Find a particular solution of the indicated linear system that satisfies the initial conditions $x_1(0) = 4$, $x_2(0) = 5$, and $x_3(0) = 6$. $\begin{bmatrix} -22 & -24 & 0\\ 20 & 22 & 0\\ -20 & -20 & 2 \end{bmatrix} x$; $x_1 = e^{-2t} \begin{bmatrix} 6\\ -5\\ 5 \end{bmatrix}$, $x_2 = e^{2t} \begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}$, $x_3 = e^{2t} \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}$ The particular solution is $x_1(t) = \Box$, $x_2(t) = \Box$, and $x_3(t) = \Box$.
Added by Remedios M.
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The given linear system is: -22x' - 240x = 20220 x' = -22x + 20220/240 x' = -22x + 84.25 So, A = -22 and f(t) = 84.25. Show more…
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