I wonder if there is any way how to made MATLAB solve the system of equation as this:
$ s\cdot x_1 = \frac {R_1L_2}{M^2-L_1L_2} x_1+ \frac {MR_2}{M^2-L_1L_2} x_2+\frac {M}{M^2-L_1L_2} x_3 - \frac {L_2}{M^2-L_1L_2} u + \frac{R_2M(M-L_1L_2-1)}{R1L2(L_1L_2-M^2)}+\frac{10}{R_1}$
$s \cdot x_2 = \frac{R_1M^2}{M(M^2-L_1L_2)} x_1 + \frac{L_1R_2}{M^2-L_1L_2}x_2 + \frac{L_1}{M^2-L_1L_2}x_3 - \frac{M}{M^2-L_1L_2} u$
$s \cdot x_3 = \frac1{C_1} x_2 + \frac{R_2}{L_1L_2-M^2}$
with the respect to $x_1$, $x_2$ and $x_3$. Others letters are parameters.
When I write intructions like this:
syms M L1 L2 R1 R2 C x1 x2 x3 C u
x=['x1', 'x2', 'x3'];
solve(['(M*x3)/(M^2 - L1*L2) - (L2*u)/(M^2 - L1*L2) + (L2*R1*x1)/(M^2 - L1*L2) + (M*R2*x2)/(M^2 - L1*L2)=s*x1-((R2*M(M-L1*L2-1))/(R1*L2(L1*L2-M^2))+10/R1)', '(L1*x3)/(M^2 - L1*L2) - (M*u)/(M^2 - L1*L2) + (R1*x1)/(M^2 - L1*L2) + (L1*R2*x2)/(M^2 - L1*L2)=s*x2', 'x2/C1=s*x3-R2/(L1*L2-M^2)'], x, 'VectorFormat')
The result is
Warning: 1 equations in 2 variables. New variables might be introduced.
In C:\Program Files\MATLAB\R2012b\toolbox\symbolic\symbolic\symengine.p>symengine at 54 In mupadengine.mupadengine>mupadengine.evalin at 97 In mupadengine.mupadengine>mupadengine.feval at 150 In solve at 160 In lap_matr at 5 Warning: Explicit solution could not be found. In solve at 169 In lap_matr at 5
ans =
[ empty sym ]
Is there any way and which one?
I do not use Matlab but let us speak about your specific system of equations.
You could notice that, from the third equation, you can extract $x_3$ as a function of $x_2$ $$x_3=\frac{{C_1} {R_2}+({L_1} {L_2}-M^2) {x_2}}{{C_1} s \left({L_1} {L_2}-M^2\right)}$$ If you plug what you obtain for $x_3$ into the second equation, you can, in a very similar manner, now extract $x_2$ as a function of $x_1$. Putting all together into the first equation, it becomes a linear equation in $x_1$. Solve it and go backward if you want.
I suppose that you could do that without any problem with Matlab : solve the third equation for $x_3$, solve the second equation for $x_2$ knowing $x_3$, solve the first equation for $x_1$ knowing $x_2$ and $x_3$.
I totally agree that this does not solve your problem. However, such an approach will avoid giving you the monstreous formulas you would get solving directly for $x_1$, $x_2$ and $x_3$.
Added later
You would get monstreous formulas because of your coefficients. What I suggest is that you first rewrite your equations as (for example) $$x_1 = a_1 x_1 + a_2 x_2+a_3 x_3 + a_4$$ $$x_2 = b_1 x_1 + b_2 x_2+b_3 x_3 + b_4$$ $$x_3 = c_1 x_2 + c_2$$ Now, proceed as I suggested and, hoping that I did not make any mistake, $$x_3 = c_1 x_2 + c_2$$ $$x_2=-\frac{{b_1} {x_1}+{b_3} {c_2}+{b_4}}{{b_2}+{b_3} {c_1}-1}$$ $$x_1=-\frac{{a_2} ({b_3} {c_2}+{b_4})+{a_3} (-{b_2} {c_2}+{b_4} {c_1}+{c2})-{a_4} ({b_2}+{b_3} {c_1}-1)}{-{a_1} ({b_2}+{b_3} {c_1}-1)+{a_2} {b_1}+{a_3} {b_1} {c_1}+{b_2}+{b_3} {c_1}-1}$$
Now, you could replace or not the coefficients you so introduced by their expressions and try to simplify. But I do not think this could be a good idea. Just think how you would do it in a program.