MHB Draw Trajectories for Autonomous DE: Where to Begin?

[onie mti]

when you are asked to draw a trajectory for a an autonomous DE, where does one begin
eg
given the matrix A=
1 0
0 -1


they say find e^tA then draw the trajectories. I know how to find e^tA but i cannot draw

 

[Ackbach]

Well, go ahead and compute $e^{t \mathbf{A}}$. You'll have some initial condition $\mathbf{x}_{0}$. Your solution to the DE $ \dot{ \mathbf{x}}=A \mathbf{x}$ is then $e^{t \mathbf{A}} \mathbf{x}_{0}$. Go ahead and perform the matrix multiplication $e^{t \mathbf{A}} \mathbf{x}_{0}$ and see what you get.

 

[onie mti]

Well, go ahead and compute $e^{t \mathbf{A}}$. You'll have some initial condition $\mathbf{x}_{0}$. Your solution to the DE $ \dot{ \mathbf{x}}=A \mathbf{x}$ is then $e^{t \mathbf{A}} \mathbf{x}_{0}$. Go ahead and perform the matrix multiplication $e^{t \mathbf{A}} \mathbf{x}_{0}$ and see what you get.

My eigenvalues were 2 and 1
Eigenvector associated with λ_1=2 is ■(0@1) and
λ_1= Is ■(1@0)
Where x(t) = c(1)e^(2t) ■(0@1) + c(2)e^t ■(1@0)

Now solving for e^tA
x_1(0)= ■(1@0) = c(1)e^2t ■(0@1) + c(2) e^t ■(1@0)
Where c(1)= 0 and c(2)= 1

Then for x_2(0), c(1)= 1 and c(2) = 0
Such that, x_1(t)= c(2) ■(1@0)
x_2(t) = e^2t ■(0@1)
Therefore e^tA is: ■(c(2) ■(1@0) @e^2t ■(0@1))
I still do not understand how to do the trajectories

 

[Ackbach]

Hmm. You might want to check your calculations. With a diagonal matrix, the eigenvalues are simply the values on the diagonal. In this case, you get $\lambda= \pm 1$; the eigenvector for $\lambda=1$ is $\begin{bmatrix}1 \\ 0 \end{bmatrix}$, and the eigenvector corresponding to $\lambda=-1$ is $\begin{bmatrix}0 \\ 1 \end{bmatrix}$. As the matrix is already diagonal, you can exponentiate it fairly easily:
$$e^{t \mathbf{A}}= \begin{bmatrix} e^{t} &0 \\ 0 &e^{-t} \end{bmatrix}.$$
What is your initial condition?

 

[onie mti]

Hmm. You might want to check your calculations. With a diagonal matrix, the eigenvalues are simply the values on the diagonal. In this case, you get $\lambda= \pm 1$; the eigenvector for $\lambda=1$ is $\begin{bmatrix}1 \\ 0 \end{bmatrix}$, and the eigenvector corresponding to $\lambda=-1$ is $\begin{bmatrix}0 \\ 1 \end{bmatrix}$. As the matrix is already diagonal, you can exponentiate it fairly easily:
$$e^{t \mathbf{A}}= \begin{bmatrix} e^{t} &0 \\ 0 &e^{-t} \end{bmatrix}.$$
What is your initial condition?

I was not given any initial condition

 

[Ackbach]

I was not given any initial condition

Ok, that's fine. Let's call the initial condition something arbitrary, then: $\begin{bmatrix}x_0 \\ y_0 \end{bmatrix}$. What is the full solution to the DE?