MHB Inflection Point of the Threshold Equation

[MacLaddy1]

Hello all,

This is probably a simple question, but for some reason no matter how much I fight these figures I can't seem to make it work out correctly.

For many DE's the easiest way to pinpoint inflection points is not from the solution but from the DE itself. Find y'' by differentiating y', remembering to use the chain rule wherever y occurs. Then substitute for y' from the DE and set y''=0. Solve for y to find the inflection points (sometimes in terms of t).

$$y' = -r(1-\frac{y}{T})y$$
$$y' = -ry + \frac{ry^2}{T}$$
$$y'' = -ryy' + \frac{2ryy'}{T}$$
$$0 = -ryy' + \frac{2ryy'}{T}$$
$$ryy' = \frac{r}{T}2yy'$$
$$Tyy' = 2yy'$$
$$2 = T$$

So when T=2 the second derivative indeed equals zero. However, my book is telling me the solution is when $$y=\frac{T}{2}$$

I have substituted the first derivative back into the second, and it's a long drawn out process that never reveals the correct solution for me, and I didn't want to type it out if the solution is more obvious.

Let me know what you guys/gals think.

Thanks,
Mac

 

[MarkFL]

You are correct here:

$$y'=-ry+\frac{r}{T}y^2$$

You made an error in differentiating, you want instead:

$$y''=-ry'+\frac{2r}{T}yy'=ry'\left(\frac{2}{T}y-1 \right)$$

Can you finish?

 

[MacLaddy1]

You are correct here:

$$y'=-ry+\frac{r}{T}y^2$$

You made an error in differentiating, you want instead:

$$y''=-ry'+\frac{2r}{T}yy'=ry'\left(\frac{2}{T}y-1 \right)$$

Can you finish?

Yes, I believe I can. However, before I try would you mind showing me why that first y is canceled and the second isn't? If you could do it in Leibniz notation I would be even more appreciative. I get mixed up when dealing with Newton's notation.

Thanks,
Mac

** Just tried anyway. That made things a whole lot simpler. Thanks

 

[MarkFL]

It boils down to:

$$\frac{d}{dt}(y)=\frac{dy}{dt}=y'$$

and

$$\frac{d}{dt}\left(y^2 \right)=2y\frac{dy}{dt}=2yy'$$

 

[MacLaddy1]

It boils down to:

$$\frac{d}{dt}(y)=\frac{dy}{dt}=y'$$

and

$$\frac{d}{dt}\left(y^2 \right)=2y\frac{dy}{dt}=2yy'$$

So it still is wrt "t." Could you use the implicit formula to figure this out.
$$-\frac{F_t}{F_y}?$$ I suppose not.

 

[MarkFL]

I think the best way is to use ordinary differentiation with respect to $t$.

 

[MacLaddy1]

I think the best way is to use ordinary differentiation with respect to $t$.

Thanks again.