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!


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.