Solving a System of Differential Equations

|λI − A| = det[λ−4, 1, 1; −1, λ−5, −2; 0, −1, λ−5]

= (λ−4)[(λ−5)(λ−5) − (−2)(−1)] − 1[(−1)(λ−5) − (−2)(0)] + 1[(−1)(−1) − (λ−5)(0)]
= (λ−4)(λ²−10λ+25−2) − 1(−λ+5) + 1(1)
= (λ−4)(λ²−10λ+23) + λ − 5 + 1
= (λ−4)(λ²−10λ+23) + λ − 4
= (λ−4)(λ²−10λ+24)
= (λ−4)(λ−4)(λ−6)

6I − A = [2, 1, 1; −1, 1, −2; 0, −1, 1]

R₂ → R₂ + ½R₁: [2, 1, 1; 0, 3/2, −3/2; 0, −1, 1]
R₃ → R₃ + ⅔R₂: [2, 1, 1; 0, 3/2, −3/2; 0, 0, 0]

v₃ = t (free), v₂ = t, v₁ = −t
Choosing t = 1: v₁ = [−1, 1, 1]^T

4I − A = [0, 1, 1; −1, −1, −2; 0, −1, −1]

R₃ → R₃ + R₁: [0, 1, 1; −1, −1, −2; 0, 0, 0]
R₂ → R₂ + R₁ (swap): [−1, −1, −2; 0, 1, 1; 0, 0, 0]

v₃ = t (free), v₂ = −t, v₁ = t
Choosing t = −1: v₂ = [1, 1, −1]^T

The geometric multiplicity is 1 (only one eigenvector), while the algebraic multiplicity is 2.
Therefore, we will need a generalized eigenvector in the next step.

λ=4 is a double eigenvalue (algebraic multiplicity 2) but has only one eigenvector (geometric multiplicity 1).
Therefore, the third solution is not of the standard form e^λtv, and requires a generalized eigenvector.

x₃(t) = te^4tv₂ + e^4tw

[0, −1, −1; 1, 1, 2; 0, 1, 1] · [w₁; w₂; w₃] = [1; 1; −1]

Equation 1: −w₂ − w₃ = 1
Equation 3: w₂ + w₃ = −1 (identical to Equation 1)
Equation 2: w₁ + w₂ + 2w₃ = 1

From Equation 1: w₂ = −1 − w₃
Substituting into Equation 2: w₁ + (−1 − w₃) + 2w₃ = 1
w₁ + w₃ = 2 → w₃ = 2 − w₁
w₂ = −1 − (2 − w₁) = w₁ − 3

Choosing w₁ = 0: w₂ = −3, w₃ = 2

x₃(t) = te^4t[1;1;−1] + e^4t[0;−3;2]
= e^4t[t; t−3; −t+2]
= e^4t[t; t−3; 2−t]

3e^3t[a₁;a₂;a₃] = A·e^3t[a₁;a₂;a₃] + e^3t[1;1;0]

(3I − A)[a₁;a₂;a₃] = [1;1;0]

[−1, 1, 1; −1, −2, −2; 0, −1, −2]

From Equation 3: a₂ + 2a₃ = 0 → a₂ = −2a₃
From Equation 2: −a₁ − 2a₂ − 2a₃ = 1 → −a₁ + 4a₃ − 2a₃ = 1 → −a₁ + 2a₃ = 1
From Equation 1: −a₁ + a₂ + a₃ = 1 → −a₁ − 2a₃ + a₃ = 1 → −a₁ − a₃ = 1

From Equations 2 and 1: 2a₃ − (−a₃) = 0 → 3a₃ = 0 → a₃ = 0
a₂ = 0, a₁ = −1