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Solve:

$y_1'=12y_3$

$y_2'=y_1+13y3$

$y_3'=y_2$

(working from wolfram) http://www.wolframalpha.com/input/?i=%7B%7B0%2C0%2C12%7D%2C%7B1%2C0%2C13%7D%2C%7B0%2C1%2C0%7D%7D

so the solution is:

$y_1=3c_1e^{4x}-4c_2e^{-3x}-12c_3e^{-x}$

$y_1=4c_1e^{4x}-3c_2e^{-3x}-c_3e^{-x}$

$y_1=1c_1e^{4x}+c_2e^{-3x}+c_3e^{-x}$

now my question is: find a value y$(0)$ other than y$(0)$$=(0,0,0)$ such that y$(x)$ approaches $(0,0,0)$as $x$ approaches infinity.

I'm not even sure what that means ><

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1 Answer 1

up vote 0 down vote accepted

It should be obvious that the first term in each expression will "blow up" as $x\rightarrow\infty$. So you must choose $c_1=0$.

For the second and third terms, these both approach $0$ as $x\rightarrow\infty$, so any choice of $c_2$ and $c_3$ will satisfy the condition which you state.

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right thanks:) I deduced that much myself, but had no idea the question wanted different values for c. so y(x)=(c_1,c_2,c_3)=(0,c_2,c_3). thanks –  redrum Oct 24 '12 at 8:37
    
The values are different as they are integration constants, one for each equation. –  Daryl Oct 24 '12 at 10:42

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