[tex]y''-y'-2y=-e^t[/tex]
The corresponding homogeneous ODE is
[tex]y''-y'-2y=0[/tex]
with characteristic equation
[tex]r^2-r-2=(r-2)(r+1)=0[/tex]
with roots at [tex]r=2[/tex] and [tex]r=-1[/tex], so the characteristic solution is
[tex]y_c=C_1e^{2t}+C_2e^{-t}[/tex]
For the non-homogeneous ODE, assume a particular solution of the form
[tex]y_p=ae^t[/tex]
[tex]\implies{y_p}'=ae^t[/tex]
[tex]\implies{y_p}''=ae^t[/tex]
Substituting [tex]y_p[/tex] and its derivatives into the ODE gives
[tex]ae^t-ae^t-2ae^t=-e^t[/tex]
[tex]-2ae^t=-e^t[/tex]
[tex]\implies-2a=-1\implies a=\dfrac12[/tex]
Then the ODE has the general solution
[tex]\boxed{y(t)=C_1e^{2t}++C_2e^{-t}+\dfrac12e^t}[/tex]
