tripiverded9
2022-01-20
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zurilomk4

Answered 2022-01-20
Author has **35** answers

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 $f\left(x\right),\text{let}\text{}{r}_{1},{r}_{2},\dots ,{r}_{n}Z$ 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$(1,\frac{1+i\sqrt{3}}{2},\text{and}\text{}\frac{1-i\sqrt{3}}{2})$ , 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 $y{}^{\u2033}-2{y}^{\prime}+1=0$ . The characteristic polynomial of this equation has the same root repeated twice, $(r-1)(r-1)$ , meaning that the root $r=1$ has multiplicity 2. Thus the homogeneous solution is $y={C}_{1}{e}^{x}+{C}_{2}x{e}^{x}$ .

Since we find three roots

If you're looking for an example of a differential equation that has

Buck Henry

Answered 2022-01-21
Author has **33** answers

The roots of the characteristic equation $r}^{3}-1=\text{are}\text{}r=1\text{}\text{and}\text{}r=\frac{-1\pm i\sqrt{3}}{2$ , so the general solution of the homogeneous DE $y{}^{\u2034}-y=0$ is

$y={C}_{1}{e}^{x}+{e}^{-\frac{x}{2}}({C}_{2}\mathrm{cos}\left(\frac{\sqrt{3}}{2}x\right)+{C}_{3}\mathrm{sin}\left(\frac{\sqrt{3}}{2}x\right))$ .

The usual ansatz for finding a single solution of the non-homogeneous DE works, so I don't see what problems remain?

The usual ansatz for finding a single solution of the non-homogeneous DE works, so I don't see what problems remain?

RizerMix

Answered 2022-01-27
Author has **438** answers

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 ${x}^{k}{e}^{\alpha x}\mathrm{cos}(\beta x)\text{}\text{and}\text{}{x}^{k}{e}^{\alpha x}\mathrm{sin}(\beta x)\text{}\text{for}\text{}k=0..1$ .

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Quickly! Need help

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I was toying with equations of the type $f(x+\alpha )={f}^{\prime}\left(x\right)$ where f is a real function. For example if $\alpha =\frac{\pi}{2}$ then the solutions include the function ${f}_{\lambda ,\mu}\left(x\right)=\lambda \mathrm{cos}(x+\mu )$ . Are there more solutions?

On the other hand, if I want to solve the equation for any$\alpha$ , I can assume a solution of the form $f\left(x\right)={e}^{\lambda x}$ , and find $\lambda$ as a complex number that enables me to solve the equation...

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I was wondering: is the set of the solutions of dimension 2 because the derivative operator creates one dimension and the operator

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show that

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