Let A be a diagonalizable matrix. If c₁ and c₂ are solution to system
x⃗' (t) = A x⃗ ( t )
the possible values of c₁ is 1 and c₂ is 2 .
So, the correct option is option(a) and (e).
We have given that A is 2 × 2 diagonalizable matrix and x⃗ (t) = ( 3eᵗ + eᶜ₁t , 3e²ᵗ + eᶜ₂ᵗ) is a solution of system, x⃗ '(t) = A x⃗(ᵗ) .
A Solution of system always satisfied the equation of system.
Now, Differenating x⃗ (t) wih respect to t we get, x⃗' (t)=( 3eᵗ + c₁ eᶜ₁t , 6e²ᵗ + c₂ eᶜ₂ᵗ)
So, ( 3eᵗ + c₁ eᶜ₁t , 6e²ᵗ + c₂ eᶜ₂ᵗ)= A ( 3eᵗ + eᶜ₁t , 3e²ᵗ + eᶜ₂ᵗ) where A is diagonalizable and A = [a 0 0 b].
Then, ( 3eᵗ + c₁ eᶜ₁t , 6e²ᵗ + c₂ eᶜ₂ᵗ) = ( 3aeᵗ + aeᶜ₁t , 3be²ᵗ + b eᶜ₂ᵗ)
equating the coefficients on both sides we get, 3eᵗ + c₁ eᶜ₁t = 3aeᵗ + aeᶜ₁t 6e²ᵗ + c₂ eᶜ₂ᵗ = 3be²ᵗ + b eᶜ₂ᵗ
after equating the corresponding equations , 3 = 3a and c₁ = a and 6 = 3b and c₂ = b after solving all we get a = 1 and b = 2 which implies c₁ = 1 and c₂ = 2. Hence, the possible values are c₁ = 1 and c₂ = 2.
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Complete question:
Let A be a diagonalizable matrix. If c₁ and c₂ are solution to system x⃗ (t) = A x⃗
then check all the possible values of c₁ and c₂ and below.
(a) c₁ = 1
(b) c₁ = 3
(c) c₁ = 2
d) c₂ = 1
(e)c₂ =2
(f) c₂ = 3
