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Math Help - homogeneous linear differential equations

  1. #1
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    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*(a^2)*(e^(a*x)-1)+a*(e^(a*x)+1)-1=0...any advice?
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  2. #2
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    #2y'' - y' + y^2*(1 - y) = 0
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  3. #3
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    Correction

    Quote Originally Posted by AkilMAI View Post
    #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).

    What do your notes say about handling this type of equation?

    Pardon me. I was thinking about another meaning of homogeneous.
    Last edited by wonderboy1953; January 22nd 2011 at 11:32 AM. Reason: Correction
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  4. #4
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    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
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  5. #5
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    I got (2a^2+a)*e^(ax)+(-2a^2+a+1)=0 and I get stuck here...and it's linear homogeneous wonderboy1953
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  6. #6
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    Quote Originally Posted by AkilMAI View Post
    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?
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  7. #7
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    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?.....
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  8. #8
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    found the result...a=-1/2
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