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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.

1 Answers 1

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Because $\sin(x)$ and $\cos(x)$ solve the homogeneous equation, you don't try solutions $y_p(x) = A\cos(x) + B\sin(x)$ when the right hand side is a multiple of $\sin(x)$ or $\cos(x)$.

Instead, you count the multiplicity $m$ of the root $i$ of the characteristic polynomial $r(r+i)^2(r-i)^2$, in this case $2$, and use $y_p(x) = Ax^m\cos(x) + Bx^m\sin(x) = Ax^2\cos(x) + Bx^2\sin(x)$ The calculations might be annoying but you'll eventually be able to solve for $A$ and $B$.

Another thing you can do is try $y_p(x) = Ax^2e^{ix}$, and solve for $A$. The real part of this will give the particular solution for $\cos(x)$ on the right-hand side, and the imaginary part will give the particular solution for $\sin(x)$. The calculations will likely be a lot easier. Note $A$ will be a complex number this time.

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    @Zarax: Thanks, I see. They must of skipped this in the text I primarily use.2011-07-05