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Calc problem

  1. Jan 9, 2005 #1
    We have a set of problems for hw. I am stuck on 1 where I know the answer but cant seem to get it.

    If dy/dt=ky and k is a nonzero constant then y could be
    a. 2e^kty b. 2e^kt c. e^kt d. kty+5 e. 1/2ky^2 +1/2

    I know the answer is b but i cant get that answer
    Here is my work
    S=integral sign


    How do u get a 2 in there for choice b
    Last edited: Jan 9, 2005
  2. jcsd
  3. Jan 9, 2005 #2


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    Gold Member

    You forgot about the constant of integration:
    [tex]\frac{dy}{dt}=ky [/tex]
    [tex] \frac{dy}y=kdt [/tex]
    [tex] \int{\frac{dy}y} = \int{kdt} [/tex]
    [tex] \ln y = kt +C [/tex]
    [tex] e^{\ln y} = e^{kt + C} [/tex]
    [tex] y = e^{kt}\cdot e^C [/tex]
    eC is also a constant, so it can be written as C1 if you like.The value of C1 will depend on the initial conditions. Unless there's a typo in your answer list, I can see two answers that are of this form:
    b. [tex]y=2e^{kt}[/tex]
    and c. [tex] y = e^{kt} [/tex]

    I hope that helps.
  4. Jan 9, 2005 #3
    thanks forgot the C and yea choice c was a typo it should be e^kt +3
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