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: 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.
A: 
Rule to find the particular solutions: To find the particular solutions of a differential equations with constant coefficient of the form
$$f\left(y^{(n)},y^{(n-1)},\cdots,y'',y',y\right)=e^{αx}~G(x)~\tag1$$
where $~G~$ is a polynomial of $~x~.$ Now
$(a)~~$If $~e^{αx}~$ isn’t a solution of the complementary equation$$f\left(y^{(n)},y^{(n-1)},\cdots,y'',y',y\right)=0~,\tag2$$ then the particular solution of $(1)$ is of the form $~y_p=e^{αx}~Q(x)~,$ where $~Q~$ is a polynomial of the same degree as $~G~$.
$(b)~~$If $~e^{αx}~$ is a solution of equation $(2)$ but $~xe^{αx}~$ is not, then $~y_p=xe^{αx}Q(x)~,$ where $~Q~$ is a polynomial of the same degree as $~G~.$
$(c)~~$If both $~e^{αx}~$ and $~xe^{αx}~$ are solutions of equation $(2)$, then $~y_p=x^2e^{αx}Q(x)~,$ where $~Q~$ is a polynomial of the same degree as $~G~.$
and so on.


Here complementary function is $$c_{1}e^{-2t} + c_{2}e^{2t} + c_{3}\cos(2t) + c_{4}\sin(2t)~.$$
Clearly here $~e^{2t}~$ is in the complementary function but $~te^{2t}~$ is not, i.e., it is a similar case that of option $(b)$. Therefore the particular integral is  $u_p=te^{2t}Q(t)=at^2e^{2t} +b te^{2t}~,$ as here $~G(t)=t~$ so take $Q(t)=at+b~.$
Now $~u_p~$ must satisfy $~(D^4-16)u=te^{2t}~,$ so putting the value of $~u_p~$ in the equation we have $~a=\frac{1}{64}~$ and $~b=-\frac{3}{128}~.$
Therefore the complete solution of the given differential equation is $$u(t)=c_{1}e^{-2t} + c_{2}e^{2t} + c_{3}\cos(2t) + c_{4}\sin(2t) + \frac{1}{64}~t^2e^{2t} -~\frac{3}{128}~te^{2t}~.$$

For more such rule you must see the note The Method of Undetermined Coefficients.
