DE homework problem

  1. Consider dy/dt = 2*(abs(sqrt(y)))

    1.Show that y(t)=0 is a solution for all t.
    I did this part

    2.Find all solutions (hint, give solution like y(t)=... for t>=0, y(t)=... t<0).
    He told us in class that t=0 isn't necessarily the point we should be concerned with

    3.Why doesn't this contradict the uniqueness theorem?
    I have a feeling it's because our DE isn't differentiable at y=0, but my main problem is number 2.

    I graphed this DE on the computer, so assuming I typed it in right I know what it looks like. I also tried splitting the DE up into cases for part 2, but it seems that I would have to perform the same integral twice which doesn't really make sense.
  2. jcsd
  3. TD

    TD 1,021
    Homework Helper

    Could you show your work then? There should be a small difference...
    Remember the definition of the absolute value.
  4. dy/dt = 2*sqrt(abs(y)) = 2*sqrt(y) y>=0, 2*sqrt(-y) y<0

    isn't 2*sqrt(-y) when y<0 = 2*sqrt(y)
  5. I have y(t) = (t-C)^2 when y>=0. I get the same thing when y<0 as well, by seperation of variables. I use t-C rather than t+C thanks to a hint from my professor from yesterday's lecture. So, is this the solution I am looking for?
  6. TD

    TD 1,021
    Homework Helper

    The initial problem was "abs(sqrt(y))" and now you write "sqrt(abs(y))", which one is it?
  7. HallsofIvy

    HallsofIvy 41,256
    Staff Emeritus
    Science Advisor

    "isn't 2*sqrt(-y) when y<0 = 2*sqrt(y)"

    No, it's not. For example if y= -4, 2*sqrt(-y)= 2*sqrt(4)= 4 but
    2*sqrt(y)= 2*sqrt(-4)= 4i.
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