Finding a solution basis of differential equation

Asked Aug 09 '14 at 18:17

Active Aug 10 '14 at 05:10

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Find a solution basis of $$y'=\left[ \begin{matrix}3&-4&-2\\2&-3&-2\\0&0&1\\ \end{matrix} \right]y \,\text{ and find the solution } \Phi \text{ with } \Phi(0) = (1,1,1).$$

I'm preparing for an exam for which I should know how to solve this type of problem. However, I'm not sure what the general approach is.

I know we're going to need the characteristic equation of the matrix and the corresponding eigenspaces, so I figured that out:

The characteristic equation is $$p(t) = -(t-1)^2(t+1),$$ and the eigenspaces are $$E_1 = [(2,1,0), (1,0,1)] \text{ and }E_{-1}=[(1,1,0)].$$

Now my question is the following: Could you please guide me through the rest of this problem, making clear the general approach for these problems?

Sorry for the broad question. Thank you.

edited Aug 10 '14 at 05:10

asked Aug 09 '14 at 18:17

rehband

1,921

There are several ways to attack this. Are you familiar with matrix exponentials? – Git Gud Aug 09 '14 at 18:18
Yes, I have encountered that before :) – rehband Aug 09 '14 at 18:21
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Take a look at this answer. – Git Gud Aug 09 '14 at 18:23
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Check this. – Mhenni Benghorbal Aug 09 '14 at 18:26
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I just noticed your work. Assuming what you did is correct, you're nearly done. Note the following. If $(\lambda, v)$ is an eigenpair of $A$ then $(e^{\lambda t}v)'=\lambda e^{\lambda t}v=e^{\lambda t}Av=A(e^{\lambda t}v)$, that is, $t\mapsto e^{\lambda tv}$ is a solution to the differential equation. Since you got three linearly independent vectors, the functions $t\mapsto e^{\lambda _it}v_i$ where $(\lambda _i, v_i)$ are eigenpairs, are linearly independent and hence form a basis of solutions. – Git Gud Aug 09 '14 at 18:34
Did you mean to write $t\mapsto e^{t\lambda_i}v_i$, Git Gud? – rehband Aug 09 '14 at 18:37
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@rehband At the end? Yes, I just fixed it. Thanks. – Git Gud Aug 09 '14 at 18:37
So this gives us the following basis ${t\mapsto e^{t}(2,1,0), t\mapsto e^{t}(1,0,1), t\mapsto e^{-t}(1,1,0)}$ ? – rehband Aug 09 '14 at 18:39
Now $\Phi$ is a linear combination of those three basis vectors and we can find its coefficients using the initial value condition by solving a linear system of equations right? Thanks very much for the help btw :) – rehband Aug 09 '14 at 18:44
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See this. – Mhenni Benghorbal Aug 09 '14 at 18:54
Many thanks Mhenni Benghorbal! – rehband Aug 09 '14 at 19:05
@rehband That's right. But you should check for typos in the question. The eigenvalues of that matrix aren't what you say they are. See this. – Git Gud Aug 09 '14 at 19:16
Indeed, I forgot some minus signs. Fixed it now, thank you. – rehband Aug 10 '14 at 05:10

Finding a solution basis of differential equation

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