Which solution for this DE is preferred?

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  • #1
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Are they both correct? Would the first or second solution be preferred?
 

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  • #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
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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.
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
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Are they both correct? Would the first or second solution be preferred?
No, the second is not correct.
 

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  • #6
phyzguy
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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
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.
 
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  • #8
HallsofIvy
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The characteristic equation for the D.E. you give, [itex]y''- k^2y= 0[/itex] is [itex]r^2- k= 0[/itex] which is equivalent to [itex]r^2= k^2[/itex] and has roots k and -k. You, apparently, miswrote the equation as [itex]y''- k^2y'= 0[/itex], which has characteristic equation [itex]r^2- k^2r= r(r- k^2)= 0[/itex] and has roots 0 and [itex]k^2[/itex].
 

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