Question

$x'=\begin{pmatrix} 3 & 2 &4 \\ 2 & 0 & 2\\ 4 & 2 & 3 \end{pmatrix}x$

I found the general solution:$x=c1\begin{pmatrix} 2\\ 1\\ 2\end{pmatrix}e^{-8t}+c2\begin{pmatrix} -1\\ 0\\ 1\end{pmatrix}e^{-t}+c3\begin{pmatrix} -1\\ 2\\ 0\end{pmatrix}e^{-t}$

But I need to answer this: Determine all initial conditions for which solutions to x'=Ax are bounded. Describe the surface in which these solutions live.

Answer 1

Doubt in this then comment below. I will explain you..

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please thumbs up for this solution..thanks..

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here x(t) have 3 parts ...

And for large value of t , we know that firsst part becomes infinite ..

So for bounded aolution , we not ned first part ..that means we want c1c1=0 ..

2 +8+ / = ) C + C2 etc Geto Note so as ta. est po as to s for bounded soen, we want 1,20 Soen bounded W. let Xlola (3) then 1-(2 – 3 263 So (= bz b2 = 263 (3= 62, -(2- Cabi 2 It initial conditions - 63 - bub [bi + b 2 + b3 =0 / I satisfy bitby + bgão Then (60) X is bounded I be so s Alt is bounded