= Solution
The <work> done along the trajectory is
$$
W=\int_{t_a}^{t_b}F(x(t),t)\dot x(t)\,dt
=\int_a^bF(x,t(x))\,dx.
$$
By <Newton's second law>, $F=m\ddot x$, so
$$
W=m\int_{t_a}^{t_b}\ddot x\dot x\,dt
=\left[\frac12m\dot x^2\right]_{t_a}^{t_b}.
$$
This is the <work-energy theorem>: the work equals the change in <kinetic energy>.
A force is <conservative force>[conservative] when its work between two points is independent of the path, equivalently when it is the negative gradient of a <potential energy>. Here a potential with $V(0)=0$ is
$$
V(x)=F_0\lambda(1-e^{-|x|/\lambda}).
$$
Conservation of <energy> for $x>0$ gives
$$
\frac12mv^2+F_0\lambda(1-e^{-x/\lambda})=\frac12mv_0^2.
$$
Thus
$$
v(x)=\sqrt{v_0^2+v_e^2(e^{-|x|/\lambda}-1)},
\qquad
\boxed{v_e=\sqrt{\frac{2F_0\lambda}{m}}}.
$$
If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the <escape velocity> $v_0=v_e$, it decreases as $v_e e^{-x/(2\lambda)}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the <turning point>
$$
x_{\max}=-\lambda\log\left(1-\frac{v_0^2}{v_e^2}\right),
$$
where $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.
For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the <period of an oscillation>[period] is
$$
\int_0^{x_{\max}}\frac{dx}{v(x)}
=\frac{\lambda}{v_e}\int_w^1\frac{du}{u\sqrt{u-w}}
=\frac{2\lambda}{v_e\sqrt w}\cos^{-1}\sqrt w,
$$
where $u=e^{-x/\lambda}$. Therefore
$$
\boxed{T=\frac{8\lambda}{v_e\sqrt{1-v_0^2/v_e^2}}
\cos^{-1}\sqrt{1-v_0^2/v_e^2}}.
$$
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