When to multiply the assumed particular solution by $t$ (Method of undetermined coefficients) I'm given the DE : $$y''-y = 8te^{t}$$
First I solve the corresponding homogenous equation : 
$$r^2-1 = 0 \implies r_1= 1 \ r_2= -1$$
So The general solution of the corresponding homogenous equation looks like : $$y_c = c_1e^t + c_2e^{-t}$$
Now I want to find a particular solution so I assume $Y_1(t)$ solves the DE 
$$Y_1(t) = (At + B) e^t$$
Calculating we end up with :
$$Y_1''(t) = (2A+B)e^t + Ate^t$$
However then the DE takes a form : 
$$(2A+B)e^t = 8te^t$$
Which does not make sense. So I believe I should have multiplied the assumed particular solution by $t$ beforehand. However I know that I do this multiplication when the particular solution is in the homogenous equation. However it is not. So I am confused.
In General, I have a problem about when to multiply the assumed particular equation by t. For instance consider :
$$f(x,y,y',y'') = xe^xsinx$$
Where $f$ is linear in all its input variables. Now let us further assume the corresponding homogenous equation is in some form of this :
$$c_1e^x+c_2e^{-x}+ xc_3$$
Now How should I contruct the particular solution? Should I multiply the particular solution by $x?$ I don't think so since $sinx$ is not in the homogenous equation. However, If someone could give me an explanation on these two questions and make my mind clear about this issue I would be really glad. 
Thanks,
 A: Note that you missed something: $Be^t$ is in the homogeneous solution even though $Ate^t$ is not.
In general you need to multiply by an additional factor of $t^k$ if the coefficient in the exponential is a root of the characteristic polynomial of multiplicity $k$. In the first example, $k=1$ ($1$ is a single root of $x^2-1$).
The basic idea here is that if you have a linear differential operator with constant coefficients, it maps a polynomial of degree $d_1$ times an exponential to a polynomial of degree $d_2 \leq d_1$ times an exponential. But actually upon calculation you find that $d_2=d_1-k$, so if $k>0$ then your solution has a higher degree polynomial in it than the RHS of the equation does.
A: hint
The best way is to eliminate the exponential term by putting
$$y=ze^t $$
then
$$y'=(z'+z)e^t $$
$$y''=(z''+2z'+z)e^t $$
the equation becomes easier
$$z''+2z'=8t $$
you should look for a polynomial particular solution of the form $$z_p=at^2+bt=t (at+b) $$
with
$$2a+4at+2b=8t $$
$$\implies a=2\;\;,b=-2$$

$$\implies y_p=2t (t-1)e^t $$
  $$\implies y_c=c_1e^{-t}+c_2e^t+y_p$$

