I have the question to solve the ODE $$y^{(9)}-6y^{(8)}+9y^{(7)}=e^t+3e^{3t}$$
The part that I struggle with is I have $y=Ae^t+t^2Be^{(3t)}$, obviously I don't want to differentiate these 9 times, so my mark scheme says use the shifting lemma, but I don't really understand it.
My characteristic polynomial is $p^7(p-3)^2$ and the shifting lemma says $L(p)t^2Be^{(3t)}=e^{(3t)}(p+3)^7p^2t^2B=3e^{(3t)}$
This bit I follow, as it's multiplying my $t^2Be^{(3t)}$ by the characteristic polynomial but with +3 (Assuming this is because of the 3t?) But on the next line of working it says
$e^3t(p+3)^72B=3e^{(3t)}$
And I have absolutely no idea where the 2 in front of the B came from, all help would be appreciated
$$y^{(9)}-6y^{(8)}+9y^{(7)}=e^t+3e^{3t}$$ Using operator $D=\frac {dy}{dt}$ we can write it as $$D^7(D^2-6D+9)y=e^t+3e^{3t}$$ $$ \implies \begin{cases} D^7(D^2-6D+9)y=e^t \\ D^7(D^2-6D+9)y=3e^{3t} \end{cases} $$ The $e^t$ part is easy lets focus on the $e^{3t}$ part
$$D^7(D^2-6D+9)y=3e^{3t}$$ using $y=(Bt^2e^{3t})$ $$D^7(D-3)^2(Bt^2e^{3t})=3e^{3t}$$ Apply shifting's lemma
$$e^{3t}(D+3)^7(D)^2t^2B=3e^{3t}$$ And $D^2t^2$ means simply diferentiating twice $t^2$ $$\implies D^2t^2B=\frac {d^2t^2B}{dt^2}=2B$$ Thats why you have that 2 $$e^{3t}(D+3)^72B=3e^{3t}$$ You only need to evaluate the term thats independant of D in $(D+3)^7$ Which is by binomial theorem $C=1$ $$e^{3t}(D^7+c_1D^6+.....+c_6D+3^7)2B=3e^{3t}$$ $$e^{3t}2B \times 3^7=3e^{3t}$$ $$\implies B=\frac 1 {2\times 3^6}$$
For the A coeeficient in $Ae^t$ $$y^{(9)}-6y^{(8)}+9y^{(7)}=e^t$$ $$Ae^t-6Ae^t+9Ae^t=e^t$$ $$A-6A+9A=1 \implies A=\frac 14$$ Therefore $$\boxed{y_p=\frac14e^t+\frac 1{2 \times3^6}t^2e^{3t}}$$