Question-164227

Question Number 164227 by SANOGO last updated on 15/Jan/22

Answered by TheSupreme last updated on 15/Jan/22

| \begin{matrix} 4 & 6 & 0 \\ - 3 & - 5 & 0 \\ - 3 & - 6 & - 5 \end{matrix} | - λ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] {\begin{matrix} x \\ y \\ z \end{matrix}} = {\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}}

d e t (A - λ I) = 0

[(4 - λ) (- 5 - λ) + 18] (- 5 - λ) = 0

[- 20 - 4 λ + 5 λ + λ^{2} + 18] (- 5 - λ) = 0

(λ^{2} + λ - 2) (- 5 - λ) = 0

- (λ + 5) (λ - 1) (λ + 2) = 0

{x} = c_{1} {w_{1}} e^{4 t} + c_{2} {w_{2}} e^{t} + c_{3} {w_{3}} e^{- 2 t}

λ = 1

[\begin{matrix} 3 & 6 & 0 \\ - 3 & - 6 & 0 \\ - 3 & - 6 & - 6 \end{matrix}] {w} = {0}

3 w_{x} + 6 w_{y} = 0

- 6 w_{z} = 0

w_{2} = {1, - 2, 0}

λ = - 2

[\begin{array}{ccc} 6 & 6 & 0 \\ - 3 & - 3 & 0 \\ - 3 & - 6 & - 3 \end{array}] {w} = {0}

w_{x} + w_{y} = 0

w_{z} = 0

w_{3} = {1, - 1, 0}

Answered by pluton last updated on 15/Jan/22

p o s o n s X (t) = (\begin{matrix} \begin{matrix} x (t) \\ y (t) \end{matrix} \\ z (t) \end{matrix}) \frac{d X (t)}{d t} = A X

o u A = [\begin{matrix} 4 6 0 \\ - 3 - 5 0 \end{matrix}] i l v i e n t q u e \frac{d X (t)}{X} = A d t d o n c X (t) = X_{o} e^{t A} o u X_{0} = (\begin{matrix} \begin{matrix} x (o) \\ y (0) \end{matrix} \\ z (0) \end{matrix})

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