System of ODEs independent solutions

[bmxicle]

Homework Statement

so I'm trying to find the general solution of this problem:
\mathbf {x'} = \begin{bmatrix} 2 & 0\\0 & 2\end{bmatrix}\mathbf{x}

Homework Equations

det(A- rI) = 0

The Attempt at a Solution

det(A - rI) = det \begin{bmatrix} 2-r &amp; 0 \\ 0 &amp; 2-r \end{bmatrix} =<br /> (2-r)^{2} = 0 \Rightarrow r = 2
A - 2I = \begin{bmatrix} 0 &amp; 0 \\ 0 &amp; 0 \end{bmatrix} \ \ \
So since the nullspace is \mathbb{R}^{2} Two linearly independent eigenvectors are:
\mathbf{v_{1}} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \ \ \ \mathbf{v_{2}} = \begin{bmatrix} 0 \\1 \end{bmatrix}

So this is where i get confused :p. One solution is of course \mathbf{x_1} = e^{2t}\begin{bmatrix}1 \\ 0 \end{bmatrix} However when i tried to find a solution such that \mathbf{x_2} = te^{2t} \begin{bmatrix}1 \\ 0 \end{bmatrix}(\mathbf{a}t + \mathbf{b}) It comes out inconsistent so I'm guessing that's not what you're supposed to do in this case. Another idea i had was that \mathbf{x_2} = e^{2t} \begin{bmatrix}0\\1\end{bmatrix} is also an independent solution since the vector is linearly independent from the other solution, but I'm not entirely sure how to verify this by the wronskian for a system of equations.

 

[ehild]

One solution is of course \mathbf{x_1} = e^{2t}\begin{bmatrix}1 \\ 0 \end{bmatrix}

Another idea i had was that \mathbf{x_2} = e^{2t} \begin{bmatrix}0\\1\end{bmatrix} is also an independent solution since the vector is linearly independent from the other solution

That is the solution.

You get the "normal modes" of this two-variable system in the form of an eigenvector multiplied by the same exp(rt).

ehild

 

[bmxicle]

hmmm ok, I that makes enough sense, though need to do some more reading. Can you compute the wronskian as the determinant of the matrix of the spanning vectors, ie. W = e^{4t} \neq 0

 

[epenguin]

You do say you are only trying to find the solution. But if I have not misunderstood this is almost trivial - you have two independent elementary d.e.'s, x₁' = 2x₁, x₂' = 2x₂.

If instead you are trying to show it works out in accord with the Higher Theory, Wronskians etc. then I forget these between times and cannot help.

 

[bmxicle]

yes well looking at it now it definitely is almost trivial. I'm just a little hazy as to what constitutes an indepedent solution to a system of equations I guess.