Final answer:
To find the general solution to the system, we first need to find the eigenvalues and eigenvectors of the matrix. The eigenvalues are 3 and 2, and the corresponding eigenvectors are [1, -2] and [1, -1]. The general solution to the system is x(t) = c₁e^(3t)[1, -2] + c₂e^(2t)[1, -1]. The equilibrium solution 0 is unstable because one of the eigenvalues is positive.
Step-by-step explanation:
To find the general solution to the system, we need to first find the eigenvalues and eigenvectors of the matrix. The eigenvalues can be found by solving the characteristic equation det(A - I) = 0, where A is the matrix and is the eigenvalue.
In this case, the characteristic equation is (5 - )(1 - ) - (-1)(4) = 0. Solving this quadratic equation, we find that the eigenvalues are ₁ = 3 and ₂ = 2.
Next, we need to find the eigenvectors corresponding to each eigenvalue. For ₁ = 3, we solve the equation (A - 3I)x = 0, where A is the matrix and I is the identity matrix. Similarly, for ₂ = 2, we solve the equation (A - 2I)x = 0. Solving these systems of equations, we find the eigenvectors v₁ = [1, -2] and v₂ = [1, -1].
The general solution to the system is x(t) = c₁e^(3t)[1, -2] + c₂e^(2t)[1, -1], where c₁ and c₂ are constants. To determine the stability of the equilibrium solution, we look at the eigenvalues. If all eigenvalues have negative real parts, the equilibrium solution is stable. If at least one eigenvalue has a positive real part, the equilibrium solution is unstable.
In this case, since one of the eigenvalues is 3, which is positive, the equilibrium solution 0 is unstable.