According to hartman's chapter V.1
for the initial value problem $$y'=f(t,y);\ y(t_0)=y_0 \tag1$$ here $f(t,y)$ is defined on an open $(t,y)$-set $E$ with the property that if $(t_0,y_0)\in E$ then $(1)$ has unique solution $y(t)=\eta(t,t_0,y_0)$ which is then defined on a maximal t-inteval $(\omega_-,\omega_+)$ which are depends on $(t_0,y_0)$.
according to Hartman $(1)$ can be reduced $$y'=f(t-t_0,y-y_0);\ y(0)=0\tag2$$ according to the change of variable $t,y\rightarrow t-t_0,y-y_0$
here I wonder this is right change of variable.
Actually I think $$y'=f(t+t_0,y+y_0);\ y(0)=0\tag3$$
is correct. Because I think at least, $y'(0)=f(t_0,y_0)$ should be satisfied, in order for both ode systems to be equivalent.
Why hartman's change of variable is right?