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Differential Equation (Very peculiar initial condition)

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve: [itex]dy/dy = y^3[/itex] given the initial condition y(0)=0

    3. The attempt at a solution
    [itex]\int \frac{dy}{y^3} = \int dt[/itex]
    [itex]\frac{-1}{2y^2} = t + c[/itex]
    [itex]y^2 = \frac{-1}{2(t+c)}[/itex]
    [itex]y = ± \sqrt{ \frac{1}{2(-t-c)}}[/itex]

    This equation isn't going to support the initial condition, so can someone tell me where I've gone wrong in calculating the integral?
     
  2. jcsd
  3. Mar 28, 2012 #2
    You have:

    [tex]\frac{dy}{dt}=y^3,\quad y(0)=0[/tex]

    so immediately I want to divide out by [itex]y^3[/itex]. But I can do that only if [itex]y\neq 0[/itex]. Poke-a-poke then. However, if [itex]y=0[/itex], then I can't. But if y=0, then what?
     
  4. Mar 28, 2012 #3
    So if y'=y^3 = 0, then clearly y=0, and it is the constant zero, which makes sense since such a function has y'=0 as well.
     
  5. Mar 28, 2012 #4
    I think you can say that a little better: if y=0, then [itex]\frac{dy}{dt}=0[/itex] so that [itex]y=k[/itex] is a solution to the DE so that the solution to the IVP:

    [tex]y'=y^3,\quad y(0)=0[/tex]

    is y=0.
     
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