Let $f,g: \mathbb{R} \longrightarrow \mathbb{R} $ continuous being Lipschitzian. Prove that the system
$x'=f(x)\,\,\,\,, \, x(t_0)=x_0$;
$y'=g(x)y \,\,\,\,, \, y(t_0)=y_0$
has unique solution in any interval where it is defined. You can withdraw hypothesis of $f$ being Lipschitzian and obtaining the same conclusion?
I'm having trouble finding a method to solve this problem. because I have tried to show that f will be contraction, but this method can put constraints on the domain. Anyone have a better idea?
Since $f$ is Lipschitz continuous in $\mathbb{R}$, the Cauchy problem $$ x' = f(x), \quad x(t_0) = x_0 $$ admits a unique global solution $x(t)$, $t\in \mathbb{R}$.
If you define $a(t) := g(x(t))$, the second equation is a linear equation with continuous coefficients (here it is enough $g$ continuous). Hence the Cauchy problem $$ y' = a(t) y, \quad y(t_0) = y_0 $$ admits a unique global solution $y(t)$, $t\in\mathbb{R}$.