give the calculation step by step as like as full answer in 2 steps and explanation block in every steps.
And final answer summary
Definition of Parameters
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give the calculation step by step as like as full answer in 2 steps and explanation block in every steps. 
And final answer summary
Definition of Parameters 
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VIPSolutions · Accepted Answer

**Step 1: Find the General Solution to dx/dt = Ax**

Given the system:  
$$
\frac{dx}{dt} = Ax, \quad x(0) = x_0
$$
where  
$$
A = \begin{bmatrix} -1 & -2 \ 2 & -2 \end{bmatrix}, \quad x_0 = \begin{bmatrix} 1 \ 5 \end{bmatrix}
$$

**Explanation:**  
The general solution for a linear system $\frac{dx}{dt} = Ax$ is  
$$
x(t) = e^{At}x_0
$$
To compute $e^{At}$, diagonalize $A$ or use its eigenvalues and eigenvectors.

First, find the eigenvalues $\lambda$:

$$
\det(A - \lambda I) = 0
$$
$$
\left| \begin{matrix} -1-\lambda & -2 \ 2 & -2-\lambda \end{matrix} \right| = 0
$$
$$
(-1-\lambda)(-2-\lambda) - (2)(-2) = 0
$$
$$
(\lambda + 1)(\lambda + 2) + 4 = 0
$$
$$
\lambda^2 + 3\lambda + 2 + 4 = 0
$$
$$
\lambda^2 + 3\lambda + 6 = 0
$$
$$
\lambda = \frac{-3 \pm \sqrt{9 - 24}}{2} = \frac{-3 \pm i\sqrt{15}}{2}
$$
Let $ \alpha = -\frac{3}{2} $, $ \beta = \frac{\sqrt{15}}{2} $, so eigenvalues are $ \lambda_{1,2} = \alpha \pm i\beta $.

**Step 2: Write the Solution Using Eigenvalues and Initial Condition**

The solution for a real 2x2 matrix with complex eigenvalues is:
$$
x(t) = e^{\alpha t} \left[ \cos(\beta t) I + \frac{\sin(\beta t)}{\beta}(A-\alpha I) \right] x_0
$$

**Explanation:**  
For a real system with complex eigenvalues, the solution can be expressed without explicit complex numbers using this formula. Here, $I$ is the identity matrix.

Given:
- $ \alpha = -\frac{3}{2} $
- $ \beta = \frac{\sqrt{15}}{2} $
- $ A - \alpha I = A + \frac{3}{2}I = \begin{bmatrix} -1+1.5 & -2 \ 2 & -2+1.5 \end{bmatrix} = \begin{bmatrix} 0.5 & -2 \ 2 & -0.5 \end{bmatrix} $

So,
$$
x(t) = e^{-\frac{3}{2} t} \left[ \cos\left(\frac{\sqrt{15}}{2} t\right) I + \frac{2}{\sqrt{15}} \sin\left(\frac{\sqrt{15}}{2} t\right) \begin{bmatrix} 0.5 & -2 \ 2 & -0.5 \end{bmatrix} \right] \begin{bmatrix} 1 \ 5 \end{bmatrix}
$$

**Final Answer Summary**

The solution to the initial value problem  
$$
\frac{dx}{dt} = \begin{bmatrix} -1 & -2 \ 2 & -2 \end{bmatrix} x, \quad x(0) = \begin{bmatrix} 1 \ 5 \end{bmatrix}
$$
is  
$$
x(t) = e^{-\frac{3}{2} t} \left\{ \cos\left(\frac{\sqrt{15}}{2} t\right) I + \frac{2}{\sqrt{15}} \sin\left(\frac{\sqrt{15}}{2} t\right) \begin{bmatrix} 0.5 & -2 \ 2 & -0.5 \end{bmatrix} \right\} \begin{bmatrix} 1 \ 5 \end{bmatrix}
$$
where $I$ is the 2x2 identity matrix. This gives a real-valued solution in closed form.

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