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This is probably a simple question, but for some reason no matter how much I fight these figures I cant seem to make it work out correctly.

$$y' = -r(1-\frac{y}{T})y$$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).

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

$$y'' = -ryy' + \frac{2ryy'}{T}$$

$$0 = -ryy' + \frac{2ryy'}{T}$$

$$ryy' = \frac{r}{T}2yy'$$

$$Tyy' = 2yy'$$

$$2 = T$$

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