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Some calc 2

  1. Feb 26, 2007 #1
    I have to solve the differential equation and let C represent an arbitrary constant.


    I cant seem to figure it out i wind up getting z=-2e^(t^2)+e^(C)
  2. jcsd
  3. Feb 26, 2007 #2
    Ok, divide by [tex]e^z[/tex] on both sides to get

    [tex]e^{-z} \frac{dz}{dt} = -e^t [/tex]

    then integrate with respect to t,

    [tex]-e^{-z} = k - e^t [/tex]

    re-arrange to get

    [tex]z(t) = - \ln{(k + e^t)}[/tex].

    (k is a constant).
    Last edited: Feb 26, 2007
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