Let $a=a(x,y)$, $b=b(x,y)$, $c=c(x,y)$ and $d=d(x,y)$ be $C^1$ scalar functions on an open and connected set $U\subset \mathbb{R}^2$ and fix $(x_0,y_0,z_0)\in \mathbb{R}^3$. Consider the system of autonomous ODE's $$\begin{cases}\frac{\partial}{dt}x=a(x,y)\\\frac{\partial}{dt}y=b(x,y)\\\frac{\partial}{dt}z=c(x,y)z+d(x,y)\end{cases}$$ with initial conditions $$\begin{cases}x(0)=x_0\\y(0)=y_0\\z(0)=z_0\end{cases}.$$ Because we have supposed that $a,b,c,d \in C^1(U)$, then by theorem of existence and uniqueness of solution for ODE's, the previous has a unique maximal solution, say $(x(t),y(t),z(t))$ where $t\in I$ open interval of $\mathbb{R}$ containing $0$.
On the other side, because the first two equation of the system are independent from the last one, I can just solve $$\begin{cases}\frac{\partial}{dt}x=a(x,y)\\\frac{\partial}{dt}y=b(x,y)\end{cases}+\begin{cases}x(0)=x_0\\y(0)=y_0\end{cases}\qquad [1]$$ and then substitute in the third one and finally solve $$\begin{cases}\frac{\partial}{dt}z=c(x(t),y(t))z+d(x(t),y(t))\\z(0)=z_0\end{cases}\qquad [2].$$My question is the following: solving $[1]$ I certainly find a unique maximal solution $(x_0(t),y_0(t))$ defined on some interval $\overline{I}\supseteq I$. Now, for $[2]$, I find an unique maximal solution $z_0(t)$ defined on $I_0\subseteq \overline{I}$.
Is $I_0=I=\overline{I}$ and $(x(t),y(t),z(t))=(x_0(t),y_0(t),z_0(t))$??? I think the answer is yes, but I don't know how to prove it.
A linear ODE like the one for $z$ has solutions on the full domain interval where the coefficients are continuous. Thus $I_0=\bar I$, as the functions $x,y$ are continuous on $\bar I$ by construction and the composition to the coefficient functions results again in continuous functions.