Look at
f=@(t,y)[t*y(2); 4*(y(1)^(3/2))];
and notice that if both y(1) and y(2) are 0, that the result of f will be 0.
Now look at
y(i+1,:)=y(i,:)+h*f(t(i),y(i,:))';
and see that if y(i,:) is 0 then the y passed in to f will be 0 so the result of the f call would be 0, and then to that would be added the already-zero y(i,:) . Therefore if y(i,:) is 0 in that code then your output is going to be 0. As soon as you get a 0 input pair, the result of the outputs must be 0.
Now look at
s=zeros(11,2);
s(1,1)=1;
s(1,2)=0;
and see that for all s below the first row, both pairs are 0.
Then see
sol(j+1,:)=explicit_euler2(int(j),int(j+1),s(j,:));
and see that those 0's are going to be passed in on all except the first iteration. We can see from this that after that first iteration, all of the outputs are going to be 0. Which is the behaviour you saw.
