i can try The system of ODE's y''=xy, y(0)=0, y'(0)=1 solve in runge kutta method, h=0.05 and interval is [-10 2]

function [x, u]= RK4_sys12(f, tspan, u0, n)
a=tspan(1); 
b=tspan(2);
h=(b-a)/n;
x=(a:h:b)';
u(1,:)=u0;
for i=1:n
    k1=h*feval(f, x(i),     u(i,:)     )';
    k2=h*feval(f, x(i)+h/2, u(i,:)+k1/2)';
    k3=h*feval(f, x(i)+h/2, u(i,:)+k2/2)';
    k4=h*feval(f, x(i)+h,   u(i,:)+k3)';
    u(i+1, :)=u(i, :)+ k1/6 +k2/3 +k3/3 +k4/6;
end
%x=x
%u=u
%write on command window
clf; clc;
f=inline('[u(2); x*u(1)]','x','u');
[x, u]=RK4_sys12(f, [-10,2],[0,1], 50);
plot(x,u(:,2),'k') % i can take u(2)
hold on;grid on;
plot(x,u(:,1))
%disp('   x    u(1)   u(2)')
disp([x u])

[jamila nizamani]

total number steps is (n=50 or n=100)

[jamila nizamani]

please help me........... if u need more information i can gives u but pleaseeeeeeeeeeeeee help mee