Questions: Quiz Information Upload Your Work for this Problem 1. Find real-valued fundamental solutions of the differential equation x' = Ax, where matrix A has eigenvalues and eigenvectors lambda1 = -2+i, v1 = [1, 1-i] lambda2 = -2-i, v2 = [1, 1+i] 2. Write all the calculations needed to find the solution of the initial value problem x' = Ax, x(0) = [1, 1] 3. Classify (for example, source node, sink node. etc) the solution X = 0.

Solution Steps

1. To find the real-valued fundamental solutions of the differential equation \(\mathbf{x}^{\prime}=A \mathbf{x}\), we use the given eigenvalues and eigenvectors. Since the eigenvalues are complex conjugates, the solutions will involve exponential functions with real and imaginary parts. We express the solution in terms of real-valued functions using Euler's formula.
2. For the initial value problem, we use the fundamental matrix solution obtained from the eigenvalues and eigenvectors to find the specific solution that satisfies the initial condition \(\mathbf{x}(0)=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]\).
3. To classify the solution \(\mathbf{X}=\mathbf{0}\), we analyze the real parts of the eigenvalues. Since both eigenvalues have negative real parts, the origin is a stable node (sink).

We are given the matrix \( A \) defined as: \[ A = \begin{bmatrix} -2 & -1 \\ 1 & -2 \end{bmatrix} \] and the initial condition: \[ \mathbf{x}(0) = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

The eigenvalues of the matrix \( A \) are: \[ \lambda_1 = -2 + i, \quad \lambda_2 = -2 - i \]

The fundamental matrix solution \( \Phi(t) \) is computed using the matrix exponential: \[ \Phi(t) = e^{At} \] The eigenvalues indicate that the solution will involve oscillatory behavior due to the imaginary parts, combined with exponential decay due to the negative real parts.

The solution \( \mathbf{x}(t) \) at time \( t \) is given by: \[ \mathbf{x}(t) = \Phi(t) \cdot \mathbf{x}(0) \] The computed solutions over the time interval \( t \in [0, 10] \) yield a set of values that describe the behavior of the system.

The classification of the solution \( \mathbf{X} = \mathbf{0} \) is determined by the real parts of the eigenvalues. Since both eigenvalues have negative real parts (\(-2\)), the origin is classified as a stable node (sink).

Final Answer