Problem with a differential equation

[johann1301]

Homework Statement

Show that a solution to y'=y(6-y) has a an inflection point at y=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 don't get is why the book states that y' is a factor in the calculation:

(y(6-y))'y'=(6y-y²)'y'=(6-2y)y'

 

[td21]

only showing the second derivative equal to zero is not sufficient to prove that that point is a point of inflection.

 

[johann1301]

only showing the second derivative equal to zero is not sufficient to prove that that point is a point of inflection.

Ok, can you please elaborate? What other sort of point could it be?

 

[LCKurtz]

Homework Statement

Show that a solution to y'=y(6-y) has a an inflection point at y=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 don't get is why the book states that y' is a factor in the calculation:

(y(6-y))'y'=(6y-y²)'y'=(6-2y)y'

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.

 

[johann1301]

Ok, I am going to 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?

only showing the second derivative equal to zero is not sufficient to prove that that point is a point of inflection.

You still have to observe why ##y''## changes sign at such a point, giving a change in concavity, and therefore giving an inflection point.

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, I am strugguling to see what other sort of point it could be? How can it not be a point of inflection? Are there any examples?

 

[LCKurtz]

Ok, I am going to 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, I am strugguling to see what other sort of point it could be? How can it not be a point of inflection? Are there any examples?

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.

 

[johann1301]

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.

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?

 

[LCKurtz]

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##?

 

[johann1301]

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##?

Since i apparently don't understand the chain rule as well as i thought, i don't really know if its possible to argue that it changes sign. I certainly can not, but if it is possible? I don't know... I am only a rookie when comes to differential equations. Perhaps its possible if you solve it?, which isn't something i know how to do yet...

If it is possible, the problem isn't flawed, only the solution given in the book?

 

[LCKurtz]

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##.

 

[johann1301]

Ok, let's 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'' wouldn't 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.

 

[LCKurtz]

Yes. That's a bit wordy, but you have the correct argument.

 

[johann1301]

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?

 

[LCKurtz]

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?

Yes.