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Initial Value Problem / Differential Equation

  1. Mar 12, 2012 #1
    Thanks for clicking!

    So, I've got a problem here that I'm stuck on. I need to find the general solution to

    y' = (y3 + 6y2 + 9y)/9

    I found this to be

    ln|y| + (3/(y+3)) - ln|y+3| = x + c

    but I would appreciate it if you would check my work. Anywho, once I have the general solution I need to solve it for y(2) = -3. Obviously, plugging -3 in for y creates a problem in the second and third terms above.

    Is there a way to rewrite the general solution or solve it in such a way that it won't contain any undefined terms when evaluated like it does now?

    Thanks guys.

    EDIT: Shoot, sorry for the non-descriptive title. I accidentally clicked post before finishing it. I meant it to say "[Calc II] Initial Value Problem / Differential Equation." My bad.
     
    Last edited by a moderator: Mar 12, 2012
  2. jcsd
  3. Mar 12, 2012 #2
    I got the same general solution, what do you mean by undefined terms?
     
  4. Mar 12, 2012 #3
    I mean that when I plug -3 in for y, I get a zero in the denominator of the second term and ln(0) for the third term, both of which are undefined. I was wondering if there is any way to rewrite this or to solve this differently so -3 will not cause this discontinuity.
     
  5. Mar 12, 2012 #4

    tiny-tim

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    hi enkrypt0r! :smile:
    erm :redface:look at the equation! :rolleyes:

    if y = -3, what is y' ? :wink:
     
  6. Mar 12, 2012 #5
    Hi, Tim!

    I never thought to look at the original equation! This evaluates to 0 = 0. I'm afraid I don't know what this says about the differential equation. Does this mean that it's undefined, as my attempts to evaluate it have shown?
     
  7. Mar 12, 2012 #6

    tiny-tim

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    oooh :cry:think! :smile:

    if y' = 0, then … ? :wink:
     
  8. Mar 12, 2012 #7
    Ummm... I'm feeling stupid here, haha. Sorry. I'm brainstorming but coming up with nothing beyond the obvious. We could integrate both sides and see that y is a constant, but we already know that. I'm sorry, I'm at a loss.
     
  9. Mar 12, 2012 #8

    Mark44

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    If y = 0 or y = -3, then y' = 0, which says that y is a constant.

    In your work, the tacit assumption is that y != 0 and y != -3.
     
  10. Mar 12, 2012 #9
    Yeah, y=0 and y=-3 are "equilibria" solutions to the de. This sort of de is known as autonomous, because there are no t's on the right side of the equation. So, no matter what t is, the slope at y=0 and y=-3 is always zero.

    So, why can't we plug in to solve the initial value problem?
     
  11. Mar 12, 2012 #10
    If you look at the direction field, I think you might find that more than one solution curve would satisfy the initial condition. Thus, there is no particular value of c that would net you a unique solution curve. Someone correct me if this is wrong, because I have never seen this case before! That's my analysis though.
     
  12. Mar 13, 2012 #11

    Mark44

    Staff: Mentor

    For the initial value problem, y' = (1/9)(y3 + 6y2 + 9y), y(2) = -3, we see that y' = (1/9)y(y + 3)2.

    If y = -3, then y' = 0, so y(x) [itex]\equiv[/itex] C. Since y(2) = -3, C = -3, so the solution to the IVP is y(x) [itex]\equiv[/itex] -3. Are you saying that you think there are other solutions to this initial value problem?
     
  13. Mar 13, 2012 #12

    tiny-tim

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    hi enkrypt0r! :smile:

    (just got up :zzz:)

    the solution comes in three parts …

    y = -3 (for all t) is a solution

    y = 0 (for all t) is a solution

    and the other solutions are all the solutions of

    dy/(y3 + 6y2 + 9y) = dt/9

    (you multiplied by 0/0 when you got that equation, if y = -3 or 0, so you illegitmately lost those solutions :wink:)

    if you draw a graph of some specimen solutions, you'll probably find that they "asymptote up to" the vertical lines at -3 and 0,

    and (i'm guessing here :redface:) that one of y = 0 and v = -3 is stable and the other is unstable :smile:
     
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