solving equation in matlab

Hi. I want to find the l and mu , by the code below. I have linked the image that indicates l and mu how they used in equation. Could some one give me some hints , that the could below , could be optimal in speed ? Any suggestions would be appreciated.

https://docs.google.com/open?id=0B4Coq3-GewH9YTJMaU02bmFaN2s

disp( 'Identifikation: Berechnung der Faltungsprodukte ...' )
par.x1 = Fcn_myconv2( {v0, v0} ) ;%y0y0
par.x2 = Fcn_myconv2( {v1, v0} ) ;%y1y0
par.x3 = Fcn_myconv2( {v0, -1*time.*v1} ) ;%y1^'y2
par.x4 = Fcn_myconv2( {-1*time.*v0, v1} ) ;%y2^'y1
par.x5 = Fcn_myconv2( {v1, v1} ) ;%y1y1

% Berechnung der Signale
disp( 'Identifikation ...' )
dif.y0 = - 1*Fcn_integ( par.x3, 1, dt ) - 1*Fcn_integ( par.x2, 2, dt ) + 1*Fcn_integ( par.x4, 1, dt ) ;
dif.y1 = - 1*Fcn_integ( par.x5, 0, dt ) ;
dif.y2 = + 1*Fcn_integ( par.x1, 2, dt ) ;

% Integration -- Gleichungssystem
factor1 = zeros( 2, length(v0) ) ;

factor1(1,:) = Fcn_integ( dif.y1, 0, dt ) ;
factor1(2,:) = Fcn_integ( dif.y1, 1, dt ) ;
factor2 = zeros( 2, length(v0) ) ;

factor2(1,:) = Fcn_integ( dif.y2, 0, dt ) ;
factor2(2,:) = Fcn_integ( dif.y2, 1, dt ) ;
rhs = zeros( 2,length(v0) ) ;

rhs(1,:) = Fcn_integ( dif.y0, 0, dt ) ;
rhs(2,:) = Fcn_integ( dif.y0, 1, dt ) ;

% Loesen des Gleichungssystem
x = zeros(length(v0),2) ;

for k = 1:length(v0)

      A = [ factor1(:,k) factor2(:,k)];

      b = -rhs(:,k);

      x(k,:) = A\b ;

end

[Walter Roberson]

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