Question

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

Answer 1

\[\left[ \frac{ f(x) - Ae ^{-x} }{ e ^{x} } \right]'=0\]

Answer 2

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.

Answer 3

f '' - f=0

Answer 4

Just use the same approach.

Answer 5

why from:
B-Ae^(-2x)
u get
2Ae^(-2x)=0
????

Answer 6

Because I divided by e^x.
Just start it again.

Answer 7

no u didnt understand!!....

Answer 8

I see u have B-Ae^(2x)

Answer 9

I differentiated.

Answer 10

for f '' - f=0 the only difference is that is B^2 instead of B right?

Answer 11

No. B is an arbitrary constant.

Answer 12

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?

Answer 13

f '' - f=0
General solution is Be^(-x) + Be^x

Answer 14

@matheusoliveira Continue with what I just told you.

Answer 15

its Be^(-x) + Ce^x the constants need to be different

Answer 16

how so?
i dont think they need to...

Answer 17

Sorry, they are different.

Answer 18

so we get [(Be^x + Ce^-x -Ae^-x)/e^x]'=2Ae^(-2x) -2Ce^(-2x)=0

Answer 19

A/e^2x=C/e^2x so A=C right?

Answer 20

Yeah I guess