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Homework Help: Solve a differential equation for the initial conditions,

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve y' = (y^3)(t^2) for the initial condition y(0)=0 and state in which interval in 't' this solution exists.

    3. The attempt at a solution

    First I divided both sides by y^3 and then subtracted t^2 from each as well,

    -t^2 + y'(y^-3) = 0

    then solved,

    (-t^3)/3 + (y^-2)/-2 = C

    and then uh oh for y = 0 so chugging ahead and leaving 'C' alone for now,

    y^-2 = (C + (t^3)/3)/-2


    y = [-2/(C+(t^3)/3)]^(1/2)

    Therefore (C + (t^3)/3) ≤ 0

    I am sure (as usual) I am missing something simple :P
  2. jcsd
  3. Sep 29, 2013 #2
    I cannot follow this. According to the question, y is zero when t is zero. What do you get for C?
  4. Sep 29, 2013 #3


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    For [itex]y(0) \neq 0[/itex] it is the case that the solution only exists when [itex]3C + t^3 < 0[/itex]. But for [itex]y(0) = 0[/itex] that method doesn't work at all. And in fact [itex]y(t) = 0[/itex] is a solution of [itex]y' = y^3t^2[/itex] which satisfies [itex]y(0) = 0[/itex].
  5. Sep 29, 2013 #4
    Sorry, let me explain where I am coming from. We have,

    (-t^3)/3 + (y^-2)/-2 = C

    With the (1/-2y^2) term I get infinity for 'C'. That's my sticking point, so obviously I',ve goofed this up and there is another way to solve for 'C'?
  6. Sep 29, 2013 #5
    I'm sorry but I am not quite following what you are saying. Should I not solve for the general form?
  7. Sep 29, 2013 #6


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    You've done that, and shown that there is no choice of C for which y(0) = 0. At that point you need to find another approach.

    In any event, if you have the initial condition [itex]y(0) = D[/itex], and it is obvious on inspection that [itex]y' = 0[/itex] when [itex]y = D[/itex], then the solution is [itex]y(t) = D[/itex] and you can save yourself some work.
  8. Sep 29, 2013 #7
    It's the not working in the general form that is bothering me but I think I have this one figured out.
    Here is what I sent to my Professor,

    I feel satisfied with this although I also like the explanation I just saw from the Professor.
    Last edited: Sep 30, 2013
  9. Sep 29, 2013 #8

    What (s)he was trying to state is that y = 0 is an equilibrium solution of the differential equation. What would this imply about the equation?
    Last edited: Sep 30, 2013
  10. Sep 30, 2013 #9
    I understand what they were getting at now, ignoring that we didn't have a solution for the general form in order to pursue the differential equation is where I was getting stuck. Both are solved at this point so I am satisfied unless someone else has something to add?
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