1. ## Autonomous System question

I have this question and I don't know how to answer it, at all.

x''+x=0 can be solved in two steps which involves solving the equations (dx/dz)^2 + x^2 = c^2

The former equation have unique solutions whereas the latter does not. How do you reconcile these two statements? Does every solution of the latter satisfy the former?

This chapter is dealing with autonomous systems of which I know nothing about..

2. Originally Posted by sarah232
I have this question and I don't know how to answer it, at all.

x''+x=0 can be solved in two steps which involves solving the equations (dx/dz)^2 + x^2 = c^2

The former equation have unique solutions whereas the latter does not. How do you reconcile these two statements? Does every solution of the latter satisfy the former?

This chapter is dealing with autonomous systems of which I know nothing about..
The equations presented are the same meaning that they have the same solution. Let me explain. The first one was:
$\displaystyle x''+x=0$
and has as solution:
$\displaystyle x(z)=A \cdot sin(z)+B \cdot cos(z)$
This is easily found because it is a homogeneous second order DE with constant coefficients. Now if we multiply this DE with
$\displaystyle 2\cdot x'$
we get:
$\displaystyle 2\cdot x' \cdot x'' + 2 \cdot x' \cdot x=0$
which can be rewritten as:
$\displaystyle \frac{d}{dz}\left((x')^2+(x)^2\right)=0$
or, after integrating:
$\displaystyle (x')^2+(x)^2=c^2$
with c the integration constant. This equation has the same solution as the first one, indeed, extracting the derivative gives:
$\displaystyle x'=\pm \sqrt{c^2-x^2}$
and this can be integrated again to give:
$\displaystyle arcsin\left(\frac{x}{c}\right)=\pm z+K$
or:
$\displaystyle x=c\cdot sin(\pm z+K)$

Can you take it from here to show that this is the same solution as the first one?

3. Awesome, thanks. I can finish from there.