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

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The most obvious solution (but not the one you're looking for) \[\text{Let } f = {1\over g}\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0well I don't think that works!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey watchmath, i got disconnected earlier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the program just froze in the middle of convo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets see, use product rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(fg)'= f'g' and (fg)''=f''g'' but (uv)' = u'v + uv' , , so ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec , dont give answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i suspect theres an e^x in there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(fg)' = f' g + f g' = f' g'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(fg) ' = f' (g / g' ) + f/f'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since i divided both sides by g ;

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0errr, (fg)' = f' g + f g' = f' g' f ' = f ' ( g/g') + f

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so f = f' ( 1  g / g ' )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nevermind, lets use the second assumption

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(fg)'' = ( f ' g' ) ' = f '' g ' + f ' g''

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we have two equations fg ' + f ' g  f ' g' = 0 f '' g' + f' g''  f'' g'' = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral f '' *g'' = (fg)' , integral (fg)' = fg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets integrate by parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i need a hint, im stumped

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0consider fg ' + f ' g  f ' g' = 0 as a differential equation in g :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0WAIT, FROM THE FIRST equation*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g/g' + f/f' = 1, g'/g'' + f' / f'' = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what kind of diferential equation is it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so ( f  f') g ' + f' g = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know how to solve that, only if they are constant, but lets assume they are

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then (ff')^2 + f' = 0

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Well that equation can be made into something of the type: y'+p(t)y=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0There are! If you just want to find some you can guess one! What kind of functions are the obvious guess?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i have f' ( g/g' 1 ) + f  0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f' ( g/g' 1 ) + f = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see 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
 5 years ago
Best ResponseYou've already chosen the best response.0Here is one of the example \(f=e^{3x/2},g=e^{3x}\)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Now I believe you can easily characterize which eponential functions \(f=Ae^{Bx},g=Ce^{Dx}\) satisfy the requirement.
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