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Probably simple

  1. Feb 24, 2008 #1
    I am working on a diff. eq. assignment, and have encountered the simple, seperable differential equation:

    dy/dt = y^3 with the initial condition y(0)=1

    I am posting in this forum, because I feel my trouble is with my integration.

    When seperated, I am integrating this equation as [tex]\int1/y^3[/tex]=[tex]\int1dt[/tex].

    After doing so, I end up with: -1/2y^2=t+C

    After solving for y, I come up with: y=[tex]\sqrt{1/2t}[/tex]+C

    The only problem with this is I cannot solve for C, since the initial condition would require division by zero.

    Anyone have an clue as to where I went wrong?
  2. jcsd
  3. Feb 24, 2008 #2
    Just solve for the constant here: -1/2y^2=t+C.
  4. Feb 24, 2008 #3
    If I do it then, I end up with C=-1/2y^2-1 which when plugged back into the general solution, results in y disappearing.

    Can't be right..
  5. Feb 24, 2008 #4
    Look, you are not followint Mystic998's point and suggestion at all!
    YOu have the initial condition y(0)=1, right?
    so as Mystic998 suggested

    -1/2y^2=t+C.=> -1/2 *1^2=c=> c=-1/2
    and remember y(0)=1, means the value of y when t=0
  6. Feb 24, 2008 #5
    I'm an idiot, thank you.
  7. Feb 24, 2008 #6
    Oh no, i am sure you are not, it happens to all of us sometimes not to notice these minor things!!
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