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Homework Help: About differential question

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    solve dy/dt+y^3=0 , y(0)=y0 and determine how the interval in which the solution exists depends on the initial value y0


    2. Relevant equations
    Theorem: Let the fuctions f and df/dy(partial differentiation) be continuous in some rectangle a<t<b , c<y<d containing (t0,y0). Then in some interval t0-h<t<t0+h contained in a<t<b, there is a unique solution y=∅(t) of the initial value problem dy/dt=f(t,y), y(t0)=y0

    3. The attempt at a solution
    According to the theorem above, by solving dy/dt+y^3=0 y should only have one solution (y) in some interval.

    dy/dt+y^3=0

    i) when y≠0
    -y^(-3)dy=dt and integrate both sides
    0.5y^(-2)=t+c (c is a constant)

    by initial condition y(0)=y0
    0.5y^(-2)=t+0.5y0^(-2) and multiply both sides by 2
    y^(-2)=2t+y0^(-2)
    y^2=1/(2t+y0^(-2))
    =y0^2/(2ty0^(2)+1)
    Thus, y= ±y0/√(2ty0^(2)+1) (∞>t>-1/2y0^2 since numbers inside root must be bigger than 0)

    ii) y=0
    y=0 holds true for all t.

    Then there is a contradiction
    1. y has three solutions when ∞>t>-1/2y0^2 which are y= ±y0/√(2ty0^(2)+1), y=0.

    did i do something wrong?
     
    Last edited: Mar 24, 2012
  2. jcsd
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