How to solve a System of 8 ODEs in Matlab where for 4 ODEs initial value is at height z=0 and for rest initial value is at z = H - matlab

Question

I have a reactor in which solid is charged from top and gas is charged from the bottom.
I have provided the initial values of them (i.e I know the initial value of solid entities at the top and initial values of gases at the bottom). To solve the heat and mass balance equations I wrote 8 ODEs.
But how can I solve them simultaneously?
Some ODEs initial value is at z=0, and for some other initial value z it is at z = H; where H is the height of the furnace.
function f = odefun( Z,Y )
% FeO + CO -> Fe + CO2 ---(1)
% C + 1/2 O2 -> CO ---(2)
% C + O2 -> CO2 ---(3)
% C + CO2 -> 2CO ---(4)
Fg = Y(1); %volumetic flow rate of gas Nm3s-1 //initialize
Cfeo = Y(2); % initial FeO concentration mol/m3
Cc = Y(3); % initial C concentration mol/m3
yco = Y(4); % mole fraction of CO initially
yco2 = Y(5); % mole fraction of CO2 initially
yo2 = Y(6); % mole fraction of O2 initially
Tg = Y(7); % temp of gas K
Ts = Y(8); % temp of solid K
Fs = 0.044; %Mass flow rate of solid m3/s //initialize and constant
P = 5; %total pressure atm
R = 8.314; %gas constant
M = 104; %molecular mass //constant
keqm1 = 3.33; %equilibrium constant for rx 1
keqm2 = 3 * 10^10; %equilibrium constant for rx 2
keqm3 = 4.75 * 10^20; %equilibrium constant for rx 3
Az = 4; %area of the bed m2 //constant
e = 0.4; %voidage //constant
O = 0.6; %Avg shape factor of solid particle //constant
dp = 0.02; %diameter of particles // constant m
Cpg = 293000; %specific heat capacity of gas (JKg-1K-1) //constant
Cps = 440000; %specific heat capacity of solid (JKg-1K-1) //assume constant
mu = 0.005; %viscosity of gas Kg m-1 s-1 //assume constant
a = 10; %surfce area of DRI per unit volume of bed
Ac = 20; % surfce area of coal per unit volume
hgs = 200; %heat transfer coefficient b/w solid and gas J m-2 s-1 K-1
H1 = -16000; %enthalpy of reaction 1 KJ mol-1
H2 = -110520; %enthalpy of reaction 2 KJ mol-1
H3 = -393509; %enthalpy of reaction 3 KJ mol-1
H4 = -282989; %enthalpy of reaction 4 KJ mol-1
C0 = 28/22.4; % Kg/m3
C1 = 44/22.4;
C2 = 32/22.4;
Dg = (C0 * yco + C1 * yco2 + C2 * yo2); %Density of Gas Kg m-3
Db = (1-e)*(Cfeo*72 + (Y(2) - Cfeo)*56 + Cc*12 )*1000 ; %Density of Bed Kg m-3
Re = (Dg * Fg * e * dp)/(mu*Az) ; %Reynolds number
f1 = (150 * (1-e)^2 * mu)/(e^2 * Dg * dp^2); %friction factor parameter
f2 = (1.75*(1-e))/(e^3 * Dg * dp);
dFgdZ = -(Fg/(P * Az^2))*(f1*Az*Fg + f2 * Fg^2); %volumetric flow rate of gas Nm3 s-1
kp1 = 2.5*exp(-8857.6/Ts); % chemical rxn rate constant s-1
kp2 = 6.52*10^(5) * exp(-22000/Ts) * Ts^(0.5);
kp3 = kp2;
kp4 = 4*10^12 * exp(-40000/Ts);
Vg = yco + yco2; %mass fraction of volatile evolve
kf = 2.0*Re^(-0.336)* Vg; %mass transfer coefficient m/s
Po2 = yo2; % partial pressure atm
Pco = yco;
Pco2 = yco2;
Pco2eqm = 0.2; % Change
Pre2 = (Po2 - (Pco^(2)/keqm2^(2))); % driving force atm
Pre3 = (Po2 - (Pco2^(2)/keqm3^(2)));
Pre4 = (Pco2 - Pco2eqm);
R1 = a*(Fs/Y(1))^(1/3)*kp1*(yco-yco2/keqm1)*P/(R*Ts); %reaction rate constant m-3s-1
R2 = (Ac/((6/(dp*kp2))+(1/kf))) * (9.869*10^(-6)*Pre2)/(M*R*Ts);
R3 = (Ac/((6/(dp*kp3))+(1/kf))) * (9.869*10^(-6)*Pre3)/(M*R*Ts);
R4 = (Ac/((6/(dp*kp4))+(1/kf))) * (9.869*10^(-6)*Pre4)/(M*R*Ts);
dCfeodZ = (R1*Az*(1-e))/Fs; %for FeO Cfeo Kg/s
dCcdZ = ((R2+R3+R4)*Az*(1-e))/Fs; %for C
dycodZ = 22.4*Az*(R2*(yco/2 - 1)+(yco - 2)+R1)/Fg; %for CO
dyco2dZ = 22.4*Az*(yco2*(R2/2 + R4)+R4-R1-R3)/Fg; %for CO2
dyo2dZ = 22.4*Az*(R2/2 + R3 + yo2*(R2/2 + R4))/Fg; %for O2
%Heat transfer equations
S = (6*(1-e)*Az)/(dp*O) ; %surface area of the solid
B = (6*(1-e)*hgs)/(dp*O);
A = 16*R1+12*(R2+R3+R4);
E = H1*R1+ H2*R2+ H3*R3+ H4*R4;
dTgdZ = (22.4/(Fg*Cpg))*(Az*Cpg*Tg*(R2/2 + R4)+ S*(Tg - Ts)*hgs );
dTsdZ = (Az*(1-e)/(Fs*Cps*Db))*(A*Cps*Ts + B*(Tg - Ts) - E);
f = [dFgdZ;dCfeodZ;dCcdZ;dycodZ;dyco2dZ;dyo2dZ;dTgdZ;dTsdZ];
end
These is the function with ODES and the initial values of Fg(z=0), Cfeo(z=H),Cc(z=H) yco(z=0),yco2(z=0),yo2(z=0),Tg(z=0),Ts(z=H) are provided kindly help me out to solve this problem.