Jacobian Linearisation of Non Linear System Can any one please solve the below problem. It is related to Jacobian Linearisation of Non Linear System



I have only got till here

 A: You can try something like this: 
$$ \dfrac{dx_1}{dt}=f_1(u,x_1,x_2)$$
$$ \dfrac{dx_1}{dt}=f_2(u,x_1,x_2)$$
$$ \dfrac{d\Delta x_1}{dt}=(\dfrac{df_1}{du})_0\Delta u + (\dfrac{df_1}{d x_1})_0\Delta x_1 +(\dfrac{df_1}{dx_2})_0\Delta x_2 $$
$$ \dfrac{d\Delta x_2}{dt}=(\dfrac{df_2}{du})_0\Delta u + (\dfrac{df_2}{d x_1})_0\Delta x_1 +(\dfrac{df_2}{dx_2})_0\Delta x_2 $$
Notice that I use: $\delta_x=\Delta x$ notation...
Now calculate all derivations in linearization point for example some values of $(x_1)_0=\hat x_1$, so we have: 
$$ \dfrac{df_1}{du}=c_1\dfrac{dT}{du}|_{u=u_0} =b_1$$
$$ \dfrac{df_1}{dx_1}= -C_2x_2 |_{x_2=x_{20}} =a_{11}$$
$$ \dfrac{df_1}{dx_2}=-C_2x_1|_{x_1=x_{10}}=a_{12} $$
Further more, you have: 
$$ \dfrac{df_2}{du}=0 =b_2$$
$$ \dfrac{df_2}{dx_1}= \frac{c_3}{J_e}=a_{21} $$
$$ \dfrac{df_2}{dx_2}=\frac{1}{J_e} (-0,106-2c_4x_2)|_{x_1=x_{10},x_2=x_{20}}=a_{22} $$
Now you have a set of linear equations that you can use to calculate state variables, equilibrium points, and so on...
$$\left[\begin{matrix} \Delta\dot{x}_1 \\ \Delta\dot{x}_2 \end{matrix}\right]=\left[\begin{matrix} a_{11} & a_{12} &\\ a_{21} & a_{22}\end{matrix}\right]\left[\begin{matrix}\Delta x_1 \\\Delta x_2 \end{matrix}\right]+\left[\begin{matrix}b_1 \\ b_2 \end{matrix}\right]\Delta u$$
Also if you consider $v$ tp be the output variable, you get: 
$$ v=y=\left[\begin{matrix} 1 & 0\end{matrix}\right]\left[\begin{matrix}\Delta x_1 \\ \Delta x_2\end{matrix}\right]+0\Delta u$$
