4th order runge kutta algorithm gives strange results for ode

I have a boundary value problem as below. I would like to solve it as initial value problem first by implementing a numerical approximation algorithm and to fulfil boundary conditions with newton method.

y1(t)'=t*y2(t) y(0)=1 y(5)=0 y2(t)'=4*y1(t)^(3/2)

I get strange results as full of zeros and inf in solution components (y in main) when i try to solve the Ivp with 4th order runge kutta algorithm.

function ystar = rk4( t0,y0,h)
y1=zeros(11,1);
y2=zeros(11,1);
y1=y0(1);
y2=y0(2);
t=0:0.5:5;

f=@(t,y1,y2) t*y2;
g=@(t,y1,y2) 4*(y1^(3/2));

for i=1:11

      k1 = f(t(i),y1(i),y2(i));
      l1=g(t(i),y1(i),y2(i));
      k2= f(t(i)+0.5*h,y1(i)+0.5*h*k1,y2(i)+0.5*h*k1);
      l2=g(t(i)+0.5*h,y1(i)+0.5*h*l1,y2(i)+0.5*h*l1);
      k3 = f(t(i)+0.5*h,y1(i)+0.5*h*k2,y2(i)+0.5*h*k2);
      l3= g(t(i)+0.5*h,y1(i)+0.5*h*l2,y2(i)+0.5*h*l2);
      k4 = f((t(i)+h),(y1(i)+k3*h),(y2(i)+k3*h));
      l4= g((t(i)+h),(y1(i)+l3*h),(y2(i)+l3*h));

      y1(i+1)=y1(i)+(1/6)*(k1+2*k2+2*k3+k4)*h;
      y2(i+1)=y2(i)+(1/6)*(l1+2*l2+2*l3+l4)*h;
      ystar=[y1 y2];
  end
  end

main.m

y0=[1 0];
t0=[0,1];
h=0.5;
y=rk4(t0,y0,h);