Oh okay, sorry for the confusion. The characteristic polynomial comes from assuming a solution, i.e., we assume $\displaystyle x(t) = e^{wt}$ and plug this into the originial DEQ. Thus we obtain the following
$\displaystyle m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt}$
which leads us to
$\displaystyle (mw^2 + cw + k)(e^{wt}) = 0.$
This tells us the characteristic equation is
$\displaystyle mw^2 + cw + k = 0.$
When we find the roots to this equation it tells us that we have one solution containing $\displaystyle w_1$ and another solution containing $\displaystyle w_2$, i.e., we have one solution $\displaystyle x_1(t) = C_1 e^{w_1t}$ and a second solution $\displaystyle x_2 = C_2 e^{w_2 t}.$ However, since both these are solutions to the differential equation we know that a linear combination of the two is also a solution to the differential equation, and thus we obtain the general solution
$\displaystyle
x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.
$
I really hope I finally answered your question. I suppose after all this you will probably understand the problem more thoroughly.
Your absolutely correct, I made a mistake and have corrected the post. We assume this is the solution, because we want
$\displaystyle m x''(t) + c x'(t) + k x(t) = 0.$
This will only occur if $\displaystyle x'',x',x$ have the same form. Well, we know that the derivative of $\displaystyle e^t$ is itself, so this clues us into the type of solution we need, i.e., $\displaystyle e^{wt}$.
The first I understood
$\displaystyle x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L$ OK
But the second condition
$\displaystyle x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0$
How did you find this equation?
Well, I was given an example where $\displaystyle g(x) = 0$ and so I took the derivative of $\displaystyle x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
Then I plugged in $\displaystyle t= 0$, but I knew from that $\displaystyle x(0) = 0$. Putting this all together we could solve for $\displaystyle C_1,C_2$