The canonical 2nd-order LCCODE

$$y''(t)+2\zeta\omega_n y(t)+\omega_n^2 y(t)=x(t)$$

caan be converted into a 1st-order ODE system containing two 1st-order LCCODEs (see here):

$${\bf y}'=\left[\begin{array}{c}y_1' y_2'\end{array}\right]... ...t[\begin{array}{c}0 x(t)\end{array}\right] ={\bf Ay}+{\bf x}$$

The characteristic equation of the coefficient matrix ${\bf A}$ is

$$\det(\lambda{\bf I}-{\bf A}) =\det\left[\begin{array}{cc}\la... ...nd{array}\right] =\lambda^2+2\zeta\omega_n\lambda+\omega_n^2=0$$

Solving which we get the two eigenvalues and the eigenvalue matrix:

$$\lambda_{1,2}=(-\zeta\pm\sqrt{\zeta^2-1})\omega_n \;\;\;\;\;... ...egin{array}{cc}\lambda_1 & 0 0 & \lambda_2\end{array}\right]$$

Note that

$$\lambda_1\lambda_2=\omega_n^2,\;\;\;\;\;\;\;\;\;\; \lambda_2-\lambda_1=2\omega_n\sqrt{\zeta^2-1}$$

The eigenvector matrix and its inverse are:

$${\bf V}=\left[\begin{array}{cc}1 & 1 \lambda_1 & \lambda_2... ...in{array}{cc}\lambda_2 & -1 -\lambda_1 & 1\end{array}\right]$$

and we can verify thatat ${\bf A}={\bf V\Lambda V}^{-1}$.

• Homogeneous solution $x(t)=0$:

  Assuming $y_1(0)=y(0)=y_0$ and $y_2(0)=y'(0)=0$, we get the homogeneous solution:

  
  

  $${\bf y}_h(t) = e^{{\bf A}t}{\bf y}(0)={\bf V}e^{{\bf\Lambda}t}{\bf V}^{-1}{\bf y... ...ambda_1&1\end{array}\right] \left[\begin{array}{c} y_0 0\end{array}\right]$$

  $$= \frac{1}{\lambda_2-\lambda_1} \left[\begin{array}{cc} \lambda_2e... ...ambda_1t}\end{array}\right] \left[\begin{array}{c} y_0 0\end{array}\right]$$

  $$= \frac{y_0}{\lambda_2-\lambda_1} \left[\begin{array}{c} \lambda_2... ..._2t}\\ \lambda_1\lambda_2(e^{\lambda_1t}-e^{\lambda_2t}) \end{array}\right]$$

  
  
  we get the same result as shown above:
  
  $$y(t)=y_1(t)=\frac{y_0}{\lambda_2-\lambda_1}(\lambda_2e^{\lambda_1t}-\lambda_1e^{\lambda_2t})$$
  
  

• Nonhomogeneous sollution $x(t)=1$, i.e., ${\bf x}=[0,\;1]^T$:

  Assuming zero-initial condition $y_1(0)=y_2(0)=0$, i.e., ${\bf y}={\bf0}$, the non-homogeneous solution is:
  

  $${\bf y}(t)=e^{{\bf A}t}{\bf y}(0)+e^{{\bf A}t}\int_0^t e^{-... ...tau =e^{{\bf A}t}\int_0^t e^{-{\bf A}\tau} d\tau \;[0,\;1]^T$$
  
  

  
  

  $$y(t)=y_1(t)=[1,\;0]\;{\bf y}(t)=[1,\;0]\; e^{{\bf A}t}\int_0^t e^{-{\bf A}\tau} d\tau\; [0,\;1]^T$$
  
  
  where
  
  $$\int_0^t e^{-{\bf A}\tau} d\tau=-{\bf A}^{-1} e^{-{\bf A}\tau}\bigg\vert _0^t ={\bf A}^{-1}( {\bf I}-e^{-{\bf A}t} )$$
  
  
  Substituting
  
  $${\bf A}^{-1}=({\bf V\Lambda V}^{-1})^{-1}={\bf V\Lambda}^{-... ...\;\;\;\; e^{-{\bf A}t}={\bf V}e^{-{\bf\Lambda}t}{\bf V}^{-1}$$
  
  
  into the function, we get
  

  $$y(t) = [1,\;0]\;\left(e^{{\bf A}t}\int_0^t e^{-{\bf A}\tau} d\tau\right)... ...}\right) \left({\bf I}-{\bf V}e^{-{\bf\Lambda}t}{\bf V}^{-1}\right)\;[0,\;1]^T$$

  $$= [1,\;0]\;{\bf V}e^{{\bf\Lambda}t} {\bf\Lambda}^{-1}{\bf V}^{-1} ... ...ambda}t}{\bf\Lambda}^{-1}({\bf I}-e^{-{\bf\Lambda}t} ){\bf V}^{-1} \;[0,\;1]^T$$

  $$= [1,\;0]\;\left[\begin{array}{cc}1&1 \lambda_1&\lambda_2\end{ar... ... -\lambda_1&1\end{array}\right] \left[\begin{array}{c}0 1\end{array}\right]$$

  $$= \left[e^{\lambda_1t},\;e^{\lambda_2t}\right] \left[\begin{array}... ...ac{\lambda_2e^{\lambda_1t}-\lambda_1e^{\lambda_2t}}{\lambda_2-\lambda_1}\right)$$