Question

All Calculus students dream :)
Find all pair of non constant functions f, g such that
(fg)'=f'g' and (fg)''=f''g''

Answer 1

The most obvious solution (but not the one you're looking for)

\[\text{Let } f = {1\over g}\]

Answer 2

well I don't think that works!

Answer 3

Oh. true.

Answer 4

hey watchmath, i got disconnected earlier

Answer 5

the program just froze in the middle of convo

Answer 6

ok lets see, use product rule

Answer 7

(fg)'= f'g' and (fg)''=f''g''

but (uv)' = u'v + uv' , , so ...

Answer 8

one sec , dont give answer

Answer 9

i suspect theres an e^x in there

Answer 10

(fg)' = f' g + f g' = f' g'

Answer 11

(fg) ' = f' (g / g' ) + f/f'

Answer 12

since i divided both sides by g ;

Answer 13

errr, (fg)' = f' g + f g' = f' g'
f ' = f ' ( g/g') + f

Answer 14

so f = f' ( 1 - g / g ' )

Answer 15

nevermind, lets use the second assumption

Answer 16

no hints, one sec

Answer 17

(fg)'' = ( f ' g' ) ' = f '' g ' + f ' g''

Answer 18

so we have two equations
fg ' + f ' g - f ' g' = 0
f '' g' + f' g'' - f'' g'' = 0

Answer 19

integral f '' *g'' = (fg)' ,
integral (fg)' = fg

Answer 20

lets integrate by parts

Answer 21

ok i need a hint, im stumped

Answer 22

consider fg ' + f ' g - f ' g' = 0 as a differential equation in g :)

Answer 23

WAIT, FROM THE FIRST equation*

Answer 24

g/g' + f/f' = 1, g'/g'' + f' / f'' = 1

Answer 25

ok

Answer 26

what kind of diferential equation is it

Answer 27

It is linear

Answer 28

ok so ( f - f') g ' + f' g = 0

Answer 29

i dont know how to solve that, only if they are constant, but lets assume they are

Answer 30

then (f-f')^2 + f' = 0

Answer 31

Well that equation can be made into something of the type:
y'+p(t)y=0

Answer 32

ok

Answer 33

i dont think there will be any nonconstant functions satisfying the given criteria....but still searching for a proof. By the way the the problem is good!

Answer 34

There are! If you just want to find some you can guess one! What kind of functions are the obvious guess?

Answer 35

ok i have f' ( g/g' -1 ) + f - 0

Answer 36

f' ( g/g' -1 ) + f = 0

Answer 37

see g=(some constant)*g' or same for f...doesnt work...there could be more complex functions than e^kx...what are ur guesses watchmath?

Answer 38

Here is one of the example
\(f=e^{3x/2},g=e^{3x}\)

Answer 39

Now I believe you can easily characterize which eponential functions \(f=Ae^{Bx},g=Ce^{Dx}\) satisfy the requirement.

Answer 40

bye