- #1
bmxicle
- 55
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Homework Statement
so I'm trying to find the general solution of this problem:
[tex]\mathbf {x'} = \begin{bmatrix} 2 & 0\\0 & 2\end{bmatrix}\mathbf{x}[/tex]
Homework Equations
det(A- rI) = 0
The Attempt at a Solution
[tex]det(A - rI) = det \begin{bmatrix} 2-r & 0 \\ 0 & 2-r \end{bmatrix} =
(2-r)^{2} = 0 \Rightarrow r = 2[/tex]
[tex]A - 2I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \ \ \ [/tex]
So since the nullspace is [tex]\mathbb{R}^{2}[/tex] Two linearly independent eigenvectors are:
[tex]\mathbf{v_{1}} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \ \ \ \mathbf{v_{2}} = \begin{bmatrix} 0 \\1 \end{bmatrix}[/tex]
So this is where i get confused :p. One solution is of course [tex]\mathbf{x_1} = e^{2t}\begin{bmatrix}1 \\ 0 \end{bmatrix}[/tex] However when i tried to find a solution such that [tex]\mathbf{x_2} = te^{2t} \begin{bmatrix}1 \\ 0 \end{bmatrix}(\mathbf{a}t + \mathbf{b})[/tex] 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 [tex]\mathbf{x_2} = e^{2t} \begin{bmatrix}0\\1\end{bmatrix}[/tex] 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.