Time responses (position and speed) of system This is a basic question regarding state space representation and differential equations.
I want to find the time response of states $x_{1} = x$ and $x_{2} = \dot{x}$ of the following system:
$$
m\ddot{x} + k\dot{x} = 0
$$
The state space is
$$
\begin{pmatrix}
  \dot{x_{1}} \\
  \dot{x_{2}} 
 \end{pmatrix}=
\begin{pmatrix}
  0 & 1 \\
  -\frac{k}{m} & 0 
 \end{pmatrix}
\begin{pmatrix}
  x_{1} \\
  x_{2} 
 \end{pmatrix}
$$
What is the fastet and simplest way to calculate the time response of the two states $x_{1}(t)$ and $x_{2}(t)$?


*

*Just solve the differential equation for $x$ which equals $x_{1}(t)$ and then get $x_{2}(t)$ using the derivative $\dot{x_{1}}(t) = x_{2}(t)$?

*Using Laplace, transform the differential equation, solve for $X(s)$ and back transform to the time domain to get $x_{1}(t)$. Then again get $x_{2}(t)$ using the derivative?

*Or is there another possibility to get the time responses of the states directly from the state space representation?

*Or is there even another approach?

 A: fibonatic above solved your problem by inspection; I'm just  detailing why it was so easy: essentially, diagonalization of the matrix.
For starters, define $k/m\equiv \omega^2$, and you are expected to reflexively recognize the 2x2 matrix involved has paired eigenvalues $\pm i\omega$ with celebrated (un-normalized) eigenvectors $v_1 (1,i\omega)/2$ and $v_2 (1,-i\omega)/2$, respectively; that is,  $x_1=v_1+v_2$ and $x_2=i\omega(v_1-v_2)$. So your system collapses to two decoupled differential equations,
$$
\begin{pmatrix}
  \dot{v_{1}} \\
  \dot{v_{2}} 
 \end{pmatrix}=
\begin{pmatrix}
  i\omega & 0 \\
  0 & -i\omega 
 \end{pmatrix}
\begin{pmatrix}
  v_{1} \\
  v_{2} 
 \end{pmatrix}
$$
The solutions are obviously $v_1=C_1 e^{i\omega t}$ and $v_2=C_2 e^{-i\omega t}$, and you never had to solve a 2nd order differential equation, options 1 and 2---the central idea behind the exercise. 
This, then, amounts to your option 3:  $x_1=C_1 e^{i\omega t}+ C_2 e^{-i\omega t}$, and the velocity, $x_2=i\omega  (C_1 e^{i\omega t}- C_2 e^{-i\omega t})$.
