Hi,
I'm having issues with this (I'm sure very basic) example:
y'' - y' + 2y^2(1 - y) = 0
Find a value of the constant a s.t. y = (1 + e^(ax))^-1 is a solution of this equation.
I assumed I would be able to work out y'' and y', plug into the equation and solve for a but this doesn't seem to work.
Any ideas please?
y = (1 + e^(ax))^-1
y' = -ae^(ax)(1 + e^(ax))^-2
y'' = -a(^2)e^(ax)(1 + e^(ax))^-2 + 2(a^2)e^(2ax)(1 + e^(ax))^-3
Plugging this into the equation I get
-a(^2)e^(ax)(1 + e^(ax))^-2 + 2(a^2)e^(2ax)(1 + e^(ax))^-3 + ae^(ax)(1 + e^(ax))^-2 + 2(1 + e^(ax))^-2(1 - (1 + e^(ax))^-1) = 0
Sorry about the clumsiness.
I have tried different ways of manipulating this but I keep ending up with something like
e^(ax)(-(a^2) + a + 2) + e^(2ax)(a^2 + a) = 0
which I don't know how to solve.