anonymous
  • anonymous
Be f:R->R derivable up to the secound derivative of that f''-f=0 prove that there is a constant A for all x so that [(f-Ae^(-x))/(e^x)]'=0
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\left[ \frac{ f(x) - Ae ^{-x} }{ e ^{x} } \right]'=0\]
primeralph
  • primeralph
Well, for f'-f=0, solving the DE, f = Be^(x). So, (Be^(x)-Ae^(-x))/e^x = B- Ae^(-2x) The differential is 2Ae^(-2x) = 0. For this to be true for all x, A = 0.
anonymous
  • anonymous
f '' - f=0

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primeralph
  • primeralph
Just use the same approach.
anonymous
  • anonymous
why from: B-Ae^(-2x) u get 2Ae^(-2x)=0 ????
primeralph
  • primeralph
Because I divided by e^x. Just start it again.
anonymous
  • anonymous
no u didnt understand!!....
anonymous
  • anonymous
I see u have B-Ae^(2x)
primeralph
  • primeralph
I differentiated.
anonymous
  • anonymous
for f '' - f=0 the only difference is that is B^2 instead of B right?
primeralph
  • primeralph
No. B is an arbitrary constant.
anonymous
  • anonymous
but if f(x)=e^x if u derivate u get e^x but if u derivate and get Be^x the primitive is e^cx right?
primeralph
  • primeralph
f '' - f=0 General solution is Be^(-x) + Be^x
primeralph
  • primeralph
@matheusoliveira Continue with what I just told you.
anonymous
  • anonymous
its Be^(-x) + Ce^x the constants need to be different
anonymous
  • anonymous
how so? i dont think they need to...
primeralph
  • primeralph
Sorry, they are different.
anonymous
  • anonymous
so we get [(Be^x + Ce^-x -Ae^-x)/e^x]'=2Ae^(-2x) -2Ce^(-2x)=0
anonymous
  • anonymous
A/e^2x=C/e^2x so A=C right?
primeralph
  • primeralph
Yeah I guess

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