# y'''-y=x^{2} has solution ''multiplicity''?

$y{}^{‴}-y={x}^{2}$ has solution ''multiplicity''?
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zurilomk4
Your current question has a characteristic polynomial of degree 3. This means that there are 3 roots, when you count multiplicities. In general, let d be the degree of a polynomial be the distinct roots of f(x) with respective multiplicities ${m}_{1},{m}_{2},\dots ,{m}_{n}$. Then $d=\sum _{{m}_{i}}$.
Since we find three roots , and their multiplicities must add up to 3 (the degree of our characteristic polynomial), the multiplicity of each root must be 1. Thus we do not need more than $k=0$ for each root.
If you're looking for an example of a differential equation that has $m>1$, you should look at $y{}^{″}-2{y}^{\prime }+1=0$. The characteristic polynomial of this equation has the same root repeated twice, $\left(r-1\right)\left(r-1\right)$, meaning that the root $r=1$ has multiplicity 2. Thus the homogeneous solution is $y={C}_{1}{e}^{x}+{C}_{2}x{e}^{x}$.
###### Not exactly what you’re looking for?
Buck Henry
The roots of the characteristic equation , so the general solution of the homogeneous DE $y{}^{‴}-y=0$ is
$y={C}_{1}{e}^{x}+{e}^{-\frac{x}{2}}\left({C}_{2}\mathrm{cos}\left(\frac{\sqrt{3}}{2}x\right)+{C}_{3}\mathrm{sin}\left(\frac{\sqrt{3}}{2}x\right)\right)$.
The usual ansatz for finding a single solution of the non-homogeneous DE works, so I don't see what problems remain?
###### Not exactly what you’re looking for?
RizerMix
EDIT: What follows is relevant to the question as originally posted. The question was edited after I posted this answer, and this answer is no longer relevant to the revised version of the question. It's the $m=2$ case in your notation, so you need .