To solve the initial value problem:

$\mathbf{x}'(t) = \begin{bmatrix} 4 & 6 \\ 6 & 4 \end{bmatrix} \mathbf{x}(t), \quad \mathbf{x}(0) = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$

we'll follow these steps:

Step 1: Find Eigenvalues and Eigenvectors of the Matrix

1. Set up the matrix: $A = \begin{bmatrix} 4 & 6 \\ 6 & 4 \end{bmatrix}$

2. Determine the eigenvalues by solving the characteristic polynomial: $\det(A - \lambda I) = 0$

   For matrix $A$, we have: $\det \begin{bmatrix} 4 - \lambda & 6 \\ 6 & 4 - \lambda \end{bmatrix} = (4 - \lambda)^2 - 36 = \lambda^2 - 8\lambda - 20 = 0$

   Solving this quadratic equation: $\lambda^2 - 8\lambda - 20 = 0$ The eigenvalues are $\lambda_1 = 10$ and $\lambda_2 = -2$.

3. Find eigenvectors for each eigenvalue:

   • For $\lambda_1 = 10$: $\begin{bmatrix} -6 & 6 \\ 6 & -6 \end{bmatrix} \mathbf{v} = 0 \Rightarrow \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
   • For $\lambda_2 = -2$: $\begin{bmatrix} 6 & 6 \\ 6 & 6 \end{bmatrix} \mathbf{v} = 0 \Rightarrow \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Step 2: Write the General Solution

Using the eigenvalues and eigenvectors, the general solution to the system is: $\mathbf{x}(t) = c_1 e^{10t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Step 3: Apply Initial Condition

Given $\mathbf{x}(0) = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$, we can substitute $t = 0$: $\mathbf{x}(0) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$

This gives us two equations:

1. $c_1 + c_2 = 5$
2. $c_1 - c_2 = 1$

Solving these equations:

• Adding them: $2c_1 = 6 \Rightarrow c_1 = 3$
• Substituting $c_1 = 3$ into the first equation: $3 + c_2 = 5 \Rightarrow c_2 = 2$

Step 4: Write the Solution

Substitute $c_1 = 3$ and $c_2 = 2$ back into the general solution: $\mathbf{x}(t) = 3 e^{10t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + 2 e^{-2t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Expanding this, we get: $\mathbf{x}(t) = \begin{bmatrix} 3 e^{10t} + 2 e^{-2t} \\ 3 e^{10t} - 2 e^{-2t} \end{bmatrix}$

Final Answer

$\mathbf{x}(t) = \begin{bmatrix} 3 e^{10t} + 2 e^{-2t} \\ 3 e^{10t} - 2 e^{-2t} \end{bmatrix}$

Would you like more details on any part of this solution, or do you have further questions?

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1. What is the significance of eigenvalues in solving differential equations?
2. How do eigenvectors influence the solution structure?
3. Why do we use initial conditions to determine constants in the solution?
4. How does matrix exponentiation relate to solutions of linear systems?
5. How does this solution method apply to non-diagonalizable matrices?

Tip: When dealing with systems of differential equations, always check if the matrix is diagonalizable; it simplifies solving the system greatly.