Finding inflection points

[ssb]

Homework Statement

For many differential equations, the easiest way to find inflection points is to use the differential equation rather than the solution itself. To do this, we can compute $$y''$$ by differentiating $$y'$$, remembering to use the chain rule wherever $$y$$ occurs. Next, we can substitute for $$y'$$ by using the differential equation and setting $$y' = 0$$. Then we can solve for $$y$$ to find the inflection points. (Keep in mind here that solving for $$y$$ can also produce some equilibrium solutions, which may not be inflection points!)

Use the technique described above to find the inflection point for the solutions of the differential equation


$$y'=r(1-\frac{y}{L})y$$

your answer may contain $$L$$ and $$r$$



$$y = ?$$

The Attempt at a Solution

I differentiated the given equation and set it equal to zero, then I solved it for y. My answer was Lr/4 but this is wrong according to webworks.

The equation I got when I differentiated $$y'=r(1-\frac{y}{L})y$$ was $$y'' = r-((4y)/L)$$

i know the answer is $$L/2$$ but I don't know how to get there.

 

[skeeter]

y' = ry(1 - y/L)
distribute ...
y' = ry - ry²/L
differentiate ...
y" = r - 2ry/L
set y" = 0 ...
r - 2ry/L = 0
r(1 - 2y/L) = 0
1 = 2y/L
y = L/2