Eigenvalues of 2 ODES with a constant term?

Question

I have the following system and would like to convert to state space and/or matrix from to find the eigenvalues, however the constant terms are not in my analysis.

For example; $$ \ddot{x}_1 = -50x_1 + 5\dot{x}_2 - 250x_2 + 20 \\ \ddot{x}_2 = 5x_2 - x_1 +30 $$ So let $$A=\pmatrix{0& 1&0& 0\\ -50& 0& 1& -250\\ 0& 0& 0& 1\\ -1& 0& 5& 0} $$

So now

$$[A][x_1 , \dot{x_1}, x_2, \dot{x_2}]^T$$

However I seem to be "ignoring" the constant terms in the equations? Am I ok to do this?

Answer 1

Writing $x=[x_1,\dot{x_1},x_2,\dot{x_2}]^T$, your equation can be written as $$ \dot x= Ax+b, $$ where $$ A=\begin{pmatrix} 0 & 1 &0 &0\\ -50 & 0 & -250 & 5\\ 0 & 0 & 0 & 1 \\ -1 & 0 & 5 & 0 \end{pmatrix} $$ and $b=[0,20,0,30]^T$. The homogeneous equation can be easily solved after you have computed the eigenvalues and eigenvectors of $A$ and it is given by $x_H(t)=e^{tA}x(0)$. As the nonhomogeneous term is constant, you can simply write the general solution as $x(t)=x_H(t)+bt.$

Answer 2

The first order system with constant coefficients $\dot x=Ax+b$ can be simplified and solved using the exponential of $A$ as integrating factor, $$ \frac{d}{dt}(e^{-At}x(t))=e^{-At}(\dot x(t)-Ax(t))=e^{-At}b. $$ To compute the matrix exponential, you need the eigen-decomposition of $A$, without any influence of the inhomogeneous part $b$.