# Finding conditions when an initial value problem has unique solution

For each $$\alpha>0$$ find all pairs $$(x_0,y_0)\in\mathbb R^2$$ such that the following IVP has a unique solution in a neighbourhood of $$(x_0,y_0)$$ $$y'=y^\alpha,\ y(x_0)=y_0$$ I know about the Picard-Lindelof theorem which asserts uniqueness but it is only a sufficient condition. The question ask to find all such pairs.

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The function $$f(y):=y^\alpha$$ is not locally Lipschitz in $$y=0$$ for $$\alpha \in (0,1)$$. If we choose $$\alpha$$ in this interval and $$y_0\neq 0$$ then we can apply the Picard-Lindelöf theorem. The same theorem applies when $$\alpha \geq 1$$ since $$y^\alpha$$ is locally Lipschitz for all $$\alpha \geq 1$$ in all $$y\in \mathbb{R}$$.
When we lose the Lipschitz continuity we can construct $$2$$ solutions to prove non-uniqueness. For example if $$\alpha=1/2$$ and $$y(1)=0$$ we have the solutions $$y\equiv 0$$ and $$y=\frac{(x-1)^2}{4}$$. We can easily generalize this example to all $$\alpha \in (0,1)$$ and $$y(x_0)=0$$.