Resolved: "Wroian of vectorY→′=AY,A2×2 (constants over R)Y→(1)(x)=(e2x−e2x),W[Y→(1),Y→(2)](2)=e3,W[Y→(1),Y→(2)](1)=e2 (W:=wroian)."

Caroline Carey

Answered question

2022-03-07

Wro
ian of vector

{\vec{Y}}^{'} = A Y, A_{2 \times 2}

(constants over

R)

{\vec{Y}}^{(1)} (x) = (\begin{matrix} e^{2 x} \\ - e^{2 x} \end{matrix}), W [{\vec{Y}}^{(1)}, {\vec{Y}}^{(2)}] (2) = e^{3}, W [{\vec{Y}}^{(1)}, {\vec{Y}}^{(2)}] (1) = e^{2}

(W:=wro
ian).

Answer & Explanation

Ayla Ward

Beginner2022-03-08Added 1 answers

Step 1
The Wro
ian of vector-valued functions ${v_{1} (x), \dots, v_{2} (x)}$ is defined as
$W {v_{1}, \dots, v_{n}} (x) = det [\begin{matrix} v_{1} (x) & \dots & v_{n} (x) \end{matrix}] .$
I assume that $Y_{1}$ is a solution to the planar system and you want to solve for the other solution $Y_{2}$ . Following your notation, let $Y_{2} (x) = {(a e^{r x}, b e^{r x})}^{T}$ , where I assume that the coefficient matrix A has distinct real eigenvalues.
The Wro
ian of $Y_{1}, Y_{2}$ is given by
$W {Y_{1}, Y_{2}} (x) = b e^{r x} e^{2 x} + a e^{r x} e^{2 x} = (a + b) e^{(2 + r) x} .$
With the two given conditions on the Wro
ian, we have $(a + b) e^{4 + 2 r} = e^{3}$ and $(a + b) e^{2 + r} = e^{2}$ . Equivalently, we have $(a + b) e^{2 r} = e^{- 1}$ and $(a + b) e^{r} = 1$ . Solving these two equations, we get $r = - 1$ and $a + b = e .$

blanchetetck

Beginner2022-03-09Added 1 answers

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