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

ẋ = Ax, x(0) = x₀, where A = .

Which of the following statements is true?

Concept: Solution of ODE x'(t) = Ax is x(t) = c₁ue^λ₁t + c2veλ2t, where u, v are the eigenvectors corresponding to the eigenvalues λ1 and λ2 respectively and c1 and c2 are constants.

Explanation:

A = 

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

 = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

Eigenvector is u = 

Eigenvector corresponding to eigenvalue λ = 6 is given by

 = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = 

Hence solution is

x(t) = c1et + c2e6t

x(t) = 

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)| =  →  does not tends to 0

So (2) is false

e−t|x(t)| = 

Let x(0) =  then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) =  does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| =  → 0 as t → ∞, for all x0 ≠ 0.

Option (4) is correct