- #1
Turion
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Are they both correct? Would the first or second solution be preferred?
Which solution for this DE is preferred?
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Discussion Overview
The discussion revolves around the correctness of two proposed solutions for a differential equation (DE). Participants are examining the validity of each solution and debating which one might be preferred based on their characteristics and derivations.
Discussion Character
- Debate/contested
- Mathematical reasoning
Main Points Raised
- Some participants question whether both solutions are correct and seek to determine a preferred option.
- One participant asserts that the second solution is not valid and claims there exists a solution in terms of exponentials.
- Another participant suggests that the first solution can be derived from the second, indicating a potential relationship between them.
- Concerns are raised about the roots obtained from the auxiliary equation, with conflicting claims about whether they are m1=0 and m2=k^2 or m1=+k and m2=-k.
- One participant acknowledges a mistake after attempting to verify the second solution against the DE.
- A later reply clarifies the characteristic equation for the DE, pointing out a miswriting that led to confusion regarding the roots.
Areas of Agreement / Disagreement
Participants express disagreement regarding the validity of the second solution, with some asserting it is incorrect while others attempt to clarify the roots of the auxiliary equation. The discussion remains unresolved as differing views persist.
Contextual Notes
There are limitations regarding the assumptions made about the differential equation and the definitions of the solutions. The discussion reveals confusion over the correct formulation of the DE and its characteristic equation.
Are they both correct? Would the first or second solution be preferred?
- #2
phyzguy
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The second one is certainly not a solution. There is a solution in terms of exponentials, but that is not it.
- #3
Turion
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phyzguy said:
Why is the second one not a solution? If you convert the original DE into an auxiliary equation, you will get roots: m1=0 and m2=k2
- #4
Integral
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The first of your solutions can be arrived at from the second. I would say that the second is the fundamental.
- #5
- #6
phyzguy
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Turion said:
No you don't. You get roots of +k and -k. Try plugging your second solution into the DE and see if it works. You'll see that it doesn't.
- #7
Turion
- 145
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phyzguy said:
That's weird: http://www.wolframalpha.com/input/?i=m^2-k^2m=0
Going to try plugging it in.
Edit: Oh wow. You're right. I'm an idiot. Lol
- #8
HallsofIvy
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The characteristic equation for the D.E. you give, y''- k^2y= 0 is r^2- k= 0 which is equivalent to r^2= k^2 and has roots k and -k. You, apparently, miswrote the equation as y''- k^2y'= 0, which has characteristic equation r^2- k^2r= r(r- k^2)= 0 and has roots 0 and k^2.
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