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

[johann1301]

Homework Statement

Show that a solution to y' = y(6-y) has a turning point when y = 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?

 

[Orodruin]

For that you have to check the lowest order non-vanishing derivative of y' (i.e., in this case y''').

 

[johann1301]

i have to differentiate y''(x) = 6 y'(x) x' - 2 y(x) y'(x) x' ?

 

[epenguin]

Homework Statement

Show that a solution to y' = y(6-y) has a turning point when y = 3.

The Attempt at a Solution

If y has a turning point, then y'' = 0.

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.

 

[johann1301]

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.

 

[johann1301]

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 ?

 

[vela]

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.

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.

A turning point is where y'=0 and y' changes sign. An inflection point is where y''=0 and y'' changes sign.

 

[johann1301]

then i mean inflection!

 

[vela]

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 ?

That's fine. Presumably, you're differentiating with respect to $x$, so $x'=1$. Now you need to differentiate one more time to find $y'''$.

 

[epenguin]

A turning point is where y'=0 and y' changes sign. An inflection point is where y''=0 and y'' changes sign.

Sure, sorry. :s

 

[Mark44]

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.

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

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.