Position of Cart attached to Wall by Spring under Damping/Critically Damped/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:
 * $x = x_0 e^{-a t} \left({1 + a t}\right)$

Such a system is defined as being critically damped.

Proof
From Position of Cart attached to Wall by Spring under Damping: Critically Damped:
 * $x = C_1 e^{-a t} + C_2 t e^{-a t}$

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:
 * $(2): \quad x' = -a C_1 e^{-a t} + C_2 e^{-a t} - a C_2 t e^{-a t}$

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

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

Hence: