Homogeneous linear differential equations

[AkilMAI]

Homework Statement

The equation
2y'' - y' + y2(1 - y) = 0;
where y' = dy/dx
and y'' = d^2y/dx^2
represents a special case of an equation used as a model
for nerve conduction, and describes the shape of a wave of electrical activity
transmitted along a nerve fibre.

Homework Equations

the task is to find a value for the constant "a" so that y = [1 + e^(ax)]^(-1) represents the solution of the equation

The Attempt at a Solution

y'=-a*e^(a*x)/[e^(a*x)+1]^2
y''=a^(2)*e^(a*x)[e^(a*x)-1]/[e^(a*x)+1]^3
Basically after I substitute the solution into
the equation and did some calculation I've reached this point
[2*(a^2)*e^(a*x)]*[e^(a*x)-1]+[a*e^(a*x)]*[e^(a*x)+1]-e^(a*x)/[e^(a*x)+1]^3=0=>
=>2*(a^2)*(e^(a*x)-1)+a*(e^(a*x)+1)-1=0...any advice?

 

[AkilMAI]

#2y'' - y' + y^2*(1 - y) = 0

 

[LCKurtz]

I didn't check your derivatives, but assuming you have them calculated correctly, here's what I would do.

Assuming there is a solution of the type you have, your last equation must hold for all x. So I would put x = 0 and see if I could solve the resulting equation for a. Then, without taking x = 0 put your value for a in your last equation and see if it works. If so, you are home free.

And, I think this method works

 

[Dick]

That's (2a^2+a)*e^(ax)+(-2a^2+a-1)=0. For that to be equal to zero the coefficient of e^(ax) must be zero and the other term must also be zero. Before you do that, I get the other term to be (-2a^2+a+1). Could you check that?

 

[AkilMAI]

:D,thanks for the reply ,I've tried that before I posted the problem...In this case I get a=1/2 and after I substitute =>e^(2*x)+e^(1/2*x)=0

 

[LCKurtz]

:D,thanks for the reply ,I've tried that before I posted the problem...In this case I get a=1/2 and after I substitute =>e^(2*x)+e^(1/2*x)=0

You mean 1/2 in that first exponent. I agree that a = 1/2, but I get a minus sign between the two terms.

 

[Dick]

You mean 1/2 in that first exponent. I agree that a = 1/2, but I get a minus sign between the two terms.

I'm getting a=(-1/2). I think there may be a sign error in your equation. See my previous post.

 

[AkilMAI]

no, I mean e^(x)+e^(1/2*x)=0...there is only one term with I minus and I used to cancel another term(mainly -e^(1/2*x)/2)
sorry Dick but I didn't quite understand what you where trying to say...

 

[LCKurtz]

I'm getting a=(-1/2). I think there may be a sign error in your equation. See my previous post.

I found an error in my calculations. I now agree with Dick a = -1/2 works.

 

[AkilMAI]

No ,I'm sorry it is [2*(a^2)*e^(a*x)]*[e^(a*x)-1]+[a*e^(a*x)]*[e^(a*x)+1]+e^(a*x)/[e^(a*x)+1]^3=0...it is plus at the end not minus,this justifies my above result

 

[Dick]

no, I mean e^(x)+e^(1/2*x)=0...there is only one term with I minus and I used to cancel another term(mainly -e^(1/2*x)/2)
sorry Dick but I didn't quite understand what you where trying to say...

I'm not trying to say anything profound. Your equation, (2a^2+a)*e^(ax)+(-2a^2+a-1)=0 doesn't have any solutions. I get (2a^2+a)*e^(ax)+(-2a^2+a+1)=0. Which does.

 

[AkilMAI]

Dick ...I also got (2a^2+a)*e^(ax)+(-2a^2+a+1)=0 ,I copied it wrong from the notebook...but here I get stuck

 

[Dick]

Dick ...I also got (2a^2+a)*e^(ax)+(-2a^2+a+1)=0 ,I copied it wrong from the notebook...but here I get stuck

If that's going to be zero for all x, you need to set (2*a^2+a)=0 and (-2a^2+a+1)=0. Is there a value of a that works in both?

 

[AkilMAI]

(2*a^2+a)=0=>a =-1/2...(-2a^2+a+1)=0=>a =1

 

[Dick]

(2*a^2+a)=0=>a =-1/2...(-2a^2+a+1)=0=>a =1

Quadratics generally have two roots. You are only showing one for each equation.

 

[AkilMAI]

a=-1/2

 

[Dick]

a=-1/2

Sure.