I tried to solve this ode, but the solution seems to be very strange.
solve: $y'''-3y'+2y=24e^{t}$ given that $y''(0)=y'(0)=y(0)=0$.
$s^{3}Ly-3sLy+2Ly=\frac{24}{s-1}\\Ly(s^{3}-3s+2)=\frac{24}{s-1}\\Ly=\frac{24}{(s-1)^{3}(s+2)}$
partial fractions: $\frac{24}{(s-1)^{3}(s+2)}=\frac{a}{s-1}+\frac{b}{(s-1)^{2}}+\frac{c}{(s-1)^{3}}+\frac{\text{d}}{s+2}$ so: $\frac{24}{(s-1)^{3}(s+2)}=\frac{8/9}{s-1}+\frac{-8/3}{(s-1)^{2}}+\frac{8}{(s-1)^{3}}+\frac{\text{-8/9}}{s+2}$ meaning: $y=\frac{8}{9}e^{t}-\frac{8}{3}te^{t}+8t^{2}e^{t}-\frac{8}{9}e^{-2t}$
edit found a mistake: $\frac{24}{(s-1)^{2}(s+2)}=\frac{a}{s-1}+\frac{b}{(s-1)^{2}}+\frac{\text{d}}{s+2}\\24=a(s+2)(s-1)+b(s+2)+d(s-1)^{2}\\24=a(-2+s+s^{2})+b(s+2)+d(1-2s+s^{2})\\24=-2a+2b+d\\0=a+b-2d\\0=a+d$
therefore: $\frac{24}{(s-1)^{2}(s+2)}=\frac{-8/3}{s-1}+\frac{8}{(s-1)^{2}}+\frac{\text{8/3}}{s+2}\\y=-\frac{8}{3}e^{t}+8te^{t}+8t^{2}e^{t}+\frac{8}{3}e^{-2t}$