Thread: 2nd order ODE

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.

Please show your working so that it can be reviewed.

3. Originally Posted by mr fantastic Please show your working so that it can be reviewed.
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.

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

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.
Not sure where your mistake is but this is what I got after simplying

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

Search Tags

2nd, ode, order 