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

tripiverded9

tripiverded9

Answered question

2022-01-20

yy=x2 has solution ''multiplicity''?

Answer & Explanation

zurilomk4

zurilomk4

Beginner2022-01-20Added 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(x),let r1,r2,,rnZ be the distinct roots of f(x) with respective multiplicities m1,m2,,mn. Then d=mi.
Since we find three roots (1,1+i32,and 1i32), 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 y2y+1=0. The characteristic polynomial of this equation has the same root repeated twice, (r1)(r1), meaning that the root r=1 has multiplicity 2. Thus the homogeneous solution is y=C1ex+C2xex.
Buck Henry

Buck Henry

Beginner2022-01-21Added 33 answers

The roots of the characteristic equation r31=are r=1 and r=1±i32, so the general solution of the homogeneous DE yy=0 is
y=C1ex+ex2(C2cos(32x)+C3sin(32x)).
The usual ansatz for finding a single solution of the non-homogeneous DE works, so I don't see what problems remain?
RizerMix

RizerMix

Expert2022-01-27Added 656 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 xkeαxcos(βx) and xkeαxsin(βx) for k=0..1.

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