Hello, I'm hoping for some help with the following question.

Show that $\displaystyle u = \frac{1}{(1+exp[bz])^{2}} $ for b > 0 solves the following equation for a particular value of b:

$\displaystyle u'' + cu' + u(1-u) = 0 $ where $\displaystyle u' = \frac{du}{dz} $ etc.

I've got to the following point and don't know where to go from here:

$\displaystyle (4-2b^{2}-2cb) + e^{bz}(6+2b^{2}-2cb) + e^{2bz}(4b^{2}+4) + e^{3bz} = 0 $

This question's only worth about 7 marks out of 100 so I don't think it should take too long but I'm stumped!

Any help would be much appreciated