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I am having trouble expressing two first order ODEs as a second order ODE. any tips?

$$A= \begin{pmatrix}-1.005&-0.266\\ -0.1498&0.2005\\ \end{pmatrix}$$

with $dx/dt = Ax$ and $x(0) = x_0 = $$\begin{pmatrix}1\\ -2\\ \end{pmatrix}$

Namely, I know I have to take the derivative of one equation and substitute into another - but taking the derivative of one equation just leaves me with a constant. Can someone lay out a couple of steps for me to start?

Update:

I have worked through the algebra to come up with the following expression. My question is - did we create a second order ODE out of our initial ODEs? I am just trying conceptually understand the question.

$$\frac{dx_2}{dt}=-1.005\frac{dx_1}{dt}+0.0399217x_1-0.20037(\frac{dx_1}{dt}+1.005x_1) $$

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You have $$ \frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix}x_1 \\ x_2 \end{pmatrix} = A \begin{pmatrix}x_1 \\ x_2 \end{pmatrix} \textrm{,} $$ so $$ \begin{pmatrix}\frac{\mathrm{d}x_1}{\mathrm{d}t} \\ \frac{\mathrm{d}x_2}{\mathrm{d}t} \end{pmatrix} = \begin{pmatrix}-1.005 x_1 - 0.266 x_2 \\ -0.1498 x_1 + 0.2005 x_2 \end{pmatrix} \textrm{,} $$ which it might help to see as $$ \left\{ \begin{matrix} \frac{\mathrm{d}x_1}{\mathrm{d}t} = -1.005 x_1 - 0.266 x_2 \\ \frac{\mathrm{d}x_2}{\mathrm{d}t} = -0.1498 x_1 + 0.2005 x_2 \end{matrix} \right. \text{.} $$ Eliminating $x_2$ between these, we obtain $$ \left\{ \begin{matrix} \frac{\mathrm{d}x_1}{\mathrm{d}t} = -1.005 x_1 - 0.266 x_2 \\ \frac{\mathrm{d}x_2}{\mathrm{d}t} = -0.1498 x_1 + 0.2005 \frac{1}{-0.266} \left(\frac{\mathrm{d}x_1}{\mathrm{d}t} + 1.005 x_1\right) \end{matrix} \right. \text{.} $$ Differentiating the first with respect to $t$, $$ \left\{ \begin{matrix} \frac{\mathrm{d}^2x_1}{\mathrm{d}t^2} = -1.005 \frac{\mathrm{d}x_1}{\mathrm{d}t} - 0.266 \frac{\mathrm{d}x_2}{\mathrm{d}t} \\ \frac{\mathrm{d}x_2}{\mathrm{d}t} = -0.1498 x_1 + 0.2005 \frac{1}{-0.266} \left(\frac{\mathrm{d}x_1}{\mathrm{d}t} + 1.005 x_1\right) \end{matrix} \right. \text{.} $$ Then substitute the second into the first to remove all mentions of $x_2$.

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  • $\begingroup$ Thank you! But why do you use dx1/dt and not just the constants, 1 and -2? $\endgroup$
    – user9858
    Feb 23, 2020 at 23:57
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    $\begingroup$ @user9858 : You want a second order ODE. There must be derivatives. $\endgroup$ Feb 23, 2020 at 23:58
  • $\begingroup$ Ahhh ok got it, thank you. $\endgroup$
    – user9858
    Feb 23, 2020 at 23:59
  • $\begingroup$ @user9858 : Alternatively, you want an equation that is true at points other than the initial point, so don't specialize too early. $\endgroup$ Feb 23, 2020 at 23:59
  • $\begingroup$ In this case, are we actively trying to solve $x_1$ and $x_2$? $\endgroup$
    – user9858
    Feb 24, 2020 at 0:08

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