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.