How do derive equivalent complex versions of linear differential equations. I've done this before and have forgotten some of the details.  I will try my best to re-derive it.  Please help fill in the blanks.
In my Acoustics book it says:

An alternating force may be applied to a solid body with mass, $m$ so that Newton's first law holds as follows,
  $$F(t) = ma(t) \\
= m \xi''
$$
  where $\xi$ is displacement of the mass.
For sinusoidal quantities we can write this law in complex notation as
  $$
\tilde{F} = m \tilde{a} = j \omega m \tilde{v} = -\omega^2 m \xi
$$

Now when studying the transmission line equations I came across this as well - there being a complex equation corresponding to two real ones.  Here though we only have one real equation.  Not sure what to do here but I think what I want are complex tilde quantitites ($\tilde{F}$, etc) such that a solution in them yields a solution $F = Re(\tilde{F}), \ \ a = Re (\tilde{a}), etc$.  Any ideas?   And some kind of converse statement regarding real solutions implying complex solutions.
My attempt
Define $\tilde{F} = \bar{F} e^{i \omega t}$, where $F = Re(\tilde{F})$ is the real sinusoidal force applied to the mass $m$.  
Suppose that $\tilde{F} = m \tilde{\xi}''$ had a possibly complex-valued solution $\tilde{\xi}$.  Then we have that $F = Re(m \tilde{\xi}'') =m Re(\tilde{\xi}'') \implies \xi := Re(\tilde{\xi})$ is a real solution to $F = m \xi''$ if we have that $Re(\tilde{\xi}'') = Re(\tilde{\xi})''$.  But $\tilde{\xi} = u + v j$ and $d/dt (\tilde{\xi}) = u'' + v'' j$ and the real part is $u''$, which equals the real part of $\tilde{\xi}$ then the second derivative taken, so we're done there.  Thus a complex solution implies a real solution by taking real-parts of the quantities involved.
Clearly a real solution in the real equation implies a complex solution in the complexified equation because a realy solution is a complex solution.  Thus we can represent problems of this type in complex quantities and later take real parts to find a real solution if needed.  Now back to the other parts of the problem....
$\tilde{F} = m \tilde{\xi}'' \implies \tilde{\xi}' = \frac{\bar{F}}{j \omega m} e^{j \omega t} + c$
by integrating.  How did they get rid of the constants of integration here?  That's officially where I'm stuck.
 A: For SHM the displacements($\xi$) also obey the sinusoidal behaviour. Therefore I can assert that instead of treating the Force as ~$\mathrm{e}^{j\omega t}$ I can consider $\xi = A\mathrm{e}^{j\omega t}$, without loss of generality I can assume that the constant shift in displacement is zero, which you would of collected if integrating up the other way, which would require initial conditions of the system. Then proceeding to take the derivative as normal, this makes sense since the derivation you outlined is in order of derivatives i.e first derivative ($j\omega m \tilde{v}$). Does that make it clearer?
Edits:
SHM is simple harmonic motion. As for your answer I think you are confused by the fact that $\xi = u + jv$ whilst true technically, hides the fact that $\xi=A\mathrm{e}^{j\omega t}$ so you can proceed as I did above. Then you can take real or imaginary components as you like. 
If it is true that $c \neq 0$ and we then integrate to get $\xi$, a linear term in t will appear, which states that the oscillating force will have displacements that will get larger and larger with time. However, this will only occur if we have uniform force acting and not an oscillating one. 
