Given x1 = e
t

cos(4t) − 3 sin(4t)
2 cos(4t) + 2 sin(4t)

is a solution to
x
′ = Ax, find the solution with x(0) = 
7
−2

.

Given the system of equations \( \mathbf{x}' = A \mathbf{x} \) and a known solution \( \mathbf{x}_1 = e^t \begin{pmatrix} \cos(4t) - 3\sin(4t) \\ 2\cos(4t) + 2\sin(4t) \end{pmatrix} \), we need to find the general solution \( \mathbf{x}(t) \) that satisfies the initial condition \( \mathbf{x}(0) = \begin{pmatrix} 7 \\ -2 \end{pmatrix} \).

### Step 1: Recognize the form of the general solution
For a system \( \mathbf{x}' = A \mathbf{x} \), the general solution typically consists of linear combinations of particular solutions. Since \( \mathbf{x}_1(t) \) is one such solution, the general solution can be written as:
\[
\mathbf{x}(t) = c_1 \mathbf{x}_1(t)
\]
where \( c_1 \) is a constant to be determined from the initial condition.

### Step 2: Evaluate the solution at \( t = 0 \)
We substitute \( t = 0 \) into \( \mathbf{x}_1(t) \) to find \( \mathbf{x}_1(0) \):
\[
\mathbf{x}_1(0) = e^0 \begin{pmatrix} \cos(0) - 3\sin(0) \\ 2\cos(0) + 2\sin(0) \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]
Thus, at \( t = 0 \), the solution is:
\[
\mathbf{x}(0) = c_1 \mathbf{x}_1(0) = c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]

### Step 3: Apply the initial condition
We are given that \( \mathbf{x}(0) = \begin{pmatrix} 7 \\ -2 \end{pmatrix} \). Using this condition, we can solve for \( c_1 \):
\[
c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \end{pmatrix}
\]
This gives the system of equations:
\[
c_1 = 7
\]
\[
2c_1 = -2
\]
However, the system is inconsistent, meaning there might be a mistake in the setup or the expression for \( \mathbf{x}_1(t) \). Could you confirm the exact structure of \( \mathbf{x}_1(t) \) or provide any additional details?

Given x1 = e^t [cos(4t) - 3 sin(4t), 2 cos(4t) + 2 sin(4t)] is a solution to x' = Ax, find the solution with x(0) = [7, -2].

This problem explores how to solve the matrix differential equation x' = Ax with an initial condition x(0) = [7, -2]. The solution involves applying linear algebra concepts, such as systems of equations, and using the known solution x1(t). Suitable for undergraduate students.

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