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Show that if $u(t)$ solves $\dot{u} = Au$, then $v(t) = u(-t)$, solves $\dot{v} = Bv$, where $B = -A$.

Similarly, show that if u(t) solves $\dot{u} = Au$, then $v(t) = u(2t)$ solves $\dot{v} = Bv$, where $B = 2A$.

I don't know how to formally prove this. It seems obvious because since the solution $v(t) = u(-t)$ is the same as $u(t)$ but it has a scalar which is negative and a scalar can be taken out of the solution and placed outside which follows that we have $-A$. The same goes with the second part of the question.

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up vote 2 down vote accepted

Let $s=-t$ then $dv(t)/dt=du(-t)/dt=-du(s)/ds=-Au(s)=-Au(-t)=-Av(t).$

Or let $s=kt$ with $k=-1,2$ then $dv(t)/dt=du(kt)/dt=(du(s)/ds)(ds/dt)=kAu(s)=kAu(kt)=kAv(t).$

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Just substitute in to the differential equation and see that both sides are equal.

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