The question on the screen asks to convert the second-order differential equation:

$y'' - 5y' - 4y = 0$

into a vector equation for $u = (y, y')$.

Step-by-Step Solution:

1. Let $u_1 = y$ and $u_2 = y'$. This allows us to express the system of first-order equations. We know: $u_1' = y' = u_2$ $u_2' = y'' = 5y' + 4y = 5u_2 + 4u_1$

2. This gives us the system of equations: $\frac{d}{dt} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} u_2 \\ 5u_2 + 4u_1 \end{bmatrix}$

3. This system can be written in matrix form as: $\frac{d}{dt} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}$

Thus, the matrix $A$ is:

$A = \begin{bmatrix} 0 & 1 \\ 4 & 5 \end{bmatrix}$

Final Answer:

The correct choice is (d), which matches the matrix $A$.

Let me know if you would like further details or have additional questions! Here are some related questions for deeper understanding:

1. How do you convert second-order differential equations into a system of first-order equations?
2. What are the general methods for solving systems of differential equations?
3. How does the matrix form help in solving differential equations?
4. What are the eigenvalues of the matrix $\begin{bmatrix} 0 & 1 \\ 4 & 5 \end{bmatrix}$?
5. What role do matrix exponentials play in solving linear systems?

Tip: When converting higher-order differential equations to vector form, always introduce new variables to represent the derivatives systematically!