How to obtain the state space representation of a computed torque control system?

First, isolate the second order from the other terms:

$$ a , ddot{x}_1 + b , ddot{x}_2 =- c ,dot{x_1} - d ,dot{x_2} - e + u_1 \ f , ddot{x}_1 + g , ddot{x}_2 = - h ,dot{x_1} - i ,dot{x_2} - j + u_2 $$

Then, put it in a matrix form:

$$ left[begin{matrix}a&b\f&gend{matrix}right]left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right]=left[begin{matrix}- c ,dot{x_1} - d ,dot{x_2} - e + u_1\- h ,dot{x_1} - i ,dot{x_2} - j + u_2end{matrix}right]. $$

Supposing $left[begin{smallmatrix}a&b\f&gend{smallmatrix}right]$ is invertible, we can left multiply both sides by $left[begin{smallmatrix}a&b\f&gend{smallmatrix}right]^{-1}$:

$$ left[begin{matrix}a&b\f&gend{matrix}right]^{-1}left[begin{matrix}a&b\f&gend{matrix}right]left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right]=left[begin{matrix}a&b\f&gend{matrix}right]^{-1}left[begin{matrix}- c ,dot{x_1} - d ,dot{x_2} - e + u_1\- h ,dot{x_1} - i ,dot{x_2} - j + u_2end{matrix}right]. $$

Thus, we get: $$ left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right] = {1over ag-bf} left[begin{matrix}g&-b\-f&aend{matrix}right]left[begin{matrix}- c ,dot{x_1} - d ,dot{x_2} - e + u_1\- h ,dot{x_1} - i ,dot{x_2} - j + u_2end{matrix}right], $$

$$ left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right] = {1over ag-bf} left[begin{matrix}(bh-gc)dot{x_1}+(bi-gd)dot{x_2}+(bj-je)\ (fc-ah)dot{x_1}+(fd-ai)dot{x_2} + (fe-aj)end{matrix}right] +{1over ag-bf} left[begin{matrix} g&-b\ -f&a end{matrix}right] left[begin{matrix} u_1\ u_2 end{matrix}right]. $$

Using auxiliary variables: $$ left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right] = left[begin{matrix}alphadot{x_1}+betadot{x_2}+gamma\ deltadot{x_1}+epsilondot{x_2} + zetaend{matrix}right] + left[begin{matrix} eta&theta\ iota&kappa end{matrix}right] left[begin{matrix} u_1\ u_2 end{matrix}right]. $$

Here, $gamma$ and $zeta$ can represent either uncontrolled inputs or disturbances.

As uncontrolled inputs: $$ left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right] = left[begin{matrix}alphadot{x_1}+betadot{x_2}\ deltadot{x_1}+epsilondot{x_2}end{matrix}right] + left[begin{matrix} eta&theta&gamma\ iota&kappa&zeta end{matrix}right] left[begin{matrix} u_1\ u_2\ 1 end{matrix}right]. $$

As disturbances: $$ left[begin{matrix}ddot{x_1}\ddot{x_2}end{matrix}right] = left[begin{matrix}alphadot{x_1}+betadot{x_2}\ deltadot{x_1}+epsilondot{x_2}end{matrix}right] + left[begin{matrix} eta&theta\ iota&kappa end{matrix}right] left[begin{matrix} u_1\ u_2 end{matrix}right]+ left[begin{matrix} gamma\ zeta end{matrix}right]. $$

Creating the state $s=[s_1;s_2;s_3;s_4]^T=[x_1;dot{x_1};x_2;dot{x_2}]^T$:

$$dot{s} = dot{left[begin{matrix}s_1\ s_2\ s_3\ s_4end{matrix}right]} = left[begin{matrix} 0&1&0&0\ 0&alpha&0&beta\ 0&0&0&1\ 0&delta&0&epsilon end{matrix}right] left[begin{matrix}s_1\ s_2\ s_3\ s_4end{matrix}right] + left[begin{matrix} 0&0\ eta&theta\ 0&0\ iota&kappa end{matrix}right] left[begin{matrix} u_1\ u_2 end{matrix}right] + left[begin{matrix} 0\ gamma\ 0\ zeta end{matrix}right]. $$

Depending on your system, you can design a controller that rejects those disturbances; the presence of integrators in the system can help remove steady-state errors.

