## 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''

1. anonymous

The most obvious solution (but not the one you're looking for) $\text{Let } f = {1\over g}$

2. watchmath

well I don't think that works!

3. anonymous

Oh. true.

4. anonymous

hey watchmath, i got disconnected earlier

5. anonymous

the program just froze in the middle of convo

6. anonymous

ok lets see, use product rule

7. anonymous

(fg)'= f'g' and (fg)''=f''g'' but (uv)' = u'v + uv' , , so ...

8. anonymous

one sec , dont give answer

9. anonymous

i suspect theres an e^x in there

10. anonymous

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

11. anonymous

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

12. anonymous

since i divided both sides by g ;

13. anonymous

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

14. anonymous

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

15. anonymous

nevermind, lets use the second assumption

16. anonymous

no hints, one sec

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

lets integrate by parts

21. anonymous

ok i need a hint, im stumped

22. watchmath

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

23. anonymous

WAIT, FROM THE FIRST equation*

24. anonymous

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

25. anonymous

ok

26. anonymous

what kind of diferential equation is it

27. watchmath

It is linear

28. anonymous

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

29. anonymous

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

30. anonymous

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

31. watchmath

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

32. anonymous

ok

33. 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!

34. watchmath

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

35. anonymous

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

36. anonymous

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

37. 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?

38. watchmath

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

39. watchmath

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

40. anonymous

bye