Trick to find particular solution of $y'' - 2y' + y = te^t$ I have always been confused by this type of question. Given a differential equation like 
$$y'' - 2y' + y = te^t$$
The general solution is in form of $y_h + y_p$ where $y_h$ is the solution to $y'' - 2y' + y$, and $y_p$ is a particular solution. What I do not understand is that, is there any "trick" or "clever" method to find $y_p$?
For this particular question, the solution says

...the right hand side is the solution to the homogenous one, so we
  must guess a particular solution of the form $A t^3 e^t$...

Why we must guess this form? I'm sorry if this is a too-easy question. But since this is a multiple choice question, the only way I know would be to test each choice, which takes lots of time.
Further explanation: So for equation like $y'' - y =x$, I know we can do that by "inspection" because it's really simple, and it happens that $y=-x$ is a particular solution. But when the equation becomes more complicated, I'm kinda lost in terms of what to try?
 A: Let $D=\frac d{dt}$ be the differential operator. Then by the DE, we know
$$
(D-1)^2 y = te^t.
$$
Since $te^t$ is annihilated by $(D-1)^2$, we have
$$
(D-1)^4 y = (D-1)^2 (te^t) = 0.
$$
Thus, we can look for a particular solution from the equation:
$$
(D-1)^4 y = 0.
$$
Then, we can try $a t^3 e^t$, since $t^2e^t$ is annihilated by $(D-1)^3$. 
A: I think I would emphasize that the Jordan Form method for this problem says homogeneous is any combination of $e^t$ and $t e^t.$
Now, we know that, if we try some $P(t) e^t$ where $P$ is a polynomial, then the result of $y'' -2y'+y$ is some $Q(t) e^t$ with another polynomial. Writing it out, I got
$$  y'' -2y'+y = P''(t) e^t. $$
So, for the particular solution, we want $P''(t) = t.$ So we can take $P = t^3 / 6.$
A: Solving for the homogeneous equation: $y''-2y'+y=0$, let $z=y' \rightarrow z'=y''$ we then have the system of differential equations: $$\begin{cases} y'=z \\ z'=2z-y\end{cases}$$ we then have a system of the form $X_h'=AX_h$. 
Assuming you already know how to solve such systems we get that : $y_h(t)=(c_1 +tc_2)e^t$. Now for the general solution of the non homogeneous problem we need to find a particular solution. First notice that the solutions for the homogeneous system of equations are of the form: $X=Q(t)C$ for $Q(t)=\begin{pmatrix}
        e^t & te^t\\
         e^t & e^t +te^t \\ 
             \end{pmatrix} = \begin{pmatrix}
        y_1 & y_2\\
         y_1' & y_2' \\ 
             \end{pmatrix}$ , $Q(t)$ is the fundamental matrix and C a constant vector. 
Now to find the particular solution we can use the Method of Variation of Parameters: We look for a solution of the form : $X_P=Q(t)C(t)$, where now $C(t)=\begin{pmatrix}
        c_1(t) \\
         c_2(t) \\ 
             \end{pmatrix}$ is a non constant vector (hence the name of the method). So we want $X_p$ to be a solution of the non homogeneous system : $X_P=AX_P +b(t)$, where $b(t)=\begin{bmatrix}
        0  \\
         te^t  \\ 
             \end{bmatrix}$. This implies that $X_p'=Q'(t)C(t)+Q(t)C'(t)= AX_P +Q(t)C'(t)=AX_P +b(t) \Rightarrow Q(t)C'(t)=b(t)$ , from here you can find $C(t)= \int Q(t)^{-1}b(t)dt$.
Also there is a "short" way of calculating $C(t)$, using the Wronkian we have: $$c_1(t)= \int \frac  {det \begin{vmatrix}
        0 & y_2 \\
        te^t & y_2' \\
        \end{vmatrix}} {W(y_1, y_2)}dt$$
$$c_2(t)= \int \frac  {det \begin{vmatrix}
        y_1 & 0 \\
        y_1' & te^t \\
        \end{vmatrix}} {W(y_1, y_2)}dt$$ Using this method you will get $X_p=\begin{bmatrix}
        y_p(t)  \\
         y_p'(t)  \\ 
             \end{bmatrix}$, but that's fine because you only need the $y_p(t)$ component. 
I hope that helps!
A: If the right hand side is a sum of polynomial times exponential term, then the particular solution can be given as a similar sum of polynomial times exponential term, where the exponential terms stay the same. Given $p(t)e^{ct}$ then the term in the particular solution is $t^mq(t)e^{ct}$ where $m$ is the multiplicity of $c$ as a root of the characteristic polynomial of the homogeneous equation, $\deg(q)=\deg(p)$ and the coefficients of $q$ have to be determined.
In your first e, $c=1$ has multiplicity $m=2$ and $p(t)=t$ has degree $1$. Thus the trial solution is $t^2(A+Bt)e^t$.
In your second example $c=0$ is no root, thus $m=0$ and $p(x)=x$ has degree $1$, which gives $A+Bx$ as trial solution.
