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Is there any general and systematic way of reducing the higher order linear ODE to a system of first order ODE?

For example, assume we have $a_3x^{(3)}+a_2x^{(2)}+a_1x^{(1)}+a_0x=0$, then how do we convert this into matrix form(a system of first order ODE). And after we solve the system of equation, how to combine them into our final solution $x(t)$?

Thanks for helping me out!

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  • $\begingroup$ Do you know that this particular kind of DE has a very easy to find solution which doesn't require such tricks? $\endgroup$ – Git Gud Sep 22 '13 at 22:10
  • $\begingroup$ I know, plugin formula is just an easy thing. But I would like to know how to get those solutions, the process. :) $\endgroup$ – Cancan Sep 22 '13 at 22:11
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Starting with $$ x^{(n)}=f(t,x,x',x'',\ldots,x^{(n-1)}), $$ you get the system \begin{align*} x_1&=x\\ x_1'&=x'=x_2\\ x_2'&=x''=x_3\\ &\ldots\\ x_{n-1}'&=x^{(n-1)}=x_{n}\\ x_n^{'}&=x^{(n)}=f(t,x_1,x_2,\ldots,x_{n-1}) \end{align*}

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  • $\begingroup$ And one last question, after we get the solution vector, is the solution we need the first entry of that vector?($x_1$) And the matrix is pretty sparse. When we deal with this matrix, is there any trick to diagonalize it(or Jordan normal form)? $\endgroup$ – Cancan Sep 22 '13 at 22:19
  • $\begingroup$ Yes, your solution is just $x_1$. I do not quite understand why you need to diagonalize it. However, if the system is linear, the matrix form will show you where the characteristic equation appears from (this will be exactly $\det (A-\lambda I)=0$). $\endgroup$ – Artem Sep 22 '13 at 22:23
  • $\begingroup$ In my class notes, I didn't copy it clearly, but it says the fundamental solution is $\{t^{r_i}e^{\lambda_i t}$ for $r_i\in\{0,..,d_{i-1}\}\}$ but I am not sure what it means, because how do we locate the correct eigenvalue corresponding to $x_1$ (rather than $x_2$ or so) $\endgroup$ – Cancan Sep 22 '13 at 22:28
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    $\begingroup$ Because in this case actually matrix $A$ in the corresponding system is diagonalizable :) A better answer would be: We know that the dimension of the set of solutions is 2 and we already have linearly independent solutions $e^{\lambda_1t}$ and $e^{\lambda_2 t}$, hence no room for anything else. $\endgroup$ – Artem Sep 22 '13 at 22:40
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    $\begingroup$ Right. For the linear equation. For a general linear system (whose matrix does not have this special and sparse form) some additional subtleties arise. But not many ) $\endgroup$ – Artem Sep 22 '13 at 22:45

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