0
$\begingroup$

I am doing my homework and I know this is a very simple construction but I am unsure how to do this.

We are given $x''+x=0$ and are supposed to write it in the form $x'=Ax$. I see to go to $x''=-x$ but then how do I differentiate it properly? I know ultimately it ends up as a system of sines and cosines but my knowledge of solving linear systems is not that advanced and I am having trouble finding reading material on the internet. If you fully integrate the function in respect to x and t I believe it is -lnx=t. But this seems incorrect and does not give me an idea of how to construct a matrix to solve for $PDP^{-1}$. Thanks!

$\endgroup$
8
  • $\begingroup$ Introduce a new variable $u=x'$ and convert to a system of two equations. $\endgroup$
    – amd
    Oct 8, 2017 at 0:14
  • $\begingroup$ To do this would I want to make x'' = $u^2$ or something like x''+u=-x and then u=0 as my system? $\endgroup$
    – bananagurlz
    Oct 8, 2017 at 0:17
  • $\begingroup$ The goal is to eliminate explicit second derivatives of $x$. $\endgroup$
    – amd
    Oct 8, 2017 at 0:19
  • $\begingroup$ so then can I substitute $u^2$ for x''? Would it be best to convert everything in terms of the new variable and then solve that way? some u=0 and then u^2=-u'? $\endgroup$
    – bananagurlz
    Oct 8, 2017 at 0:24
  • $\begingroup$ Where is this $u^2$ coming from? $x''=u'$. $\endgroup$
    – amd
    Oct 8, 2017 at 0:24

1 Answer 1

Reset to default
0
$\begingroup$

You’re using $x$ to stand for two different things—a scalar function and a vector function—which appears to be part of what’s confusing you. Let’s call the vector function something else: The task is to write $x''+x=0$ in the form $\mathbf y'=A\mathbf y$. We do this by reducing the second derivative of $x$ to a first derivative: introduce the new function $u=x'$, so that $x''=u'$. You then have the system of first-order equations $$x'=u\\u'=-x$$ which can be written in matrix form as $$\begin{bmatrix}x'\\u'\end{bmatrix}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}x\\u\end{bmatrix}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .