Boundary conditions in ode

Hello, I have a question. It would be greatly appreciated if someone could help me out. How can I define these boundary conditions in the following code ? y'(80)=0 and y(C0)=t0.

syms y(x) x Y 
N=5;
r=0.05;
m=0.01;
p=1;
s=1;
t=0.1;
a=0.25;
b=0.25;
C0=1;
C=5;
f=(x+((x^2)+4*r*x*(1-a-b))^0.5)/(2*(1-a-b));
%t1=-(N+r+t*p*s-m);
Dy = diff(y);
D2y = diff(y,2);
t0= (((1-a)/a)*(1/r)^((a+b)/(1-a-b));
ode = y-(((1-a)/a)*(1/(r+f))^((a+b)/(1-a-b))+Dy*(C-x-m-(s^2))+0.5*D2y*(x^2)*(s^2)-(s^2)*x*Dy)/(r-m);
[VF,Subs] = odeToVectorField(ode);
odefcn = matlabFunction(VF, 'Vars',{x,Y});

tspan = [C0 C];                                     
ic = [t0 0];                                     
[x1,y1] = ode45(odefcn, tspan, ic);

figure
plot(x1,y1(:,1), 'DisplayName','(x_1,y_1_1)')

[Torsten]

Didn't I tell you that you have to use bvp4c instead of ode45 for boundary value problems ?

Further, t0 is a function of y'. So what do you mean by y(C0) = t0 ?

[John D'Errico]

@Fatemeh

You cannot use ODE45 to solve a boundary value problem. Try as hard as you want, if the conditions fall at both ends of the domain, then ODE45 is not an option. You cannot make software do what it is not written to do.

[Fatemeh]

Thank you @Torsten and @John D'Errico.

@Torsten I edited the code. Thank you for pointing out the error.

[Torsten]

@Fatemeh

After the line

odefcn = matlabFunction(VF, 'Vars',{x,Y});

use the code I gave you for your previous problem. In bcfun, just replace yb(1) by yb(2).

And there is a ( too much in the definition of t0.

[Fatemeh]

Thanks and what does yb(2) mean?

[Torsten]

It means that the second function you want to solve for (i.e. y') in the right boundary point (b) (i.e. x = 80) should be 0.

And ya(1) - t0 means that the first function you want to solve for (i.e. y) in the left boundary point (a) (i.e. x = C0) should be t0.

Here are examples to study on how to use bvp4c:

https://de.mathworks.com/help/matlab/ref/bvp4c.html