watchmath
  • watchmath
All Calculus students dream :) Find all pair of non constant functions f, g such that (fg)'=f'g' and (fg)''=f''g''
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
The most obvious solution (but not the one you're looking for) \[\text{Let } f = {1\over g}\]
watchmath
  • watchmath
well I don't think that works!
anonymous
  • anonymous
Oh. true.

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anonymous
  • anonymous
hey watchmath, i got disconnected earlier
anonymous
  • anonymous
the program just froze in the middle of convo
anonymous
  • anonymous
ok lets see, use product rule
anonymous
  • anonymous
(fg)'= f'g' and (fg)''=f''g'' but (uv)' = u'v + uv' , , so ...
anonymous
  • anonymous
one sec , dont give answer
anonymous
  • anonymous
i suspect theres an e^x in there
anonymous
  • anonymous
(fg)' = f' g + f g' = f' g'
anonymous
  • anonymous
(fg) ' = f' (g / g' ) + f/f'
anonymous
  • anonymous
since i divided both sides by g ;
anonymous
  • anonymous
errr, (fg)' = f' g + f g' = f' g' f ' = f ' ( g/g') + f
anonymous
  • anonymous
so f = f' ( 1 - g / g ' )
anonymous
  • anonymous
nevermind, lets use the second assumption
anonymous
  • anonymous
no hints, one sec
anonymous
  • anonymous
(fg)'' = ( f ' g' ) ' = f '' g ' + f ' g''
anonymous
  • anonymous
so we have two equations fg ' + f ' g - f ' g' = 0 f '' g' + f' g'' - f'' g'' = 0
anonymous
  • anonymous
integral f '' *g'' = (fg)' , integral (fg)' = fg
anonymous
  • anonymous
lets integrate by parts
anonymous
  • anonymous
ok i need a hint, im stumped
watchmath
  • watchmath
consider fg ' + f ' g - f ' g' = 0 as a differential equation in g :)
anonymous
  • anonymous
WAIT, FROM THE FIRST equation*
anonymous
  • anonymous
g/g' + f/f' = 1, g'/g'' + f' / f'' = 1
anonymous
  • anonymous
ok
anonymous
  • anonymous
what kind of diferential equation is it
watchmath
  • watchmath
It is linear
anonymous
  • anonymous
ok so ( f - f') g ' + f' g = 0
anonymous
  • anonymous
i dont know how to solve that, only if they are constant, but lets assume they are
anonymous
  • anonymous
then (f-f')^2 + f' = 0
watchmath
  • watchmath
Well that equation can be made into something of the type: y'+p(t)y=0
anonymous
  • anonymous
ok
anonymous
  • anonymous
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!
watchmath
  • watchmath
There are! If you just want to find some you can guess one! What kind of functions are the obvious guess?
anonymous
  • anonymous
ok i have f' ( g/g' -1 ) + f - 0
anonymous
  • anonymous
f' ( g/g' -1 ) + f = 0
anonymous
  • anonymous
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?
watchmath
  • watchmath
Here is one of the example \(f=e^{3x/2},g=e^{3x}\)
watchmath
  • watchmath
Now I believe you can easily characterize which eponential functions \(f=Ae^{Bx},g=Ce^{Dx}\) satisfy the requirement.
anonymous
  • anonymous
bye

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