Particular solution to a 3rd order ode (method of undetermined coefficients not working) I am trying to get the particular solution to the equation - 
$y''' + 4y'' + 5y' + 2y = e^{-t} $
We are taught the method of undetermined coefficients to solve such equations. However, one of the solutions of the homogenous equation is of the form of the particular equation (so when I substitute it, I get LHS$ = 0$, while RHS is not $0$ ). 
Please give me a hint on how should I proceed to get the particular solution.
[This question is a part of the tricky questions set given to us. The actual differential equation is a bit more complicated, but I have reduced it to the point I have got stuck in.]
 A: Since $P(D)y=(D+1)^3(D+2)y=0$ so your particular solution is $$y_p=At^2e^{-t}$$ where $A$ is an unknown constant. Note that the differential operator $D^n$ annihilates each of the functions: $$1,t,t^2,...,t^{n-1}$$ and differential operator $(D-\alpha)^n$ annihilates each of the following functions: $$e^{\alpha t},te^{\alpha t},t^2e^{\alpha t},...,t^{n-1}e^{\alpha t}$$
A: Perhaps this will grant some insight.  In a way, the left hand side can be factored like a polynomial.  So just as
$s^3+4s^2+5s+2=(s+1)(s^2+3s+2)$
if you let $u=y''+3y'+2y$, this equation can be rewritten as
$y'''+4y''+5y'+2y=(y''+3y'+2y)'+y''+3y'+2y=u'+u=e^{-t}$
So let's solve this for $u$ using integrating factors.
$u'e^t+ue^t=(ue^t)'=1$
$ue^t=t+C,u=y''+3y'+2y=te^{-t}+Ce^{-t}$
If, as you put it, this solution of the homogenous equation had not been of the form of the particular equation, the right side before integration would have been of the form $e^{nt}$.  Instead, we've had a $te^{-t}$ term introduced.
Something similar will occur when we make the substitution $v=y'+2y$, where we will again have $e^t$ as an integrating factor.
