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So I have two equations

$X' = aX + bY$

$Y' = cX + dY$

I want to convert it back to a second order equation with the form

$X'' + \alpha X' + \beta X$ with $\alpha,\beta$ in terms of a,b,c,d.

I have been racking my brain for hours trying to go backwards from a reduction of order, but just can't seem to figure it out. Any help would be much appreciated!

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If you derive the first equation, you get: $$ X ^"=aX '+bY ' $$ (if you are considering $a$ and $b$ as constants). But we have $Y'=cX+dY $, so substitute in the above equation, you get $$ X ^" = a X' +b(cX+dY).$$ Note that $Y=\frac{1}{b}(X'-aX)$ for $b \neq 0$. So, substituting again you get the final answer.

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  • $\begingroup$ ah, that was much easier than I was making it, just a question how did you get Y, I think that was where I was running into major issues. $\endgroup$ – jam Apr 20 '15 at 6:02
  • $\begingroup$ Im guessing you just solve the first equation correct? $\endgroup$ – jam Apr 20 '15 at 6:03
  • $\begingroup$ I get $Y$ from the first equation $X'=aX+bY$. $\endgroup$ – Nizar Apr 20 '15 at 6:03
  • $\begingroup$ Awesome that's what I thought thanks for the help! Storing that trick for later! $\endgroup$ – jam Apr 20 '15 at 6:06

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