Getting 1st order ODE's from a 2nd order ODE

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    2nd order Ode
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Equations:

(d^2r/dt^2) - r*(dθ/dt)^2 = -10/(r^2)

and

r^2*(dθ/dt)=1

How would I get three 1st order ODE's from this?
 
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Hint: How would you rewrite eq. 1 so that only first order ODEs were used?
 
Ok. But how would I go about doing that?
 
I've been struggling with exactly this sort of problem. So bear in mind that I might be mistaken. But maybe it'll help us both if I have a go too and thrash around for an answer, and if my suggestion is wrong, hopefully someone will correct it. I've been reading section 1.2 of Teschl's ODEs & Dynamical Systems. He says that

any system can always be reduced to a fi rst-order system by changing to the new set of dependent variables y = (x,x(1),...,x(k-1)). This yields the new first-order system

y'1 = y

y'k-1 = y(k)

y'k = f(t,y).

So, if I've understood this right, we could let y = (r,r'), so that y' = (r',r''), where r' means dr/dt. Then, following SteamKing's hint and concentrating on the first equation,

y' = (r',-rθ'2-10r-2).

But, hey, from your 2nd equation, we have θ'2 = r-2, so we can substitute for this to get an equation that doesn't depend on θ, like this

y' = (r',-r*r-2-10r-2) = (r',-r-1-10-2).

Then, you could let w = (θ,θ') and use substitution and the chain rule to get an expression for w' in terms of r. Would that be the three equations you're looking for: one for y', one for w' in terms of θ, and one for w' in terms of r? Or maybe the three would be y' and w' in terms of r, and the second of your given equations, rearranged as θ' = r-2; that makes three first order ODEs, doesn't it?
 
See the attached page. Is it that what was expected ?
 

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