# Homework Help: Problem with a differential equation

1. Jul 20, 2014

### johann1301

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. Jul 20, 2014

### td21

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

3. Jul 20, 2014

### johann1301

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

4. Jul 20, 2014

### LCKurtz

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.

5. Jul 20, 2014

### johann1301

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?

6. Jul 20, 2014

### LCKurtz

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.

7. Jul 20, 2014

### johann1301

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?

8. Jul 20, 2014

### 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 $y=3$?

9. Jul 20, 2014

### johann1301

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?

10. Jul 20, 2014

### 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 $y=3$.

11. Jul 21, 2014

### johann1301

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.

12. Jul 21, 2014

### LCKurtz

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

13. Jul 21, 2014

### 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?

14. Jul 21, 2014

Yes.