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Consider the following system of ODEs $$\begin{bmatrix} x\\ y \end{bmatrix}'= \begin{bmatrix} -2& 3\\ -1 & 2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} $$ with initial conditions $x(0)=1, y(0)=1$.


I'm used to going from a second order ODE to a system of equations using some sort of substitution, but going backwards, i.e. from a system to a single second order ODE, is not so intuitive.

How might I reduce this system to a single second order ODE in terms of $x$?

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Our system is

$$x' = -2x + 3 y \\ y' = -x + 2 y \\ x(0) = 1, ~y(0) = 1$$

From the first equation, we have

$$x' = -2x + 3 y \implies y = \dfrac{1}{3}\left(x' + 2x\right)$$

We can also differentiate the first equation and have

$$x'' = -2 x' + 3 y' = -2 x' + 3(-x + 2 y)$$

Hopefully, you can take it from there.

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  • $\begingroup$ If I'm understanding correctly, all that remains to do is to substitute $y$ into the $x''$? Is that correct? $\endgroup$ – user352541 May 9 '17 at 17:43
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    $\begingroup$ Yes that is correct, and then to use the first equation to find $x'(0)$ as a second IC in terms of $x$. $\endgroup$ – Moo May 9 '17 at 17:45

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