Particular integral of a fifth order linear ODE? I am presented with the differential equation
$y^{(5)}-y^{(1)}=x$
Finding the complementary function was OK. I obtained solutions to the characteristic equation of $\lambda = 0, \pm 1, \pm i$ so the complementary function is
$y_{CF}=a_0+a_1e^x+a_2e^{-x}+a_3e^{i}+a_4e^{-i}$
Now for the particular integral, my first thought would be to use a trial form 
$y_{PI}=b_1x+b_0$
However this clearly cannot be the solution as, plugging it in, I get $-b_1=x$. Trying $b_2x^2+b_1x+b_0$ I get $-2b_2x-b_1=x$ which works for $b_1=0, b_2=-1/2$ and I suppose the constant $b_0$ doesn't really matter because it is amalgamated into the constant in the complementary function anyway.
However I am finding it quite strange that the form of the particular integral was $b_2x^2+b_1x+b_0$, and not the general form for a linear forcing term of $b_1x+b_0$ that I have been taught to use.
I was wondering if there is a rule particular integrals when you have different orders of ODE? I can see here that my first trial would fail because there is no $y$ term on the RHS. So I suppose this trial form would also in fact fail for a second order ODE of form $ay''+by'=x$. Do you just have to make observations like this in guessing the particular integral? I have never come across this before in class.
 A: You can apply general trial functions with one simple modification to your assumptions. Normally, to solve for a particular solution $x$, you would try $y_p=b_1x+b_0$. However, this would clearly only work if you had a $y$ term on the LHS. So the choice of trial function shouldn't be of order $1$, but rather of order $1+r$, where $r$ is the order of the lowest order derivative on the LHS. For example, for $y^{(5)}-y^{(2)}=x$, you wouldn't expect a linear or quadratic trial form to work, so you would need to use a cubic equation. You don't run into this same problem when the RHS is a an exponential or trig function because the effect of any derivatives can be captured by the coefficients.
A: You would try
$$y_p=x(Ax+B)$$
Then get $$-2A-B=x,$$
$$A=-\frac12,\quad B=0$$
$$y_p=-\frac{x^2}{2}.$$
A: In this case you can just integrate the full ODE to
$$
y^{(4)}-y=\frac12x^2+c
$$
where the homogeneous solution has no terms that are simply polynomial, thus the method of undetermined coefficients works without degree corrections.
