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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''

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  1. anonymous
    • 5 years ago
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    The most obvious solution (but not the one you're looking for) \[\text{Let } f = {1\over g}\]

  2. watchmath
    • 5 years ago
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    well I don't think that works!

  3. anonymous
    • 5 years ago
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    Oh. true.

  4. anonymous
    • 5 years ago
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    hey watchmath, i got disconnected earlier

  5. anonymous
    • 5 years ago
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    the program just froze in the middle of convo

  6. anonymous
    • 5 years ago
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    ok lets see, use product rule

  7. anonymous
    • 5 years ago
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    (fg)'= f'g' and (fg)''=f''g'' but (uv)' = u'v + uv' , , so ...

  8. anonymous
    • 5 years ago
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    one sec , dont give answer

  9. anonymous
    • 5 years ago
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    i suspect theres an e^x in there

  10. anonymous
    • 5 years ago
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    (fg)' = f' g + f g' = f' g'

  11. anonymous
    • 5 years ago
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    (fg) ' = f' (g / g' ) + f/f'

  12. anonymous
    • 5 years ago
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    since i divided both sides by g ;

  13. anonymous
    • 5 years ago
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    errr, (fg)' = f' g + f g' = f' g' f ' = f ' ( g/g') + f

  14. anonymous
    • 5 years ago
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    so f = f' ( 1 - g / g ' )

  15. anonymous
    • 5 years ago
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    nevermind, lets use the second assumption

  16. anonymous
    • 5 years ago
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    no hints, one sec

  17. anonymous
    • 5 years ago
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    (fg)'' = ( f ' g' ) ' = f '' g ' + f ' g''

  18. anonymous
    • 5 years ago
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    so we have two equations fg ' + f ' g - f ' g' = 0 f '' g' + f' g'' - f'' g'' = 0

  19. anonymous
    • 5 years ago
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    integral f '' *g'' = (fg)' , integral (fg)' = fg

  20. anonymous
    • 5 years ago
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    lets integrate by parts

  21. anonymous
    • 5 years ago
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    ok i need a hint, im stumped

  22. watchmath
    • 5 years ago
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    consider fg ' + f ' g - f ' g' = 0 as a differential equation in g :)

  23. anonymous
    • 5 years ago
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    WAIT, FROM THE FIRST equation*

  24. anonymous
    • 5 years ago
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    g/g' + f/f' = 1, g'/g'' + f' / f'' = 1

  25. anonymous
    • 5 years ago
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    ok

  26. anonymous
    • 5 years ago
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    what kind of diferential equation is it

  27. watchmath
    • 5 years ago
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    It is linear

  28. anonymous
    • 5 years ago
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    ok so ( f - f') g ' + f' g = 0

  29. anonymous
    • 5 years ago
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    i dont know how to solve that, only if they are constant, but lets assume they are

  30. anonymous
    • 5 years ago
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    then (f-f')^2 + f' = 0

  31. watchmath
    • 5 years ago
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    Well that equation can be made into something of the type: y'+p(t)y=0

  32. anonymous
    • 5 years ago
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    ok

  33. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    There are! If you just want to find some you can guess one! What kind of functions are the obvious guess?

  35. anonymous
    • 5 years ago
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    ok i have f' ( g/g' -1 ) + f - 0

  36. anonymous
    • 5 years ago
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    f' ( g/g' -1 ) + f = 0

  37. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Here is one of the example \(f=e^{3x/2},g=e^{3x}\)

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

  40. anonymous
    • 5 years ago
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    bye

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