MHB Does the Existence and Uniqueness Theorem Guarantee Solutions for dy/dx = 2xy²?

[find_the_fun]

Given $$\frac{dy}{dx} =2xy^2$$ and the point $$y(x_0)=y_0$$ what does the existence and uniqueness theorem (the basic one) say about the solutions?

1) $$2xy^2$$ is continuous everywhere. Therefore a solution exists everywhere
2) $$\frac{\partial }{\partial y} (2xy^2) = 4xy$$ which is continuous everywhere. Therefore the solution is unique everywhere.

Is this all? What does the point $$y(x_0)=y_0$$ have to do with it? I actually couldn't find any fully worked examples of the existence and uniqueness theorem. Is there a way I should be writing the answers that is more mathy?

 

[Ackbach]

Given $$\frac{dy}{dx} =2xy^2$$ and the point $$y(x_0)=y_0$$ what does the existence and uniqueness theorem (the basic one) say about the solutions?

1) $$2xy^2$$ is continuous everywhere. Therefore a solution exists everywhere.

Actually not. $f(x,y)$ must be Lipschitz continuous in $y$ and continuous in $x$ in order to guarantee existence and uniqueness. $f(x,y)=2xy^2$ is continuous in $x$, but it is not Lipschitz continuous in $y$. Now, it is locally Lipschitz continuous in $y$; if you wanted to argue that on any finite interval containing $x_0$ there exists a unique solution, you'd be on solid ground.