# 2nd order ODE

• Jan 25th 2010, 12:48 PM
s0791264
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.

• Jan 25th 2010, 01:41 PM
mr fantastic
Quote:

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.

• Jan 25th 2010, 01:55 PM
s0791264
Quote:

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.
• Jan 26th 2010, 06:32 AM
Jester
Quote:

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

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