_2 = -1$.

2. Find eigenvectors for each eigenvalue:

   • For $\lambda_1 = 2$: $(A - 2I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix} \mathbf{v} = 0$ This yields the eigenvector $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

   • For $\lambda_2 = -1$: $(A + I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix} \mathbf{v} = 0$ This yields the eigenvector $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

3. Construct the general solution: Using the eigenvalues and eigenvectors, the solution can be written as: $\mathbf{x}(t) = c_1 \begin{pmatrix} 2 \\ 1 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 1 \\ 2 \end{pmatrix} e^{-t}$ This matches the form provided in the problem.

Part (b): Direction Field

The direction field given (Field 5) is chosen to represent the flow of solutions to this system. The positive eigenvalue $\lambda = 2$ suggests exponential growth in one direction, while the negative eigenvalue $\lambda = -1$ indicates exponential decay in the orthogonal direction. The field confirms this behavior, showing trajectories moving away from the origin along one direction and spiraling due to the interaction between the eigenvalues.

Part (c): Behavior as $t \to \infty$

With $c_1, c_2 \neq 0$, the dominant term as $t \to \infty$ will be the one with the positive exponent $e^{2t}$. Therefore, the solution will grow unbounded in the direction associated with $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, meaning the solution will tend towards infinity as $t \to \infty$.

Let me know if you need further explanation or details.

Follow-up Questions:

1. How are eigenvalues used to determine the stability of solutions in a differential system?
2. What impact does each eigenvalue's sign have on the behavior of solutions over time?
3. How can direction fields help visualize solutions to differential systems?
4. Why is the direction associated with the positive eigenvalue dominant as $t \to \infty$?
5. Can you describe the behavior of the solution if both eigenvalues were negative?

Tip:

For systems with real eigenvalues, the general behavior of solutions can be quickly understood by observing the sign of the eigenvalues: positive eigenvalues indicate instability (growth), while negative eigenvalues indicate stability (decay).