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Math Help - Find a differentiable function with the given properties

  1. #1
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    Find a differentiable function with the given properties

    hello, need your help

    find all function f differentiable over \mathbb{R}
    - (\forall x\in \mathbb{R}) : \big(f'(x)\big)^2-\big(f(x)\big)^2=1 \quad (1)
    -f'(0)=1 \quad(2)
    -f' is differentiable over \mathbb{R}\quad (3)
    Last edited by linda2005; April 15th 2009 at 03:12 PM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Ok not to be a pain but the word is differentiable not derivable.

    Taking the square root

    f ' ^2 -f = 1 or -1

    Differentiate -- in either case

    2f " f ' -f ' = 0

    f ' (2f"-1) = 0

    Now solve f ' = 0 (use original equation to find values of c) and 2f " - 1 = 0

    That should be enough to get you to the end
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    Ok not to be a pain but the word is differentiable not derivable.

    Taking the square root

    f ' ^2 -f = 1 or -1
    either i'm missing something or you made a mistake here
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  4. #4
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    Calculus26 it's \big(f'(x)\big)^2-\big(f(x)\big)^2=1 and not \big(f'(x)\big)^2-f(x)=1
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  5. #5
    MHF Contributor Calculus26's Avatar
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    Sorry I read it too quickly I was seeing

    [f '^2 -f]^2 = 1 for the original

    For (f ')^2 - f^2 =1

    the process is still the same

    2f ' f "- 2 f f '= 0

    f ' = 0 or f '' -f = 0

    yields f = c from which there are no solutions

    or f " -f = 0 which yield Acos(x) + Bsin(x) now apply f '(0) = 1

    and finish
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  6. #6
    MHF Contributor Calculus26's Avatar
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    I believe there are no solutions as f ' (0) =1 yields f =Acos(x) +sin(x)

    Plugging this into the original there are no values of A Solving the equation.
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  7. #7
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    Quote Originally Posted by Calculus26 View Post
    or f " - f = 0 which yield Acos(x) + Bsin(x) now apply f '(0) = 1
    sorry but you are wrong , you mean

    f " + f = 0 which yield Acos(x) + Bsin(x)

    but here we have f " - f = 0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by linda2005 View Post
    sorry but you are wrong , you mean

    f " + f = 0 which yield Acos(x) + Bsin(x)

    but here we have f " - f = 0
    yes, take f(x) = Ae^x + Be^{-x} instead
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  9. #9
    MHF Contributor Calculus26's Avatar
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    Ok I've made every possible mistake on this to now--thats what i get for hurrying

    we get f = Ae^t + Be^-t

    f ' (0) yields A+B =1

    Substituting into original equation yields -4AB =1

    Now solve --solutions are when A = (2^1/2+1)/2 or (1-2^1/2)/2

    with corresponding values for B = 1 -A

    Sorry for the mistakes
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Calculus26 View Post
    Ok I've made every possible mistake on this to now
    i'm sure you can find more mistakes to make if you tried

    Sorry for the mistakes
    it happens to the best of us, don't worry about it. mistakes were corrected before they caused much damage.

    did you double check your solutions for A?

    i'm not going to check them, i trust you
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  11. #11
    MHF Contributor Calculus26's Avatar
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    I rechecked for both A and B in the original both by hand and with Mathcad to make sure-- thanx for the kind words

    But still I always told my students to slow down --Calculus is more about patience than intelligence I'd say--too bad I don't follow my own advice.

    That's what happens a year removed from teaching--skills are slipping
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  12. #12
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    I find f(x) = \frac{e^x - e^{-x}}{2} I'm wrong or you are Calculus26( f ' (0) yields A+B =1 false we have f ' (0) yields A-B =1) thanks you guys xxx
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  13. #13
    MHF Contributor Calculus26's Avatar
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    Damn--- you're right using the initial condition A - B = 1 not A+B =1

    as I said before--my solution satisfies the equation which is where I checked my results but not the initial conditions.

    Now I quit

    3 mistakes on one problem
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  14. #14
    MHF Contributor Calculus26's Avatar
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    By the way not that it matters that much but


    is also known as the hyperbolic sine and written

    = sinh(x)

    You may see the answer written that way

    Thanx for your kindness amongst my blunders I should have stopped after

    describing the process.
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  15. #15
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Calculus26 View Post
    Thanx for your kindness amongst my blunders I should have stopped after

    describing the process.
    i told you you could find more mistakes if you tried

    that was mean. i am only messing with you Calc. but yes, if you find you made 2 blunders, just resort to describing the process and letting the OP do the groundwork. this comes with experience on forums

    take care
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