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**HallsofIvy** Oh, dear, oh dear, oh, dear! I really dislike the Laplace Transform! It can be used in two diametrically opposite ways- to "look up" solutions to non-homogeneous equations or to assert a form of a solution in very abstract theorems. The first is more easily done in other ways and the second, undergraduates never see.

I would look at that equation and immediately write down it characteristic equation: $\displaystyle r^3- r^2- r+ 1= 0$. It's easy to see, by inspection, that r= 1 is a root of that and that $\displaystyle r^3- r^2- r+ 1= (r-1)(r^2- 1)= (r- 1)^(r+ 1)$ so that r= 1 is a double root and r= -1 is a root. That tells us that the general solution to the associated homogeneous equation is $\displaystyle y= Ae^t+ Bte^t+ Ce^{-t}$.

The "non-homogeneos" part is $\displaystyle 8te^{-t}$. Since $\displaystyle te^{-t}$ is already a solution to the associated homogeneous equation, we try a solution of the form $\displaystyle y= (At^2+ Bt)e^{-t}$. Then $\displaystyle y'= (At+ B)e^{-t}- (At^2+ Bt)e^{-t}= (-At^2+ (A- B)t+ B)e^{-t}$, $\displaystyle y''= (-2At+ (A- B)e^{-t})e^{-t}- (-At^2+ (A- B)t+ B)e^{-t}$$\displaystyle =(At^2+ (B- 3A)t+ A- 2B)e^{-t}$, and $\displaystyle y'''= (2At+ B- 3A)e^{-t}- (At^2+ (B- 3A)t+ A- 2B)e^{-t}= (-At^2+ (5A- B)t+ 3B- 5A)e^{-t}$.

Putting those into the differential equation will give you equations for A and B.