Find a general solution for the given

linear system using the elimination method of Section 5.2.

$\begin{array}{l}\frac{d^{3}x}{d{t}^3}-x+\frac{dy}{dt}+y=0\\ \frac{dx}{dt}-x+y=0\end{array}$

In the elimination method you either add or subtract the equations to get an equation in one variable. When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

We will do the following question on the basis of solving linear system using the elimination method;

$\begin{array}{l}\frac{d^{3}x}{d{t}^3}-x+\frac{dy}{dt}+y=0\\ \frac{dx}{dt}-x+y=0\\ \\ D^{3}x-x+Dy+y=0\\ Dx-x+y=0\\ (D^3-1)x+(D+1)y=0\\ (D-1)x+y=0\\ \\ (D-1)(D^3-1)x+(D-1)(D+1)y-(D^3-1)(D-1)x+(D^3-1)y=0\\ (D^2-D^3)y=0\\ \\ m^2-m^3=0\\ m^3(1-m)=0\\ \Rightarrow m=0,0,1\\ \\ y\left(t\right)=C_1{e}^{m_{1}t}+C_2{e}^{m_{2}t}+C_3{e}^{m_{3}t}\\ y\left(t\right)=(C_1+t{C}_2)+C_3{e}^t\\ \\ (D^3-1)x+(D+1)y-[(D-1)x+(D+1)y]=0\\ (D^3-D)x=0\\ \\ m^3-m=0\\ m(m^2-1)=0\\ \Rightarrow m=0,\pm 1\\ \\ x\left(t\right)=A_1{e}^{m_{1}t}+A_2{e}^{m_{2}t}+A_3{e}^{m_{3}t}\\ =A_1+A_2{e}^t+A_3{e}^{-t}\end{array}$

Hence, the final answer is:

$x\left(t\right)=A_1+A_2{e}^t+A_3{e}^{-t}$