Uniqueness theorem in non authonomous ODE I am reading (the italian version of) Arnold's book on Ordinary Differential Equation. On page 29 (Chapter 1, paragraph 2: vectorial fields on the real line) problem 2 says: Given the differential equation $\dot x = v(x,t)$, where $v$ is a differentiable function, if there exists a solution $x = \varphi(t)$ which verifies the intial condition $\varphi(t_0)=x_0$, prove its unicity.
Arnold gives an indication: let $y = x-\varphi(t)$, confront $y$ with a convenient equation of the form $\dot x = kx$, $k\ne 0$.
My question is: how does the substitution $y=x-\varphi(t)$ works? Isn't $x$ equal to $\varphi(t)$? Should this substitution transform the non-authonomous equation in an authonomous one? How?
 A: The general idea of proving that two solutions is to take their difference and prove that it's zero. For nonlinear equations this is less straightforward because the difference of two solutions will not solve the same equation. But it will solve some other equation - one in which zero is also a solution. Ultimately, we want to be in the position to apply the Gronwall lemma.
So, letting $x(t)=y(t)+\varphi(t)$ we transform the given equation into 
$$\dot y=v(y+\varphi(t),t)-\varphi'(t),\quad y(0)=0\tag1$$ 
One solution of (1) is $y\equiv 0$. To show there is no other, we need an inequality of the form $|\dot y|\le k|y|$; then Gronwall finishes off the problem thanks to $y(0)=0$. 
Since $\varphi'(t)=v(\varphi(t),t)$, the mean value theorem helps: 
$$|\dot y|=  |v(y+\varphi(t),t)-v(\varphi(t),t)| \le |y| \sup \left|\frac{\partial v}{\partial x}\right|$$
So, if we can control the derivative of $v$ with respect to the first variable, we are done. The derivative isn't really necessary: the Lipschitz condition suffices. 
