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Find f(x) from following data

  1. May 30, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If f(x/y) = f(x)/f(y) f(y)≠0 and f'(1)=2 find f(x).

    2. Relevant equations

    3. The attempt at a solution
    Diff both sides wrt x
    f'(x/y)*1/y=f'(x)/f(y)
    putting x=1
    f'(1/y)*1/y=2/f(y)
     
  2. jcsd
  3. May 30, 2013 #2
    Your first step is wrong. Apply chain rule.
     
  4. May 30, 2013 #3

    utkarshakash

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    I have differentiated wrt x treating y as a constant.
     
  5. May 30, 2013 #4

    tiny-tim

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    hi utkarshakash! :smile:
    ok so far

    now you need a relation between f(1/y) and f(y) :wink:
     
  6. May 30, 2013 #5
    Ah yes, sorry about that. Now as tiny-tim said, a relation between f(y) and f(1/y) would be useful.

    Hint: Put x=1 in the original equation. :smile:
     
  7. May 30, 2013 #6

    haruspex

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    You might find the path a little more obvious if you differentiate wrt y instead. Note that you can easily deduce f(1).
     
  8. May 31, 2013 #7

    utkarshakash

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    How to find that?
     
  9. May 31, 2013 #8

    tiny-tim

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    apply the formula in the question to f(1/y)
     
  10. May 31, 2013 #9

    AGNuke

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    If I had to guess, I'd say ##f(x)=x^2##.
     
  11. May 31, 2013 #10

    utkarshakash

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    Yes the answer is correct.
     
  12. May 31, 2013 #11

    tiny-tim

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    ok, now derive it! :rolleyes:
     
  13. May 31, 2013 #12

    utkarshakash

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    f(1)=f(y)f(1/y)
     
  14. May 31, 2013 #13

    tiny-tim

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    ok, now combine that with your original equation …
    … to get a differential equation in f(1/y) :wink:

    (also, isn't it fairly obvious what f(1) is?)
     
  15. May 31, 2013 #14
    Please can you help me understand a very important concept and i.e why have we differentiated wrt x treating y as a constant .

    Isnt x and y both variable ? Why is applying chain rule to differentiate wrong as earlier suggested by Pranav-Arora ?

    Thanks
     
  16. May 31, 2013 #15

    tiny-tim

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    yes, x and y are independent variables, and so we can ∂/∂x keeping y constant, or ∂/∂y keeping x constant :wink:
    i think you've misread that …

    everybody's been applying the chain rule
     
  17. May 31, 2013 #16
    I have been trying this question from quite some time and I always ended up with the same differential equation you are talking about but I don't see a way to solve it.

    What I end up with is this:
    [tex]\frac{∂f(1/y)}{∂x}\cdot \frac{1}{y}=2f(1/y)[/tex]
    I don't see how can i solve this but if I replace y with 1/x, I do get the right answer but is it valid to replace y with 1/x?

    Thanks!
     
  18. May 31, 2013 #17
    Thanks tiny-tim for clarification...Yes I indeed misread things
     
  19. May 31, 2013 #18

    utkarshakash

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    f'(1/y)/f(1/y)=2y (Assuming x=1)

    But I haven't learned how to solve DE's.
     
  20. May 31, 2013 #19

    utkarshakash

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    Yes of course it is valid. Since the function is valid for all y belonging to ℝ.
    But can you tell me the method to solve this equation? I haven't been taught DE yet.
     
  21. May 31, 2013 #20
    I still don't get this. :confused:

    Assuming that replacing y with 1/x is valid,
    [tex]\frac{∂f(x)}{∂x}\cdot x=2f(x)[/tex]
    Rearranging,
    [tex]\frac{∂f(x)}{f(x)}=2\frac{∂x}{x}[/tex]
    You do know how to solve this now. (##\int dx/x=\ln x+c##)
     
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