1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Y' = y(6-y) has a turning point when y = 3.

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that a solution to y' = y(6-y) has a turning point when y = 3.

    3. The attempt at a solution
    If y has a turning point, then y'' = 0. I find that y'' = 6 - 2y. If i solve 0=6-2y i get y =3.

    But how do i know that this is a turning point? yes, y'' equals zero, but don't i have to know that y'' changes sign when 'passing' zero? It could be that the solution is for instance concave, then linear, then concave again as x increases? When the function/solution is linear, the y'' is zero, right?
     
  2. jcsd
  3. Dec 4, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For that you have to check the lowest order non-vanishing derivative of y' (i.e., in this case y''').
     
  4. Dec 4, 2014 #3
    i have to differentiate y''(x) = 6 y'(x) x' - 2 y(x) y'(x) x' ?
     
  5. Dec 4, 2014 #4

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Has the feel of a misquoted problem, am I right? Anyway recall that at a turning point it is y' that is 0 and y'' changes sign.
     
  6. Dec 4, 2014 #5
    When i write turning point, i mean where the function changes from concave to convex, or vica versa. In my native language we call this point a turning point.
     
  7. Dec 4, 2014 #6
    first of all...

    if i start with
    y'(x) = 6y(x) -(y(x))^2

    and try to find y''(x) i get...

    y''(x) = (6y(x))' -((y(x))^2)'

    y''(x) = 6y'(x)x' - 2y(x)y'(x)x'

    y''(x) = (6- 2y(x))(y'(x)x')

    If y''(x) = 0 the y'(x)x' = 0 or 6- 2y(x) = 0

    Is this even correct ?
     
  8. Dec 4, 2014 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    A turning point is where y'=0 and y' changes sign. An inflection point is where y''=0 and y'' changes sign.
     
  9. Dec 4, 2014 #8
    then i mean inflection!
     
  10. Dec 4, 2014 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's fine. Presumably, you're differentiating with respect to ##x##, so ##x'=1##. Now you need to differentiate one more time to find ##y'''##.
     
  11. Dec 4, 2014 #10

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Sure, sorry. :s
     
  12. Dec 4, 2014 #11

    Mark44

    Staff: Mentor

    A possibility that no one has mentioned is to just go ahead and solve the differential equation, assuming that you know a little about differential equations. The equation in this problem is separable, and is fairly easy to solve.

    By separable, I mean that it can be rewritten as ##\frac{dy}{y(6 - y)} = dx##. Split the fraction on the left side using partial fraction decomposition, and then integrate both sides.

    This notation is really cumbersome, IMO. Since it's pretty obvious that x is the independent variable, all of the y(x), y'(x) and similar terms can be written more clearly as just y, y', and so on.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Y' = y(6-y) has a turning point when y = 3.
Loading...