$$ begin{align} a , ddot{x}_1 + b , ddot{x}_2 + c ,dot{x_1} + d ,dot{x_2} + e &= u_1 tag{1} \ f , ddot{x}_1 + g , ddot{x}_2 + h ,dot{x_1} + i ,dot{x_2} + j &= u_2 tag{2} end{align} $$ Let's write Eq(1) without $ddot{x}_2$, hence: $$ begin{align} a , ddot{x}_1 + b , Big[frac{u_2-f , ddot{x}_1 - h ,dot{x}_1 - i ,dot{x}_2 - j}{g}Big] + c ,dot{x}_1 + d ,dot{x}_2 + e &= u_1 \ left[a-frac{bf}{g} right] , ddot{x}_1 + , frac{b}{g}u_2 - frac{bh}{g} ,dot{x}_1 - frac{bi}{g} ,dot{x}_2 - frac{b}{g}j + c ,dot{x}_1 + d ,dot{x}_2 + e &= u_1 \ % left[a-frac{bf}{g} right] , ddot{x}_1 = u_1 - , frac{b}{g}u_2 + frac{bh}{g} ,dot{x}_1 + frac{bi}{g} ,dot{x}_2 + frac{b}{g}j - c ,dot{x}_1 - d ,dot{x}_2 - &e \ % ddot{x}_1 = left[frac{g}{(ag-bf)} right] left[u_1 - , frac{b}{g}u_2 + frac{bh}{g} ,dot{x}_1 + frac{bi}{g} ,dot{x}_2 + frac{b}{g}j - c ,dot{x}_1 - d ,dot{x}_2 - eright] tag{3} end{align} $$ Now we do same procedure which is eliminating $ddot{x}_1$ from Eq(2) hence, $$ begin{align} f , Big[frac{u_1- b , ddot{x}_2 - c ,dot{x_1} - d ,dot{x_2} - e}{a}Big] + g , ddot{x}_2 + h ,dot{x_1} + i ,dot{x_2} + j &= u_2 \ % Big[ g - frac{fb}{a}Big] ddot{x}_2 + frac{f}{a} u_1 - frac{fc}{a} ,dot{x_1} - frac{fd}{a} ,dot{x_2} - frac{fe}{a} + h ,dot{x_1} + i ,dot{x_2} + j &= u_2 \ % Big[ g - frac{fb}{a}Big] ddot{x}_2 = u_2 - frac{f}{a} u_1 + frac{fc}{a} ,dot{x_1} + frac{fd}{a} ,dot{x_2} + frac{fe}{a} - h ,dot{x_1} - i ,dot{x_2} - j \ % ddot{x}_2 = left[frac{a}{(ag-bf)} right] left[ u_2 - frac{f}{a} u_1 + frac{fc}{a} ,dot{x_1} + frac{fd}{a} ,dot{x_2} + frac{fe}{a} - h ,dot{x_1} - i ,dot{x_2} - j right] tag{4} end{align} $$ Now let $y_1=x_1, y_2=dot{x}_1, y_3=x_2, y_4=dot{x}_2$, hence: $$ begin{align} dot{y}_1 &= y_2 \ dot{y}_2 &= left[frac{g}{(ag-bf)} right] left[u_1 - , frac{b}{g}u_2 + Big[frac{bh-cg}{g}Big] ,y_2 + Big[frac{bi-dg}{g} Big] ,y_4 + frac{bj-eg}{g}right] \ dot{y}_3 &= y_4 \ dot{y}_4 &= left[frac{a}{(ag-bf)} right] left[ u_2 - frac{f}{a} u_1 + Big[frac{fc-ha}{a} ,Big] y_2 + Big[frac{fd-ia}{a}Big] ,y_4 + frac{fe-aj}{a} right] end{align} $$

where $agneq bf$. You can proceed from here.

Before the question can be answered with mathematical terms, a bit meta knowledge may help to understand the thought system. The goal of creating a state space formula for robotics problems is a typical application of the matlab software. The students are educated how to use a commercial software package for describing real world problems.

With this pre-knowledge in mind it is much easier to identify the correct answer. It was given in the internet already under the URL https://in.mathworks.com/help/control/getstart/linear-lti-models.html The website explains, how to use the famous mathematical software for creating a state space model for a torgue control. What is provided too, is how to create the differential equations and transform them into the matrix notation.