Your code is working correctly: in fact, if you use the built-in ode45 solver, you obtain exactly the same result. My guess there is a mistake in the definition of your coefficients W, A, B, or the initial conditions. Note that if you are not asked to write your own code, ode45 is more efficient and accurate, since it is using adaptive step size with error control.
close all; clear variables;
h=0.01e-6; % step size
x = (-5e-6):h:(20e-6);
y = zeros(1,length(x));
z = zeros(1,length(x));
y(1) = 10; % y at x=-5e-6
z(1) = -1e-1; % dy/dx at x=-5e-6
dWdx = @(x) 2.3139e+13 * exp((-x.^2)/(2e-12));
W = @(x) integral(dWdx,-100e-6,x);
A = @(x) 7.8957e+03 *W(x);
B = @(x) 1/W(x)*dWdx(x);
% use built in ode45
dfdx = @(x,f)[B(x)*f(1)+ 1i*A(x)*f(2); f(1)];
[t,f] = ode45(dfdx,[x(1) x(end)],[z(1) y(1)]);
% plot
h0 = figure;
plot(t,f(:,2));
xlim([-5 20]*1e-6);
grid on; hold on;
% OP code to compare
f = @(x,y,z) z;
g = @(x,y,z) B(x)*z+ 1i*A(x)*y;
N=length(x);
for j=1:(N-1)
k0 = h*f(x(j),y(j),z(j));
L0 = h*g(x(j),y(j),z(j));
k1= h*f(x(j)+0.5*h,y(j)+0.5*k0,z(j)+0.5*L0);
L1= h*g(x(j)+0.5*h,y(j)+0.5*k0,z(j)+0.5*L0);
k2= h*f(x(j)+0.5*h,y(j)+0.5*k1,z(j)+0.5*L1);
L2= h*g(x(j)+0.5*h,y(j)+0.5*k1,z(j)+0.5*L1);
k3= h*f(x(j)+1*h,y(j)+1*k2,z(j)+1*L2);
L3= h*g(x(j)+1*h,y(j)+1*k2,z(j)+1*L2);
y(j+1)=y(j)+1/6*(k0+2*k1+2*k2+k3);
z(j+1)=z(j)+1/6*(L0+2*L1+2*L2+L3);
end
figure(h0), hold on;
% plot(x,y./max(y));
plot(x,y);
xlim([-5 20]*1e-6);
grid on; hold on;
