Position of Cart attached to Wall by Spring under Damping/Overdamped/x = x0 at t = 0

Problem Definition
Let:
 * $a^2 = \dfrac k m$
 * $2 b = \dfrac c m$

Let $b > a$.

Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.

Then the horizontal position of $C$ at time $t$ can be expressed as:
 * $x = \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)$

where $m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + 2 b + a^2 = 0$:
 * $m_1 = -b + \sqrt {b^2 - a^2}$
 * $m_2 = -b - \sqrt {b^2 - a^2}$

Such a system is defined as being overdamped.

Proof
From Position of Cart attached to Wall by Spring under Damping: Overdamped:
 * $(1): \quad x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$

where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$.

It remains to evaluate $C_1$ and $C_2$ under the given conditions.

Differentiating $(1)$ $t$ gives:
 * $(2): \quad x' = C_1 m_1 e^{m_1 t} + C_2 m_2 e^{m_2 t}$

Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:

Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$:

Then we have:

Hence: