TITLE: General solution for $y^{iv}+ 2y''+y=\cos x$
QUESTION [6 upvotes]: Here is another problem from Mathews and Walker that has given me some trouble.

1-18. Find the general solution of
  $y^{iv}+ 2y''+y=\cos x$.

Note: Thanks, everyone, for clearing up the interpretation of $y^{iv}$ as the fourth derivative of $y$ and for the clear solutions. 
I had interpreted $y^{iv}$ as $y^{\sqrt{-1} \ v}$ with $v\in \mathbb{C}$. 
Of course, this is an awful nonlinear DE!

REPLY [2 votes]: You did not try the easiest way. It is a linear equation with constant coefficients. The characteristic equation is
$$
r^4+2\,r^2+1=(r^2+1)^2=0\implies r=\pm i,\text{ roots of multiplicity $2$.}
$$
The solution of the homogeneous equation is
$$
y_h=C_1\cos x+C_2\sin x+C_3x\cos x+C_4x\sin x.
$$
Lookig at the right hans side, we know that there is a particular solution of the complete equation of the form
$$
y_p=x^2(A\cos x+B\sin x).
$$
$A$ and $B$ are found substituting $y_p$ in the equation. Its general solution is
$$
y=y_h+y_p.
$$