Problem with fsolve function in my equation systems

I'm a beginner in Matlab and I try to calculate this system:

equagenral1.m:

function F = equageneral1(x,lambda,mu,B,c,b,a,n,e,z,R,y)

F=[

     x(3)+(lambda*(mu*x(2)-x(1)-B)+c)/(b+lambda)

     x(2)+((a*n*(e+z*((1/((2*pi)^(1/2)))*y)))+lambda*mu*(x(3)-x(1)-B)-c)/(a+lambda*(mu^2))
     -a*z*((1/((2*pi)^2))*(exp((-(-1.2-(R+x(1)))^2)/2)))*(x(2)+n*(e+z*((1/((2*pi)^(1/2)))*y)))+lambda*(mu*x(2)+x(3)-x(1)-B)+z*((1/((2*pi)^2))*(exp((-(-1.2-(R+x(1)))^2)/2)))

];

This is how I try to do this:

calcul.m:

R=16.09;

syms x

y=int(exp(((-1.2-(R+x(1))).^2)/2),0,+inf);

lambda = 1.7723;

mu = 0.50;

B = 21.6;

c = 1;

b = 9.0759;

a = 0.9657;

n = 1;

e = 0;

z = 0;

R=16.09;

Rss=0;

options=optimset('MaxFunEvals',1000);

[ex,fval]=fsolve(@(x)equageneral1(x,lambda,mu,B,c,b,a,n,e,z,R,y),[0 0 0],options);

But the answer tell me that FSOLVE requires all values returned by user functions to be of data type double.

How can I fix that?

Thank you!

Paul-André

[Walter Roberson]

What variable are you doing the int() with respect to? The function mentions x(1) within itself; is that the variable of integration? If it is then the integral is infinity.

But you pass y to equageneral1 which is written in terms of x(1) and x(2) and x(3). Is the x(1) inside of y intended to refer to the value of the vector x inside of equageneral1 ? If so then y should be defined as a function and inside equalgeneral1 should be used in function form instead of as a variable.

[Paulandre]

Yes x(1) inside of y refer to the value of the vector x inside of equageneral1.

y=int(exp(((t-(R+x(1))).^2)/2)dt,0,+inf), where t=-1.2

If I put this equation in function form instead of as a variable, an other error appear.

How can I insert the int in equagenral1 without error?

Thank for helping me

Pa

[Paulandre]

You can also try like this:

equageneral1.m:

function F = equageneral1(y,lambda,mu,B,c,b,a,n,e,z,R,M) F=[

   y(3)+(lambda*(mu*y(2)-y(1)-B)+c)/(b+lambda)
   y(2)+((a*n*(e+z*((1/((2*pi)^(1/2)))*M)))+lambda*mu*(y(3)-y(1)-B)-c)/(a+lambda*(mu^2))
   -a*z*((1/((2*pi)^2))*(exp((-(-1.2-(R+y(1)))^2)/2)))*(y(2)+n*(e+z*((1/((2*pi)^(1/2)))*M)))+lambda*(mu*y(2)+y(3)-y(1)-B)+z*((1/((2*pi)^2))*(exp((-(-1.2-(R+y(1)))^2)/2)))
];

and calcul:

calcul.m:

R=16.09;

syms y

x=-1.2;

M=int(exp(((x-(R+y(1))).^2)/2),0,+inf);

lambda = 1.7723;

mu = 0.50;

B = 21.6;

c = 1;

b = 9.0759;

a = 0.9657;

n = 1;

e = 0;

z = 0;

R=16.09;

Now i think that int() the variable of the integration is now x. But I have the same error.