I found this question and I cannot seem to answer it correctly and its kinda bothering. I am not seeing what I am not getting right with this particular problem. I took the same route as the OP and found the individual particular solutions of the RHS and added them together as a linear combination but to my surprise, get something totally different. Can someone look at this and let me know what I may be doing wrong.
Original question is linked here: Solving Diff. Eq.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ {\textbf{Method of Undetermined Coefficients}}$
${\bf{SOLUTION:}}$
$y(x)=y_{h}~+~y_{p}$
${\text{Differential Equation:}}~~~~~~~~~~~~~~~~~$ $y^{(5)}+2y^{(3)}+y'=2x+\sin(x)+\cos(x)$.
${\text{Homogeneous Case:}}~~~~~~~~~~~~~~~~~~~~$ $y^{(5)}+2y^{(3)}+y'=0$.
${\text{Characteristic Polynomial:}}~~~~~~~~~$ $r^5+2r^3+r=0$.
${\text{Solved Roots of polynomial:}}~~~~~~$ $\bigg[\{r\rightarrow 0\},\; \{r\rightarrow -i\},\; \{r\rightarrow -i\},\; \{r\rightarrow i\},\; \{r\rightarrow i\}\bigg]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
${\text{General Form of the Homogeneous Solution}}$
$ y_{h}(x)=C_{1}e^{r_{1}x}+e^{ax}\Big(C_{2}\cos(bx)+C_{3}\sin(bx)+C_{4}x\cos(bx)+C_{5}x\sin(bx) \Big) $
${\text{Homogeneous Solution to the Differential Equation}}$ $y_{h}(x)=C_{1}+C_{2}\cos(x)-C_{3}\sin(x)+C_{4}x\cos(x)-C_{5}\sin(x);~~\Big(~\because \sin(-x)=-\sin(x)~\Big).$
Now we shall seek a particular solution.
${\text{Non-Homogeneous Case:}}~~~~~~~~~~~~~~~~$ $y^{(5)}+2y^{(3)}+y'=2x$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
$ f(x)=2x $
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$Let,
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \begin{array}{llll} y_{p}(x)=Ax+b \\ y_{p}'(x)=A \\ y_{p}''(x)=0 \\ y_{p}^{(3)}(x)=0 \\ y_{p}^{(4)}(x)=0 \\ y_{p}^{(5)}(x)=0 \end{array}$
Substituting derivatives into differential equation:
$(0)+2(0)+(A)=2x$.
After equating the undetermined coefficient ${\underline{A}}$ we get:
$ \begin{array}{l} A=~0 \end{array} $
Making our particular solution to become,
$ y_{p}(x)=0. $
(2)$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y^{(5)}+2y^{(3)}+y'=\sin(x)$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
$ f(x)=\sin(x) $
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$Let,
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \begin{array}{llll} y_{p}(x)=A\sin(x)+B\cos(x) \\ y_{p}'(x)=A\cos(x)-B\sin(x) \\ y_{p}''(x)=-A\sin(x)-B\cos(x) \\ y_{p}^{(3)}(x)=-A\cos(x)+B\sin(x) \\ y_{p}^{(4)}(x)=A\sin(x)+B\cos(x) \\ y_{p}^{(5)}(x)=A\cos(x)-B\sin(x) \end{array}$
Substituting derivatives into differential equation:
$A\cos(x)-B\sin(x)+2\Big(-A\cos(x)+B\sin(x)\Big)+\Big(A\cos(x)-B\sin(x)\Big)=\sin(x)$.
After equating the undetermined coefficients ${\underline{A}}$ and ${\underline{B}}$ we get:
$ \begin{array}{l} A=~0 \\ 0\cdot B=1~~ ????~~ {\text{Huh}} \end{array} $
Making our particular solution to become,
$ y_{p}(x)=0~~?? $
I guessed it will be the same situation for the $\cos(x)$ on the RHS when finding the particular solution, though I could be missing an important fact.
Thanks.