Method of undetermined coefficients for ODEs to. find particular solutions I have hit a conceptual barrier. So let's say we had the following ODE:
$$\frac{d^{4}u}{dt^{4}} - 16u = te^{2t}.$$
The general solution of the associated homogeneous equation is:
$$u_h(t) = c_{1}e^{-2t} + c_{2}e^{2t} + c_{3}\cos(2t) + c_{4}\sin(2t)$$
Now to guess the particular solution, I was following the reasoning presented in class:
We try to guess $e^{2t}$ but it is part of the homogeneous solution, so we guess $te^{2t}$ but since this is the RHS, we go one power high, and our guess is $At^{2}e^{2t} + Bt^{}e^{2t}$.
I really just do not understand the reasoning behind this. Why do we care what the RHS is to increase powers? Why do we go one power higher than the RHS? Also how are these "guesses" being made?

A simple way of seeing what is behind these guesses is to use the annihilator polynomial method. You want to solve an equation of the form $P(D) y = f(t)$, where $P$ is a polinomial in the differentiation operator. (for instance, the dif. eq. $y'''+y''-y = e^t$ would be written as $(D^3+D^2-1)y = e^t$). If you are able to find a polynomial $Q(D)$ such that $Q(D) f(t)=0$, the original equation can be reduced to
$$
P(D) y = f(t) \Rightarrow Q(D)P(D) y = 0.
$$
So, you reduce the original equation to an homogeneous equation with higher degree (of coarse this is not possible for all $f$, just for the ones that can be a solution to an homogeneous equation).
The solution to this higher degree but homogeneous problem is obtained and decomposed in two parts: i. the general solution to the original homogeneous equation; ii. the rest.
The "rest" is what you should use as a particular solution.
