(a) The system of differential equations can be written in matrix-vector notation as follows:

[ x' ] = [ 4 1 ] [ x ]
[ y' ] [ 2 5 ] [ y ]

(b) The characteristic equation is obtained by subtracting λ from the diagonals of the matrix and setting the determinant equal to zero:

det([4-λ, 1], [2, 5-λ]) = 0
(4-λ)(5-λ) - (2)(1) = λ^2 - 9λ + 18 = 0

The solutions to this quadratic equation are the eigenvalues of the matrix. They can be found using the quadratic formula:

λ = [9 ± sqrt((9)^2 - 4*1*18)] / (2*1)
λ = [9 ± sqrt(81 - 72)] / 2
λ = [9 ± sqrt(9)] / 2
λ = [9 ± 3] / 2
λ1 = 6, λ2 = 3

The eigenvectors are found by plugging the eigenvalues back into the equation (A - λI)v = 0 and solving for v:

For λ1 = 6:
(4-6)v1 + v2 = 0
2v1 + (5-6)v2 = 0
which simplifies to:
-2v1 + v2 = 0
2v1 + -v2 = 0
The solution to this system is v1 = [1, 2].

For λ2 = 3:
(4-3)v1 + v2 = 0
2v1 + (5-3)v2 = 0
which simplifies to:
v1 + v2 = 0
2v1 + 2v2 = 0
The solution to this system is v2 = [1, -1].

(c) The general solution of the system is a linear combination of the eigenvectors, scaled by the exponential of the eigenvalues:

x(t) = c1 * e^(6t) * [1, 2] + c2 * e^(3t) * [1, -1]

(d) The fundamental matrix Φ(t) is a matrix whose columns are the solutions to the system. Since the initial condition is the identity matrix, the fundamental matrix at t=0 is:

Φ(0) = [1, 1]
[2, -1]

(e) The equilibrium point of the system is the origin (0,0). Since both eigenvalues are positive, the origin is an unstable node. The phase portrait would show trajectories moving away from the origin along the directions of the eigenvectors.

Question

(a) The system of differential equations can be written in matrix-vector notation as follows:

[ x' ]   =   [ 4  1 ] [ x ]
[ y' ]       [ 2  5 ] [ y ]

(b) The characteristic equation is obtained by subtracting λ from the diagonals of the matrix and setting the determinant equal to zero:

det([4-λ, 1], [2, 5-λ]) = 0
(4-λ)(5-λ) - (2)(1) = λ^2 - 9λ + 18 = 0

The solutions to this quadratic equation are the eigenvalues of the matrix. They can be found using the quadratic formula:

λ = [9 ± sqrt((9)^2 - 4*1*18)] / (2*1)
λ = [9 ± sqrt(81 - 72)] / 2
λ = [9 ± sqrt(9)] / 2
λ = [9 ± 3] / 2
λ1 = 6, λ2 = 3

The eigenvectors are found by plugging the eigenvalues back into the equation (A - λI)v = 0 and solving for v:

For λ1 = 6:
(4-6)v1 + v2 = 0
2v1 + (5-6)v2 = 0
which simplifies to:
-2v1 + v2 = 0
2v1 + -v2 = 0
The solution to this system is v1 = [1, 2].

For λ2 = 3:
(4-3)v1 + v2 = 0
2v1 + (5-3)v2 = 0
which simplifies to:
v1 + v2 = 0
2v1 + 2v2 = 0
The solution to this system is v2 = [1, -1].

(c) The general solution of the system is a linear combination of the eigenvectors, scaled by the exponential of the eigenvalues:

x(t) = c1 * e^(6t) * [1, 2] + c2 * e^(3t) * [1, -1]

(d) The fundamental matrix Φ(t) is a matrix whose columns are the solutions to the system. Since the initial condition is the identity matrix, the fundamental matrix at t=0 is:

Φ(0) = [1, 1]
           [2, -1]

(e) The equilibrium point of the system is the origin (0,0). Since both eigenvalues are positive, the origin is an unstable node. The phase portrait would show trajectories moving away from the origin along the directions of the eigenvectors.

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