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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:

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

$y_2=4c_1e^{4x}-3c_2e^{-3x}-c_3e^{-x}$

$y_3=1c_1e^{4x}+c_2e^{-3x}+c_3e^{-x}$


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.

Could someone explain what the question is asking?

Thanks! (note: I asked a similar question earlier, but this has corrected the error and changed the question)

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1 Answer 1

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If you select $c_1,c_2,c_3$ in your answer, then plugging in $x=0$ will give you the value of $y(0)$. The question asks you to select $c_1,c_2,c_3$ in such a way that $y(0)\neq 0$ (which as you may verify just means that you should avoid taking all three constants to be $0$) and in such away that letting $x\to+\infty$ makes $y(x)$ tend to $(0,0,0)$.

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