Answer: The required general solution is[tex]y(x)=Ae^{-x}+e^x(B\cos x+C\sin x).[/tex]Step-by-step explanation: We are given to find the general solution of the following differential equation :[tex]y^{\prime\prime\prime}-y^{\prime\prime}+2y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]Let y = y(x) and [tex]y=e^{mx}[/tex] be an auxiliary solution of equation (i).Then, we have[tex]y^\prime=me^{mx},~~~y^{\prime\prime}=m^2e^{mx},~~~y^{\prime\prime\prime}=m^3e^{mx}.[/tex]Substituting these values in equation (i), we have[tex]m^3e^{mx}-m^2e^{mx}+2e^{mx}=0\\\\\Rightarrow (m^3-m^2+2)e^{mx}=0\\\\\Rightarrow m^3-m^2+2=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2(m+1)-2m(m+1)+2(m+1)=0\\\\\Rightarrow (m+1)(m^2-2m+2)=0\\\\\Rightarrow m+1=0~~~~~\Rightarrow m=-1[/tex]and[tex]m^2-2m+2=0\\\\\Rightarrow (m^2-2m+1)+1=0\\\\\Rightarrow (m-1)^2=-1\\\\\Rightarrow m-1=\pm\sqrt{-1}\\\\\Rightarrow m=1\pm i~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{where }i^2=-1][/tex]So, we get[tex]m_1=-1,~~m_2=1+i,~~m_3=1-i.[/tex]Therefore, the general solution of the given equation is given by[tex]y(x)=Ae^{m_1x}+e^{1\times x}(B\cos 1x+C\sin 1x)}\\\\\Rightarrow y(x)=Ae^{-x}+e^x(B\cos x+C\sin x).[/tex]Thus, the required general solution is[tex]y(x)=Ae^{-x}+e^x(B\cos x+C\sin x).[/tex]