The problem asks to find the general solution of the differential equation:

$\frac{dY}{dt} = AY$

where:

$A = \begin{bmatrix} 7 & 10 \\ -5 & -8 \end{bmatrix}$

This is a system of first-order linear differential equations. The general solution to such a system involves finding the eigenvalues and eigenvectors of matrix $A$, and then using them to construct the solution.

Step 1: Find the eigenvalues of matrix $A$

To find the eigenvalues, we solve the characteristic equation:

$\det(A - \lambda I) = 0$

where $I$ is the identity matrix and $\lambda$ is an eigenvalue. For this matrix:

$A - \lambda I = \begin{bmatrix} 7 - \lambda & 10 \\ -5 & -8 - \lambda \end{bmatrix}$

The determinant is:

$\det(A - \lambda I) = (7 - \lambda)(-8 - \lambda) - (-5)(10)$

Expanding this:

$\det(A - \lambda I) = (7 - \lambda)(-8 - \lambda) + 50$

$= \lambda^2 + \lambda - 56 + 50 = \lambda^2 + \lambda - 6$

The characteristic equation is:

$\lambda^2 + \lambda - 6 = 0$

Factoring this:

$(\lambda - 2)(\lambda + 3) = 0$

Thus, the eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = -3$.

Step 2: Find the eigenvectors

For each eigenvalue, we solve $(A - \lambda I)v = 0$ to find the corresponding eigenvector.

For $\lambda_1 = 2$:

$A - 2I = \begin{bmatrix} 5 & 10 \\ -5 & -10 \end{bmatrix}$

Solving $(A - 2I)v = 0$, we get:

$5x + 10y = 0 \quad \Rightarrow \quad y = -\frac{1}{2}x$

Thus, the eigenvector corresponding to $\lambda_1 = 2$ is:

$v_1 = \begin{bmatrix} 1 \\ -\frac{1}{2} \end{bmatrix}$

For $\lambda_2 = -3$:

$A + 3I = \begin{bmatrix} 10 & 10 \\ -5 & -5 \end{bmatrix}$

Solving $(A + 3I)v = 0$, we get:

$10x + 10y = 0 \quad \Rightarrow \quad y = -x$

Thus, the eigenvector corresponding to $\lambda_2 = -3$ is:

$v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Step 3: General Solution

The general solution is of the form:

$Y(t) = k_1 e^{\lambda_1 t} v_1 + k_2 e^{\lambda_2 t} v_2$

Substituting the eigenvalues and eigenvectors:

$Y(t) = k_1 e^{2t} \begin{bmatrix} 1 \\ -\frac{1}{2} \end{bmatrix} + k_2 e^{-3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

This matches the third option on the list:

$Y = k_1 e^{2t} \begin{bmatrix} 1 \\ -\frac{1}{2} \end{bmatrix} + k_2 e^{-3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Answer:

The correct answer is the third option.

Would you like more details or have any further questions?

Here are 5 related questions:

1. How do you derive the characteristic equation of a matrix?
2. Why is finding the eigenvalues important in solving systems of differential equations?
3. How do you find eigenvectors after obtaining eigenvalues?
4. Can this method be extended to larger matrices?
5. What are the conditions under which a system of differential equations has distinct solutions?

Tip: Always check the eigenvalues and eigenvectors carefully, as they form the foundation of the solution!