# Thread: homogeneous linear differential equations

1. ## homogeneous linear differential equations

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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. #2y'' - y' + y^2*(1 - y) = 0

3. ## Correction

Originally Posted by AkilMAI
#2y'' - y' + y^2*(1 - y) = 0
This is the type of equation which is based only on the y variable (btw, this equation isn't homogeneous).

Pardon me. I was thinking about another meaning of homogeneous.

4. 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.....
it is of the type y(n) + an−1(x)y(n−1) + · · · + a1(x)y0 + a0(x)y = 0 ...where the coefficients are now allowed to depend on x. Equations of this form have the
property that if y1 and y2 are solutions, then so is Ay1 + By2 for any constants A and
B, so that solutions can be built up by taking linear combinations....the above equations is similair since it equals to 0

5. I got (2a^2+a)*e^(ax)+(-2a^2+a+1)=0 and I get stuck here...and it's linear homogeneous wonderboy1953

6. Originally Posted by AkilMAI
I got (2a^2+a)*e^(ax)+(-2a^2+a+1)=0 and I get stuck here...and it's linear homogeneous wonderboy1953
And second order. Have you tried an auxilary equation?

7. 2y'' - y' + y^2*(1 - y) = 0.....2y'' - y =>2a^2 -a but how can I transform the y^2*(1 - y) into an auxilary equation?.....

8. found the result...a=-1/2