As you correctly note, the solution is different when the applied forces are not equal. **The bar is not in static equilibrium: Both static and dynamic forces deform the bar in motion.**   These concepts are illustrated by superposition.
[![Superposition][1]][1]
$$\delta = \delta_\text{static} + \delta_\text{dynamic} \qquad = \frac {F_\text{less}L}{EA} + \frac{(F_\text{more} - F_\text{less})L}{2AE}$$

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>During changes in acceleration (when $\frac{\mathrm{d\;a(t)}}{\mathrm{dt}} \neq 0$), forces and accelerations within the [damped](https://en.wikipedia.org/wiki/Damping) solid body are transient, where $a(x,t)$, until they reach steady-state, where $\frac{\partial{\;a(x,t)}}{\partial{x}} = 0$.  
[![Dynamic System][2]][2]

><sup>Transient deformations in solid bodies are illustrated by a mass/spring system, where each mass element can be thought to represent differential mass element.</sup>


[Newton's Second Law](https://www.grc.nasa.gov/www/K-12/airplane/newton2.html) requires that the bar (of mass $M$) accelerate in the direction of $F_{net}$. 
$$\sum F \;\text{on bar:} \qquad F_\text{more}-F_\text{less} = Ma \qquad \Rightarrow \;\therefore a = \frac{F_\text{more} - F_\text{less}}{M}$$

The derivation of deformation is shown when a single force acts on the bar. 
 
> **Dynamic Deformation:**

>A Free Body Diagram is taken at an arbitrary cross-section of the bar, where the mass of the split body is $m = (\frac{M}{L})x$.  Summation of forces acting on $m$ is solved for $T(x)$.  
[![Summation of Force][3]][3]
$$\sum F \;\text{on split body:} \qquad F_{o} - T = ma$$
$$F_{o} - T = \overbrace{\left(\frac{M}{L}x\right)}^\text{m} \overbrace{\left(\frac{F_{o}}{M}\right)}^\text{a} = \frac{F_{o}}{L}x \qquad \Rightarrow \qquad T = F_{o} - \frac{F_{o}x}{L}$$
$$\therefore T = F_{o}\left(1-\frac{x}{L}\right)$$
The static axial deformation ($\delta = \frac{FL}{AE}$) written in differential form:  
$$\mathrm{d \delta} = \frac{T \mathrm{dx}}{AE} = \frac{[F_{o}(1-\frac{x}{L})]\mathrm{dx}}{AE}$$
Integrate differential deformation over the length of the bar to determine total deformation:
$$\delta = \int_0^L \mathrm{d \delta} \;  = \frac{F_{o}}{AE} \int_0^L 1-\frac{x}{L} \mathrm{dx}  \implies \;  \delta = \frac{F_{o}L}{2AE}$$

**The derivation can be generalized to include both forces, where integration of $T(x) = F_\text{more} - \dfrac{F_\text{more}-F_\text{less}}{L}x$ results in the same solution given by superposition.**

$$\therefore \delta = \; \overbrace{\frac{F_{more}L}{2AE} + \frac{F_{less}L}{2AE}}^\text{Integration} \;=\; \overbrace{\frac{F_{less}L}{AE} + \frac{(F_{more}-F_{less})L}{2AE}}^\text{Superposition}$$
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References:

* [Stress and Deformation Analysis of Linear Elastic Bars in Tension](http://www.mae.ncsu.edu/ssml/Materials/MAE314/MAE%20314%20chpt%203.pdf)
* [Introduction to Elasticity](http://ocw.mit.edu/courses/materials-science-and-engineering/3-11-mechanics-of-materials-fall-1999/modules/elas_1.pdf)




  [1]: https://i.sstatic.net/vxaNU.jpg
  [2]: https://i.sstatic.net/wMl2z.gif
  [3]: https://i.sstatic.net/zo3CZ.jpg