Chapter 1 - First-Order Differential Equations - 1.3 The Geometry of First-Order Differential Equations - Problems - Page 32: 15

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We are given: $y'=xy^{\frac{1}{2}}$ $\frac{\partial f}{\partial y}=\frac{x}{2\sqrt y}$ The equation is not continuous along the entirety of the (x,y) plane. Thus the given initial-value problem doesn't have a unique solution.