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It's Only a Minus Sign

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

Particle A

Qualitative behaviour: moves in positive direction with an increasing velocity. $$\frac{dx}{dt} = x \Rightarrow \int \frac{1}{x}dx = \int dt \Rightarrow x = Ae^t$$ $x(0) = 1$ so $A = 1$ and the solution is $x(t) = e^t$.

Particle B
Qualitative behaviour: moves in negative direction with a decreasing velocity, gradually approching origin. $$\frac{dx}{dt} = x \Rightarrow x = Ae^{-t}$$ $x(0) = 1$ so $A = 1$ and the solution is $x(t) = e^{-t}$.
Particle C
Qualitative behaviour: moves in positive direction with an increasing velocity. $$\frac{dv}{dt} = x \Rightarrow \frac{d^2x}{dt^2} - x = 0 \Rightarrow x = Ae^t + Be^{-t}$$ If particle starts at origin with velocity $v$ then $x(t) = \frac{v}{2}(e^t - e^{-t})$.
Particle D
Qualitative behaviour: oscillates around origin. $$\frac{dv}{dt} = -x \Rightarrow \frac{d^2x}{dt^2} + x = 0 \Rightarrow x = Ae^{it} + Be^{-it} = C\sin{t} + D\cos{t}$$ If particle starts at origin with velocity $v$ then $x(t) = v\sin{t}$.
Particles C and D have symmetrical equations of motion, in the sense that starting at origin with negative velocity would result in the same motion but in the opposite direction.
Observe that if we start particle C at $+1$ with velocity $-1$ then our equation of motion is $x(t) = e^{-t}$, so the particle eventially stops at the origin. Are there any other initial conditions that can be imposed on particles C or D which force the particle to stop eventually?