Suppose x(t) is the solution of the following initial value problem in ℝ²

ẋ = Ax, x(0) = x₀, where A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\).

Which of the following statements is true?

Concept: 

Explanation:

A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\)

tr(A) = 5 + 2 = 7 and det(A) = 10 - 4 = 6

Eigenvalues are given by

λ² - tr(A)λ + det(A) = 0

λ2 - 7λ + 6 = 0

(λ - 1)(λ - 6) = 0

λ = 1, 6

Eigenvector corresponding to eigenvalue λ = 1 is given by

\(\left[\begin{array}{ll}4 & 4 \\1 & 1\end{array}\right]\)\(\left[\begin{array}{ll}u_1 \\u_2\end{array}\right]\) = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

Eigenvector is u = \(\begin{bmatrix}-1 \\1\end{bmatrix}\)

Eigenvector corresponding to eigenvalue λ = 6 is given by

\(\begin{bmatrix}-1 & 4 \\1 & -4\end{bmatrix}\)\(\left[\begin{array}{ll}v_1 \\v_2\end{array}\right]\) = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = \(\left[\begin{array}{ll}4 \\1\end{array}\right]\)

Hence solution is

x(t) = c1\(\begin{bmatrix}-1 \\1\end{bmatrix}\)et + c2\(\left[\begin{array}{ll}4 \\1\end{array}\right]\)e6t

x(t) = \(\begin{bmatrix}-c_1e^t+4c_2e^{6t} \\c_1e^t+c_2e^{6t}\end{bmatrix}\)

e^t → ∞ as t → ∞ also e6t → ∞ as t → ∞

So x(t) is not bounded solution for any x0 ≠ 0

(1) is false

e−6t|x(t)| = \(\begin{bmatrix}-c_1e^{-5t}+4c_2 \\c_1e^{-5t}+c_2\end{bmatrix}\) → \(\begin{bmatrix}4c_2 \\c_2\end{bmatrix}\) does not tends to 0

So (2) is false

e−t|x(t)| = \(\begin{bmatrix}-c_1+4c_2e^{5t} \\c_1+c_2e^{5t}\end{bmatrix}\)

Let x(0) = \(\begin{bmatrix}1 \\-1\end{bmatrix}\) then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) = \(\begin{bmatrix}-1 \\1\end{bmatrix}\) does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| = \(\begin{bmatrix}-c_1e^{-9t}+4c_2e^{-4t} \\c_1e^{-9t}+c_2e^{-4t}\end{bmatrix}\) → 0 as t → ∞, for all x0 ≠ 0.

Option (4) is correct