State-Space Equation
Using nodal analysis, along with the obvious relationships between states in the parallel circuits, and also assuming ALL component values are unity (since I just want to show the method, rather than do a complete symbolic solution), I get the following set of equations:

$$\small x_1-x_4=x_3 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)$$
$$\small \dot x_2=x_3-x_2 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(2)$$
$$\small \dot x_1+\dot x_3=V_s-2x_1-x_2-x_3 \:\:\:\:\:(3)$$
$$\small \dot x_4-\dot x_3=x_2+x_3-2x_4 \:\:\:\:\:\:\:\:\:\:\:\:(4)$$

From (1):
$$\small \dot x_1=\dot x_3+\dot x_4 $$
Substitute into (3):
$$\small 2\dot x_3+\dot x_4=V_s-2x_1-x_2-x_3\:\:\:\:\:\:\:\:\:$$

Re-arranging (4):

$$\small -\dot x_3+\dot x_4=x_2+x_3-2x_4 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$

Solving simultaneously:

\$\hspace{50mm} \small \dot x_3=\left( V_s-2x_1-2x_2-2x_3+2x_4\right)/3\$

\$\hspace{50mm}\small \dot x_4=\left( V_s-2x_1+x_2+x_3-4x_4\right)/3\$

Also:

\$\hspace{50mm}\small \dot x_1=\left(2V_s -4x_1-x_2-x_3-x_4\right)/3\$

and equation (2) is:

\$\hspace{50mm}\small \dot x_2=x_3-x_2\$

These last four expressions give the state-space equation:

\begin{gather}
\small
 \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \\ \dot x_4\end{bmatrix}
 = 
 \frac{1}{3}
  \begin{bmatrix}
   -4 &amp;-1 &amp;-1 &amp;-1 \\
   0&amp;-3&amp;3&amp;0 \\-2&amp;-2&amp;-2&amp;2 \\-2&amp;1&amp;1&amp;-4
   \end{bmatrix}
\begin{bmatrix} \ x_1 \\ x_2 \\  x_3 \\ x_4\end{bmatrix} +
\begin{bmatrix} \ 2/3 \\ 0 \\  1/3 \\ 1/3\end{bmatrix}V_s
\end{gather}

with eigenvalues:

\$\hspace{50mm}\small \lambda_1=-5.42\$

\$\hspace{50mm}\small \lambda_2=-3.61+j2.3\$

\$\hspace{50mm}\small \lambda_3=-3.61-j2.3\$

\$\hspace{50mm}\small \lambda_4=-0.36\$