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I need to solve the system of the DE:
$$ \begin{cases} \frac{dx}{dt} = 2x-4y, \\ \frac{dy}{dt} =x+2y+z,\\ \frac{dz}{dt} = 3y+2z, \end{cases} $$ First, I rewtite the coefficents in the matrix form: $$A = \begin{bmatrix} 2 & -4 & 0\\ 1 & 2 & 1\\ 0 & 3 & 2 \end{bmatrix}$$
Then, I find $$det(A-\lambda I) = \begin{vmatrix} 2 - \lambda & -4 & 0\\ 1 & 2 - \lambda & 1\\ 0 & 3 & 2 - \lambda \end{vmatrix} =-(\lambda-2)(\lambda-(2-i))(\lambda-(2+i)) $$ For $\lambda =2$ the approach is not very difficult - I will solve the system $Ev =0$ where $E=A-\lambda I$ and $v = \begin{bmatrix} x\\ y\\ z \end{bmatrix}$.
But this approach is much for difficult more complex $\lambda$.

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  • $\begingroup$ What does the general solution look like for a $2\times2$ coefficient matrix with complex eigenvalues? Extrapolate from that. $\endgroup$ – amd Mar 4 at 7:02
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If we stay within real matrices, we cannot diagonalize the coefficient matrix, but we can block-diagonalize it.For $\lambda=2$, an eigenvector is $$\begin{bmatrix}-1\\0\\1\end{bmatrix}$$ For $\lambda=2+i$, an eigenvector is $$\begin{bmatrix}-4\\i\\3\end{bmatrix}=\begin{bmatrix}-4\\0\\3\end{bmatrix}+\begin{bmatrix}0\\1\\0\end{bmatrix}i$$ Let $$P=\begin{bmatrix}-1&-4&0\\0&0&-1\\1&3&0\\\end{bmatrix}$$ Let $$\check D=\begin{bmatrix}2&0&0\\0&2&-1\\0&1&2\end{bmatrix}$$ Then $P^{-1}AP=\check D$ Let $$\begin{bmatrix}x\\y\\z\end{bmatrix}=P\begin{bmatrix}u\\v\\w\end{bmatrix}$$ Then $$\begin{bmatrix}du/dt\\dv/dt\\dw/dt\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&-1\\0&1&2\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix}$$ $$u=ae^{2t},v=e^{2t}(b \cos t+c \sin t),w=e^{2t}(-c \cos t+b \sin t)$$. You can take it from here.

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  • $\begingroup$ Could you please comment oh how you got eigenvector for $\lambda = 2+i$. You just took the second column of $A-\lambda I $ matrix? $\endgroup$ – student Mar 4 at 14:24
  • $\begingroup$ @student: Work out A-(2+i)I. This matrix has determinant 0, so a vector in the solution space, i.e. an eigenvector, can be obtained by going along a row of A-(2+i)I and taking co-factors.. $\endgroup$ – P. Lawrence Mar 5 at 17:02

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