Differential equation, reduce order with three variables. I have this problem i never seen before. Most of the time when you are reducing orders you just have to set z1' = x1, z1'' = x1' and so on. But here i have a wild h' and h. How do i approach this?
Problem
Edit
\begin{align}
m_1z_1''&+c_1(z_1'-z_2')+k_1(z_1-z_2)=0
\\
m_2z_2''&+c_1(z_2'-z_1')+c_2(z_2'-h')+k_1(z_2-z_1)+k_2(z_2-h)=0
\end{align}
I have solved it this far.
\begin{align}
z_1'&=z_3  \\
z_2'&=z_4  \\
z_3'&=(-c_1(z_3-z_4)-k_1(z_1-z_2))/m_1  \\
z_4'&=(-c_1(z_4-z_3)-c_2(z_4-h')-k_1(z_2-z_1)-k_2(z_2-h))/m_2
\end{align}
 A: Assuming that $h$ is some input, possibly given only as data series that has to be interpolated to be a continuous function, the computation of the derivative $\dot h$ has a rather large uncertainty relative to the uncertainty of the interpolation itself. So one strategy is to avoid using $\dot h$ in the first order system.
This can be achieved by setting $w_2=\dot z_2-\frac{c_2}{m_2}h$, and then using
$$
\dot z_2=w_2+\frac{c_2}{m_2}h
$$
as one of the differential equations and to replace $\dot z_2$ everywhere.
Then the $\dot h$ in the second order equation gets absorbed into $\dot w_2$,
$$
m_2\dot w_2+c_1(w_2+\tfrac{c_2}{m_2}h−w_1)+c_2(w_2+\tfrac{c_2}{m_2}h)+k_1(z_2−z_1)+k_2(z_2−h)=0
$$
would then be the reduced second equation in the first-order system, using $\dot z_1=w_1$ and $w_1$, $w_2$ as names instead of the also possible $z_3,z_4$. It no longer contains a derivative of $h$, however $h$ will also occur now in the transformed first equation.
This idea of removing the derivatives of inputs was also explored in

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