3rd-order Runge-Kutta and 3rd order Adams-bashforth

Ordinary Differential Equation Water tank flow rate problem:

I attached a picture of the problem I need to solve using 3rd-order Runge-Kutta for the first h2 and h3 and points 3 to 1501 using the 3rd order Adams-Bashforth method.

I'm having trouble running the code for both to solve the given dh/dt equation (in the picture).

Here is what I have so far:

%Start. clear

%Define the size of the step:

h=0.5; x=0:h:2; t=0:150;

%Define variables:

V1=0.001; %Velocity 1, m^3/s.

V2=0.0008; %Velocity 2, m^3/s.

p=999.97; %Density, kg/m^3.

C=pi/5; %s^-1.

Apipe=0.002; %Area of pipe (tank outlet pipe), m^2.

Atank=0.2; %Area of tank, m^2.

g=9.81; %Gravity, m/s.

x=0.5; %Height of fluid in tank, m.

y(i)=0; %Initial

a2=0.5; a3=1; b21=0.5; b31=-1; b32=2; c1=1/6; c2=4/6; c3=1/6; y=zeros(size(x));

%Runge-Kutta to estimate h2 and h3:

for i=1:3

     %Evaluate slope at xi,yi.

     K1=(V1*p+V2*p*cos(C*t(i))-p*Apipe*sqrt(2*g*h))/(p*Atank);

     %Evaluate slope at the first intermediate point.

     xp=x(i)+a2*h;

     yp=y(i)+b21*K1*h;         

     K2=

     %Evaluate slope at the second intermediate point.

     xp=x(i)+a3*h;

     yp=y(i)+b31*K1*h+b32*K2*h;

     K3=yp+;

     slope=c1*K1+c2*K2+c3*K3;

    %Calculate the next y.

    y(i+1)=y(i)+h*slope;

end

%Plot height of water (h) vs. time, for 0=<t=<150.

figure

plot(x,y)

%3rd Order Adams-Bashforth for remaining points (i=3 to 1501):

y(i) = (V1*p+V2*p*cos(C*t)-p*Apipe*sqrt(2*g*h))/(p*Atank); %dh/dt.

for i = 3:1501

    y(i+1) = y(i)+h/12*(23*f(i)-16*f(i-1)+5*f(i-2));

end

%Plot height of water (h) vs. time, for 0=<t=<150.

figure

plot(x,y)

I really appreciate any help! Let me know if something is missing or hard to read.