Let ${du\over dt} = u, \ u(0)=1$. Let the step size be $\tau = 1/3$. Let $0=t_0
Given $U^0=u(0) =1$, we find $U^{n+1}$
$U^{n+1}=U^n+\tau f(t,U^N) \\ U^1=U^0+\tau f(t_0,U^0) \\ U^1 =1+\frac{1}{3}\cdot 1$
I tried to understand but Im still stuck, how does $f(t_0,U^0)=1$? I know $f(t_0,U^0)=f(0,1)$, but how does $f=(0,1)=1$?