Consider the IVP $$\frac{dx}{dt}=1+x^2, \ \ \ x(0)=1.$$ I am trying to show that $f(x)=1+x^2$ is Lipschitz and hence the above IVP has a unique solution.
My attempt:
I considered $f(x)=1+x^2$. Now if $f$ is Lipschitz, then $\exists L\in\mathbb{R}, \ \text{such that} \ \forall x,y\in\mathbb{R}$, $$|f(x)-f(y)|\leq L|x-y|.$$ Now, \begin{align} L&\geq \frac{|1+x^2-1-y^2|}{|x-y|} \\ &=\frac{|(x-y)(x+y)|}{|x-y|} \\ &=|x+y| \end{align}
I'm a bit unsure of how to proceed. How can I use the initial value conditions to further my answer?
All you need for uniqueness is local Lipschitz property of $f(x)=1+x^2$: for any $x,y \in [a,b]$, by the Lagrange Theorem, $$|f(x)-f(y)|=|f'(t)||x-y|\leq L|x-y|$$ where $L=\max\{2t : t\in [a,b]\}=2\max(|a|,|b|)$ (or, according to you approach, we can also take $L=2\max(|a|,|b|)\geq |x|+ |y|\geq |x+y|$).
By the way, in this case, uniqueness follows by solving the ODE after separating the variables: $$\int_1^{x(t)}\frac{dx}{1+x^2}=\int_0^{t}dt\implies \arctan(x(t))-\frac{\pi}{4}=t\implies x(t)=\tan\left(t+\frac{\pi}{4}\right).$$