I need help with my Dynamical systems homework, I honestly don't even really know where to start on it. Any help would be appreciated
Show that if $x(t)$ is a solution to $x'(t) = x(t)^2 - 1 ,$ then so is $z(t) = -x(t)^{-1}$ and $y(t) = -x(-t).$ What type of symmetries are these and how are they reflected in the phase plane?
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As a Riccati equation, you can directly solve this via $x(t)=-\frac{u'(t)}{u(t)}$ where $$ u''(t)-u(t)=0\implies u(t)=Ae^{-t}+Be^t \implies x(t)=\frac{Ae^{-t}-Be^{t}}{Ae^{-t}+Be^{t}} $$ Then $$ z(t)=-y(-t)=\frac{Be^{-t}-Ae^{t}}{Be^{-t}+Ae^{t}}, $$ has the same form, and so does $$ y(t)=x(t)^{-1}=\frac{Ae^{-t}-(-B)e^{t}}{Ae^{-t}+(-B)e^{t}}. $$ Both symmetries combined give a fourth solution $$ w(t)=y(t)^{-1}=-x(-t)^{-1}=\frac{Be^{-t}-(-A)e^{t}}{Be^{-t}+(-A)e^{t}}. $$
I think $z(t)=-x(t)^{-1}$ doesn't satisfy $z'(t)=z(t)^2-1$. Notice that $$ z'(t)=\frac{d}{dt}(-x(t)^{-1})=\frac{x'(t)}{x(t)^2}=\frac{x(t)^2-1}{x(t)^2}=1-\frac{1}{x(t)^2}=1-z(t)^2. $$ Did you mean $z(t)=x(t)^{-1}$?
For $y(t)=-x(-t)$, we have that $$ y'(t)=\frac{d}{dt}(-x(-t))=x'(-t)=x(-t)^2-1=y(t)^2-1. $$