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I am trying to write the 2nd-order linear ODE: $u'' + 2u' + 5u = 0$ as an equivalent 1st-order system specifically using the substitution $v(t) = u'(t)$.

That is, something of the form $\begin{bmatrix} u'\\v' \end{bmatrix} = \left[\Rule{0pt}{1.6em}{0pt}\right. \ \ \ \ \left]\Rule{0pt}{1.6em}{0pt}\right. \begin{bmatrix} u\\v \end{bmatrix}$

So I know that $v' = u'' = -2u' + 5u$, $u = \dfrac{-u - 2u'}{5}$ and that $v' + 2v + \dfrac{-u - 2u'}{5} = 0$.

However, I don't know where to go from here.

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Let $u'=v$, then $u''=v'=-2u'-5u=-2v-5u$. Putting everything together gives us:

\begin{cases} u' & =v\\ v' & =-2v-5u\\ \end{cases}

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