The version of theorem I am using.
Let $t_0 \in \mathbb{R},a>0,R>0,y_0 \in \mathbb{R}^n$ and $f:[t_0,t_0 + a] \times B_R(y_0) \rightarrow \mathbb{R}^n$ (jointly) continuous, $|f(t,y)| \leq M\quad (t\in [t_0,t_0+a] y \in B_R(y_0)) $ and $\alpha = \min(a,\frac{R}{M})$ Assume also that there is an $L>0$ such that the Lipschitz condition. $$ |f(t,y)-f(t,z)|\leq|y-z|\quad (t\in [t_0,t_0+a] y \in B_R(y_0)) $$ holds, then the IVP $$ u'(t) = f(t,u(t)) ,\quad u(t_0) = y_0 $$ has a unique solution on $[t_0,t_0+a]$
The IVP I am investigating.
$$\frac{du}{dx} = \frac{1}{u} + x, u(0) = 1$$
$$\text{Does the IVP have a unique solution on } (0,\infty)?? $$
I think I have misused the theorem, because I have said.
We can define $f[0,a] \times B_{\frac{1}{2}}(1) \rightarrow \mathbb{R}^n$, noting that we can take $a$ as big as we like. and
$$|f(x,u)|\leq 2+a$$
also $$|f(x,u)- f(x,v)|\leq4|u-v|$$ So $f$ is lipschitz on the defined intervals.
But since we can take a as large as we like then the result follows
Should this be true for any $R>0$??