1. ## 2nd order ODE

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.

2. Originally Posted by s0791264
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.

3. Originally Posted by mr fantastic
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

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.

4. Originally Posted by s0791264
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

$\displaystyle \frac{(a+1)e^{ax}(ae^{ax}-a+2)}{\left(e^{ax}+1\right)^3}.$