Explained: "Find the solution of the system x′=x−2y "

Deangelo Hardy

Answered question

2022-04-07

Find the solution of the system

x^{'} = x - 2 y and y^{'} = 4 x + 5 y

Answer & Explanation

Ausspruchx807

Beginner2022-04-08Added 6 answers

Step 1
The solution for such systems is as follows:

Let
$= [\begin{matrix} 1 & - 2 \\ 4 & 5 \end{matrix}]$
We have $λ_{1} = 3 + 2 i, λ_{2} = \bar{λ_{1}} = 3 - 2 i$ and
$v_{1} = [\begin{matrix} - 1 + i \\ 2 \end{matrix}] = [\begin{matrix} - 1 \\ 2 \end{matrix}] + i [\begin{matrix} 1 \\ 0 \end{matrix}]$
and $v_{2} = [\begin{matrix} - 1 - i \\ 2 \end{matrix}] = [\begin{matrix} - 1 \\ 2 \end{matrix}] + i [\begin{matrix} - 1 \\ 0 \end{matrix}]$
Step 2
Now we construct the matrix $P$ from the imaginary and real parts of the eigenvectors;
$P = [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] \Rightarrow P^{- 1} = \frac{1}{13} [\begin{matrix} 3 & 2 \\ - 2 & 3 \end{matrix}]$
and

$P^{- 1} A P = [\begin{matrix} 3 & - 2 \\ 2 & 3 \end{matrix}]$
And the final solution

$[\begin{matrix} x (t) \\ y (t) \end{matrix}] = P e^{3 t} [\begin{matrix} \cos 2 t & - \sin 2 t \\ \sin 2 t & \cos 2 t \end{matrix}] P^{- 1} [\begin{matrix} x_{0} \\ y_{0} \end{matrix}]$
and

$[\begin{matrix} x (t) \\ y (t) \end{matrix}] = \frac{e^{3 t}}{13} [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] [\begin{matrix} \cos 2 t & - \sin 2 t \\ \sin 2 t & \cos 2 t \end{matrix}] [\begin{matrix} 3 & 2 \\ - 2 & 3 \end{matrix}] [\begin{matrix} x_{0} \\ y_{0} \end{matrix}]$

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