MATLAB: How to solve ODE for varying input using ode23

MATLABode23ordinary differential equaion for varying input.

I'm trying to solve a RC circuit using ode23. For my case the input is varying with time. Here u is my input vector.
% function
function vdot = diff3(t,v)
u = [1 2 3 4 5] ; 
vdot(1) = -40*v(1) + 20*v(2) + 20*u(t);
vdot(2) = 20*v(1) - 20*v(2) - v(3);
vdot(3) = 0.1*v(2) -1000*v(3);
vdot = [vdot(1);vdot(2);vdot(3)];
end
% main program
tspan = 1:5;
x0 = [0 0 0]; % initial conditions
[t,x] = ode23('diff3', tspan, x0);
tt = length(t);
for i = 1:tt
 vo(i) = x(i,1) - x(i,2);
end
plot(t, vo)
title('Transient analysis of RLC')
xlabel('Time, s'), ylabel('Output voltage')
In case of fixed value of u = 5, there is no error. As my input is varying MATLAB is throwing some error. _
Attempted to access u(1); index must be a positive integer or logical.
Error in diff3 (line 9)
vdot(1) = -40*v(1) + 20*v(2) + 20*u(t);
Error in ode23 (line 256)
    f(:,2) = feval(odeFcn,t+h*A(1),y+f*hB(:,1),odeArgs{:});
Error in rcckt (line 5)
[t,x] = ode23('diff3', tspan, x0);_
So how can I solve ODE for varying input? Any kind of suggestion will be really helpful. Thanks in advance!

Best Answer

% function
function vdot = diff3(t,v)
u = [1 2 3 4 5] ;
tu = [1 2 3 4 5];
u_at_t = interp1(tu,u,t); 
vdot(1) = -40*v(1) + 20*v(2) + 20*u_at_t;
vdot(2) = 20*v(1) - 20*v(2) - v(3);
vdot(3) = 0.1*v(2) -1000*v(3);
vdot = [vdot(1);vdot(2);vdot(3)];
end
Best wishes
Torsten.
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