if you have a differential equation of the form(adsbygoogle = window.adsbygoogle || []).push({});

x' = Ax

where A is the coefficient matrix, and you get a triple eigenvalue with a defect of 1. (meaning you get v1 and v2 as the associated eigenvector). How do you get v3 and how do you set up the solutions?

I tried finding v3 such that [itex](\mathbf{A}-\lambda \mathbf{I})^2 \mathbf{v_3} = \mathbf{0}[/itex] but not exactly sure what to do after that. I got another v_2 (called it v_2') by using [itex](\mathbf{A}-\lambda \mathbf{I}) \mathbf{v_3} = \mathbf{v_2'}[/itex] but again not sure how to set up solution.

my answer was

[tex]\mathbf{x_1(t)} = \mathbf{v_1} e^{\lambda t}[/tex]

[tex]\mathbf{x_2(t)} = \mathbf{v_2'} e^{\lambda t}[/tex]a

[tex]\mathbf{x_3(t)} = (\mathbf{v_2'} + \mathbf{v_3}t)e^{\lambda t}[/tex]

but it is not what is in the back of the book. I also tried using the original v_2:

[tex]\mathbf{x_1(t)} = \mathbf{v_1} e^{\lambda t}[/tex]

[tex]\mathbf{x_2(t)} = \mathbf{v_2} e^{\lambda t}[/tex]a

[tex]\mathbf{x_3(t)} = (\mathbf{v_2} + \mathbf{v_3}t)e^{\lambda t}[/tex]

Can anyone help?

Thanks in advance.

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# Homework Help: Defective eigenvalues?

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