Final answer:
A unique solution for the differential equation (x² + y²)y' = y² exists in every region of the xy-plane except at the origin, according to the Existence and Uniqueness Theorem. Option D is correct.
Step-by-step explanation:
To determine the region of the xy-plane where the differential equation (x² + y²)y' = y² has a unique solution whose graph passes through a specific point (x₀, y₀), we need to assess where the equation satisfies the conditions of the Existence and Uniqueness Theorem.
The differential equation can be written in the form y' = f(x, y), where f(x, y) = y² / (x² + y²) for all x and y except for the point where x² + y² = 0, which is at the origin (0, 0).
Because f(x, y) and its partial derivative with respect to y are continuous everywhere except at the origin, the Existence and Uniqueness Theorem tells us that a unique solution exists at every point in the xy-plane except the origin. Thus, the correct answer is D. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
The given differential equation is (x² + y²)y' = y². We need to determine the region in the xy-plane where this equation has a unique solution that passes through the point (x₀, y₀).
Let's analyze the equation. Notice that if y = 0, the equation becomes 0 = 0, so any point where y = 0 is a solution. However, our point of interest is (x₀, y₀), and we want a unique solution that passes through this point.
In this case, a unique solution exists in the region D. consisting of all points in the xy-plane except the origin. In this region, the solution will not pass through the point (0,0), ensuring that it passes through our given point (x₀, y₀).