MHB Non-linear differential equation

Poirot1
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system is $x'=y+xf(r^2)$ and $y'=-x+yf(r^2)$. where $r^2=x^2+y^2$

(i) prove that $\frac{dr^2}/{dt}=2r^2f(r^2)$. My solution ( I won't write out details): use chain rule and the fact that rr'=xx'+yy'.

(ii)assume $f(r^2)$ has N zeroes. determine the number of fixed points and periodic solutions the system has and write about the stability of fixed points.

This one I think I did the first (I will give details) but can't do the others

Solution: $r^2{\theta}'=xy'-yx'$and if (x,y) is fixed point, then xy'-yx'=0. If you work this out you get fixed point implies -r^2=0 so x=y=0 is only fixed point. In the solution however it has a different justification, namely that ${\theta}'=-1$ but I don't understand how this means (0,0) is only fixed point.
 
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You already know that x'= y+ xf(r^2) and y'= -x+ yf(r^2) so that xy'- yx'= x(-x+ yf(r^2))- y(y+ xf(r^2)= -x^2+ xyf(r^2)- (y^2+ xyf(r^2)= -(x^2+y^2)= -r^2= 0 if and only if r= 0.
 
ok what about the question about periodic solutions
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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