Not sure if this is the right place to post this cos its off my module called Applied Maths but seemed more of a calculus topic...

$\displaystyle y^{''} + y^{'} + 2y^2 (1 - y) = 0$

Find the value of the constantaso that $\displaystyle y = (1 + e^{ax})^2$ is a solution of the above equation.

Now i just plugged the y = ... bit into the equation but it ended as

... $\displaystyle = \frac{1}{(1 + e^{ax})^2} (1 - a^2 e^{ax} - \frac{2}{(1 + e^{ax})} + \frac{2a^2 e^{2ax}}{(1 + e^{ax})})$

and figured something had gone wrong somewhere as i cant solve it from there.

Any help appreciated