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If dy/dt = ky and k is a nonzero constant, y could be

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    If dy/dt = ky and k is a nonzero constant, than y could be

    a. 2e^kty
    b. 2e^kt
    c. e^kt + 3
    d. kty + 5
    e. .5ky^2 + .5

    Correct answer is b. 2e^kt

    2. Relevant equations



    3. The attempt at a solution

    I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 2, 2010 #2
    Look at the derivative of e^kt and see how that relates to e^kt.
     
  4. May 2, 2010 #3

    rock.freak667

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    Homework Helper

    The 'correct' thing to do would be

    dy/dt = ky

    ∫ dy/y = ∫ k dt


    But since you have a multiple choice, you could quickly differentiate each one if you wanted to.
     
  5. May 2, 2010 #4
    Hm.. this is what I got when I did ∫ dy/y = ∫ k dt

    lny = kt
    y = e^kt

    Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.
     
  6. May 2, 2010 #5

    rock.freak667

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    Homework Helper

    you'd get

    ln y = kt+A such that y=ekt+A=Bekt where B is a constant. So B can be any number, such as '2'.
     
  7. May 2, 2010 #6
    Ah, I forgot about that. Thanks!
     
  8. May 2, 2010 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If this means y=-2e^(kty) then its derivative is dy/dt= 2ke^(kty)(ky+ kt(dy/dt)).
    dy/dt(1- 2kt e^(kty)= 2k^y e^(kty). No, that differential equation is not satisfied.

    If y= 2e^(kt), then dy/dt= 2(ke^(kt))= k(2e^(kt))= ky and satisfies the equation.
    (I am assuming the parentheses combine these as you wanted- that you did not intend 2(e^k)t.)

    If y= e^(kt)+ 3, then y'= ke^(kt) which is not ky= k(e^(kt)+ 3)

    If y= kty+ 5, then dy/dt= ky+ kt dy/dt so dy/dt(1- kt)= ky and dy/dt= (k/(1-kt) y. That does not satisfy the given equation.

    If y= .5y^2+ .5, then dy/dt= 1.0 y dy/dt so dy/dt(1- y)= 0. Either dy/dt= 0 or y= 1. That does not satisfy the given equation.

     
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