$1\mathrm{.}1$ Laplace Transfomation solve

$1\mathrm{.}2$ Laplace transformaiton Matrix solve

$1\mathrm{.}3$ Inverse Laplace Transfomration solve

For

$F{x}_1=m_s{x}_1^{}$

$m_s{x}_1^{}=-K_1(x_1-x_2)-C_1((x_1^{})-(x_2^{}))$

$m_s{x}_1^{}+K_1(x_1-x_2)+C_1((x_1^{})-(x_2^{}))=0$

Fo$2\mathrm{.}1.$Modeling the mechanical system :

Figure $8-$Model of Car suspension systemr $M_u$

$F{x}_2=m_u{x}_2^{}$

$m_u{x}_2^{}=K_1(x_1-x_2)+C_1((x_1^{})-(x_2^{}))--K_2(x_2)-C_2((x_2^{}))$

$m_u{x}_2^{}-K_1(x_1-x_2)-C_1((x_1^{})-(x_2^{}))+K_2(x_2)+C_2((x_2^{}))=0$

Equation $1$:

$m_s{x}_1^{}+C_1((x_1^{})-(x_2^{}))+K_1(x_1-x_2)=0$

Equation $2$:

$m_u{x}_2^{}-C_1((x_1^{})-(x_2^{}))+C_2((x_2^{}))-K_1(x_1-x_2)+K_2(x_2)=0$ Ignore Z the base is constan