Hard Differential eq on Matrix...?

I'll write x,y,z for x_1, x_2, and x_3, respectively.

We want to solve
dx/dt = y - x = -x + y + 0z
dy/dt = z - y = 0x - y + z
dz/dt = x - z = x + 0y - z.
with initial conditions x(0) = -1, y(0) = 0, and z(0) = 1.

In matrix form, this is given by
[x]'.....[-1..1..0]..[x]
[y]..=..[0..-1..1]..[y]
[z]......[1..0..-1]..[z].
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Now we diagonalize the matrix.

First, we find its eigenvalues.
|-1-λ...1...0|
|0...-1-λ...1| = 0
|1...0...-1-λ|

==> (-1 - λ) [(-1 - λ)^2 - 0] - 1 [0 - 1] + 0 = 0
==> (1 + λ)^3 - 1 = 0
==> λ^3 + 3λ^2 + 3λ = 0
==> λ(λ^2 + 3λ + 3) = 0
==> λ = 0 or λ = (-3 ± i√3)/2.

Next, we find the corresponding eigenvectors.

For λ = 0, we row reduce
[-1...1...0]
[0...-1...1]
[1...0...-1]

We obtain
[1...0...-1]
[0...1...-1]
[0...0...0]

So, an eigenvector for λ = 0 is
[1]
[1]
[1].
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For λ = (-3 - i√3)/2, we row reduce
[(1 + i√3)/2...1...0]
[0...(1 + i√3)/2...1]
[1...0...(1 + i√3)/2]

This row reduces to
[1...0...(1 + i√3)/2]
[0...1...(1 - i√3)/2]
[0...0.....0].

Thus, an eigenvector for λ = (-3 - i√3)/2 is
[-(1 + i√3)]
[-(1 - i√3)]
[2].
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As for the general solution, ve^(λt) is a solution.
In the case where λ is complex, we extract its real and imaginary parts.

[-(1 + i√3)] * e^[t(-3 - i√3)/2]
[-(1 - i√3)]
[2]

This equals
[-(1 + i√3)] * e^(-3t/2) * [cos(t√3/2) - i sin(t√3/2)], by Euler's Formula
[-(1 - i√3)]
[2]

which equals e^(-3t/2) *
[-cos(t√3/2) - √3 sin(t√3/2)]. +....[sin(t√3/2)) - √3 cos(t√3/2))]
[√3 sin(t√3/2) - cos(t√3/2)]...+ i..[sin(t√3/2) + √3 cos(t√3/2)]
[2 cos(t√3/2)]......................+...[-2 sin(t√3/2)]

Thus, the general solution is given by [x,y,z]^t = Au + e^(-3t) (Bv + Cw), where
u = [1, 1, 1]^t
v = [-cos(t√3/2) - √3 sin(t√3/2), √3 sin(t√3/2) - cos(t√3/2), 2 cos(t√3/2)]^t
w = [sin(t√3/2)) - √3 cos(t√3/2)), sin(t√3/2) + √3 cos(t√3/2), -2 sin(t√3/2)]^t.

^t stands for transpose; it makes a row vector into a column vector.

Now, we use x(0) = -1, y(0) = 0, and z(0) = 1:

[-1, 0, 1]^t = Au + (Bv + Cw), where
u = [1, 1, 1]^t
v = [-1, -1, 2]^t
w = [-√3, √3, 0]^t.

Thus, we have
-1 = A - B - C√3
0 = A - B + C√3
1 = A + 2B

Solving for A, B, C yields
A = 0, B = 1/2, C = 1/(2√3).

Thus, we have
[x,y,z]^t = e^(-3t/2) (v/2 + C/(2√3)).

(You can equate like entries to find x(t), y(t), z(t) in scalar form.)
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I hope this helps!

P.S.: I apologize if there are any typos in this. This calculation is messy and tedious!