Uniqueness of ODE by checking Lipschitz condition There are many questions on this site about this topic, but as far as I could find, all omitted the step I am not certain about. $I \subseteq \mathbb{R},u_0 \in \mathbb{R}$
The IVP is given as $$\begin{cases}
      u'(t)=2tu(t)=f(t,u(t)) & \text{for}\ t\in I \\
      u(0)=u_0
    \end{cases} $$
I want to first check for existence and uniqueness, so I want to know if it satisfies the Lipschitz Condition in the second variable. I know that if $$\frac{\partial f}{u(t)} $$ is bounded, then by the MVT the Lipschitz Condition is satisfied. 
But how do I differentiate $\frac{\partial f}{u(t)} $? 
I looked at questions like these: Derivative of a function with respect to another function., but I can't make that work for my example and in examples often times only the solution is given, which I can not follow.
 A: Focus on $f:[a,b] \times \Bbb R \to \Bbb R $ and forget for a moment about the IVP itself. 
One standard condition for uniqueness of ODE solutions is that if $f$ is Lipschitz in its second argument you have uniqueness of solutions wherever they overlap. That is to say there is a positive constant $C$ such that $|f(t,y) - f(t,x)| \le C |y-x|$. Notice that we don't have to consider potential solutions to check this condition. 
In this case a direct argument shows that the condition is satisfied. $$|f(t,x) - f(t,y)| = |2t(x-y)| \le |2t||x-y| \le |2b||x-y|$$ 
Setting $C = |2b|$ shows that $f$ is Lipschitz in its second argument. 
A: A first order IVP is of the form, $$y'=f(x,y(x))\;\;\;,y(x_0)=y_0$$
Here $x$ denotes the independent variable whereas, $y$ is the dependent one.
Computationally, $\displaystyle \frac{\partial f(x,y(x))}{\partial y(x)}$ and $\displaystyle \frac{\partial f}{\partial y}$ are one and the same as, $y$ in itself is a variable. 
So for computational ease, we replace $y(x)$ by $y$ and proceed with usual partial differentiation.
