I was able to find part (a), and I got 4 and -1 for the eigenvalues and from these values I got eigenvectors of [1,1] and [-3,2], but I don't know what to do for part (b) and (c)
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2You don't know how to show that $(1,1)$ and $(-3,2)$ are linearly independent? That's usually covered long before eigenvectors and eigenvalues. Sounds like something you're going to have to page backwards to learn. – Gerry Myerson May 20 '13 at 13:06
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1A related problem. – Mhenni Benghorbal May 20 '13 at 13:15
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Note this, to construct the general solution of the system, the two solutions have to be linearly independent. See here for the case of a single differential equation. – Mhenni Benghorbal May 20 '13 at 13:26
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@MhenniBenghorbal for the link you sent me, what does c1 and c2 represent? – George Randall May 20 '13 at 13:42
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@GeorgeRandall: $c_1$ and $c_2$ are constants. – Mhenni Benghorbal May 20 '13 at 13:54
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1Another related problem. – Mhenni Benghorbal May 20 '13 at 14:02
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1Thank you for the useful links! For part (b), they're linearly independent because the vectors are not multiples of each other? – George Randall May 20 '13 at 14:09
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1@GeorgeRandall: Yes, you are right. – Mhenni Benghorbal May 20 '13 at 14:14
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@MhenniBenghorbal What do I do with the two differential equations given to me, do I sub in the general equation I found, into y. y=c1e^4[1,1] + c2e^−t [−3,2]. – George Randall May 20 '13 at 14:22
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@GeorgeRandall: This is your general solution you were asked to find of the system of differential equations. By the way don not forget to up vote the answers if you benefit from them. – Mhenni Benghorbal May 20 '13 at 14:29
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1@GeorgeRandall: Note that, the system of differentia equations can be written as $$ \pmatrix{\frac{dx(t)}{dt} \ \frac{dy(t)}{dt} } = \pmatrix{1&3\2&2}\pmatrix{{x(t)} \ {y(t)} } . $$ So, that's why you had to find the eigenvalues and the eigenvectors of the matrix. – Mhenni Benghorbal May 20 '13 at 14:41
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Let ${\bf y}=(x,y)$. Then the coupled system can be written as $${\bf y}'=A{\bf y}$$ The general solution is $${\bf y}=c_1e^{4t}(1,1)+c_2e^{-t}(-3,2)$$ assuming that you did the eigenvector/eigenvalue calculations correctly.
Gerry Myerson
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