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Problem with a differential equation

  1. Jul 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that a solution to y'=y(6-y) has a an inflection point at y=3.

    3. The attempt at a solution

    If y has an inflection point, then y''=0. I know that y'=y(6-y), and therefore i know that y''=(y(6-y))'=(6y-y2)'=6-2y

    So, if y''=0, and y''=6-2y then 0=6-2y => y=3.

    Solved.

    But the answer in the back of my book writes the following:

    "At the inflection point, y''=0. We derive each side in the equation. When we derive the right side - according to the chain rule - we should get:

    (y(6-y))'y'=(6y-y2)'y'=(6-2y)y'

    If y=3, then both the right side and thus y'' equals zero"

    What i dont get is why the book states that y' is a factor in the calculation:

    (y(6-y))'y'=(6y-y2)'y'=(6-2y)y'
     
    Last edited: Jul 20, 2014
  2. jcsd
  3. Jul 20, 2014 #2

    td21

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    only showing the second derivative equal to zero is not sufficient to prove that that point is a point of inflection.
     
  4. Jul 20, 2014 #3
    Ok, can you please elaborate? What other sort of point could it be?
     
  5. Jul 20, 2014 #4

    LCKurtz

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    Remember that the independent variable is, for example, ##x## and ##y'=\frac{dy}{dx}##. So to calculate ##y''## you need the chain rule ##y''=\frac {dy'}{dx} = \frac{dy'}{dy}\frac{dy} {dx}##. That is why ##y''=(6-2y)y'## which tells you that any point on a solution with ##y## coordinate of ##3## might be an inflection point. You still have to observe why ##y''## changes sign at such a point, giving a change in concavity, and therefore giving an inflection point.
     
  6. Jul 20, 2014 #5
    Ok, im gonna try to understand the part about the chain rule in a bit, but first:

    why is it not certain that it is a point of inflection?
    If i understand you right; even though the second derivative of a function is equal to zero, it does not necesseraly mean that there is a point of inflection in the original function? If this is the case, im strugguling to see what other sort of point it could be? How can it not be a point of inflection? Are there any examples?
     
  7. Jul 20, 2014 #6

    LCKurtz

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    Sure. Look at the graph of ##y = x^4## at ##x=0##. The first and second (and third) derivatives are zero but the graph doesn't have a change of concavity there. Unlike ##y=x^3##, which does.
     
  8. Jul 20, 2014 #7
    Ok, no i see how it is possible.

    Does this mean that the original problem is flawed, or at least imperfect, since the solution in the book only uses that y''=0 to show that it has a point of inflection?

    In other words, is this (see below) a sufficient solution to the problem?

    "At the inflection point, y''=0. We derive each side in the equation. When we derive the right side - according to the chain rule - we should get:

    (y(6-y))'y'=(6y-y2)'y'=(6-2y)y'

    If y=3, then both the right side and thus y'' equals zero"


    or perhaps since the book uses the chain rule in a way i did not do, it is sufficient?
     
  9. Jul 20, 2014 #8

    LCKurtz

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    No, it is not sufficient as I pointed out. You have to show a change of concavity which is a change of sign of ##y''##. But you have ##y'' = (6-2y)y'##. Can you argue from this and from what you know about ##y'## that ##y''## changes sign as ##y## passes through ##y=3##?
     
  10. Jul 20, 2014 #9
    Since i apparently dont understand the chain rule as well as i thought, i dont really know if its possible to argue that it changes sign. I certainly can not, but if it is possible? I dont know... I am only a rookie when comes to differential equations. Perhaps its possible if you solve it?, which isnt something i know how to do yet...

    If it is possible, the problem isnt flawed, only the solution given in the book?
     
  11. Jul 20, 2014 #10

    LCKurtz

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    The problem isn't flawed. You have formulas for both ##y'## and ##y''##. Look at them. It isn't difficult to see if ##y''=0## when ##y=3## or whether ##y''## changes sign when ##y=3##.
     
  12. Jul 21, 2014 #11
    Ok, lets see if this is sufficient...

    If y has an inflection point when y=3, then y''=0 when y=3, and y'' has to change sign when y passes the value 3.

    First i find an expression for y'' using the chainrule:
    Since i know that y'=y(6-y)=6y-y2. Then y''=(6y-y2)'=(6y)'-(y2)'=6y'-2yy'=(6-2y)y'

    So the expression for y'' is (6-2y)y'. If i assume that y does have an inflection point, then y''=0, and therefore

    0=(6-2y)y' => y'=0 or y=3.

    If y'=0 then y=0 or y=6. Since these values are different than the value i am trying to prove, i discharge them. I only use y=3.

    So y'' does in fact equal 0 when y equals 3!

    But does y'' change sign when y passes the value 3?

    Lets look at the expression y''=(6-2y)y' which i found earlier...

    Its easy to see that the factor (6-2y) changes sign when y passes the value 3, but what if y' altso changes sign when y passes 3? then y'' wouldnt change sign at all...

    If we look at the original equation y'=y(6-y)=6y-y2 we see that y' only changes sign when y=0 or Y=6, not when y=3.

    Therefore we know that y'' changes sign when y passes 3. Since we altso know that y=3 when y''=0, we have shown that a solution to y'=y(6-y) has a an inflection point at y=3.
     
  13. Jul 21, 2014 #12

    LCKurtz

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    Yes. That's a bit wordy, but you have the correct argument.
     
  14. Jul 21, 2014 #13
    Thanks for the help. I understod why y''=(6-2y)y' as soon as i looked at y as a function rather then as a simple variable.

    But can I still call y a variable? Even though it is a function? Is it called a dependent variable?
     
  15. Jul 21, 2014 #14

    LCKurtz

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    Yes.
     
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