# Convert two first order ODEs into second order ODE

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.

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

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 $$x_2$$.
• In this case, are we actively trying to solve $x_1$ and $x_2$? Feb 24, 2020 at 0:08