$$y'''-3y''+3y'-y=t^{-2} e^{t}$$
I already have the solution of the homogeneous equation solved, but I cannot solve for the general solution. I realize it is supposed to be in the form $u_{1}y_{1}+u_{2}y_{2}+u_{3}y_{3}$, but I do not know how to obtain the $u$'s.
Hint: observe that: $$ y''' - 3y'' +3y' -y = \frac{\mathrm d^3}{\mathrm d x^3}y - 3\frac{\mathrm d^2}{\mathrm d x^2}y + 3\frac{\mathrm d}{\mathrm dx}y -y \stackrel{\text{somewhat informally}}{===}(\frac{\mathrm d}{\mathrm d x} - 1)^3 y $$ Formally, we can write this as $$ \left(\frac{\mathrm d}{\mathrm d x} - 1\right) \underbrace{\left(\frac{\mathrm d}{\mathrm d x} - 1\right) \overbrace{\left(\frac{\mathrm d}{\mathrm d x} - 1\right) y}^{u}}_v $$ This can be solved by introducing additional variables: \begin{align} &\left(\frac{\mathrm d}{\mathrm d x} - 1\right) y = u\\ &\left(\frac{\mathrm d}{\mathrm d x} - 1\right) u = v\\ &\left(\frac{\mathrm d}{\mathrm d x} - 1\right) v = t^{-2}e^t. \end{align} Note that $$ \left(\frac{\mathrm d}{\mathrm d x} - 1\right)v = v' - v $$ and so on.
You could also find the parameters $u_i$ by plugging $u_i y_i$ into your equation and solving that for $u_i$ for every $i$. This is more general approach, but it seemingly needs more calculations. Also, those parameters ostensibly call for using the former approach.
Hint: For the homogeneous equation make the ansatz $$y_h=e^{\lambda t}$$
