Answer: In summary, to determine the region of the xy-plane for which the given differential equation has a unique solution passing through the point (x₀, y₀), we solved the differential equation, analyzed the behavior of its solutions, and found the specific value of C that satisfies the initial condition. The region is defined by the equation (1/2)y² - xy = xy + (1/2)x² + (1/2)y₀² - (1/2)x₀².
Step-by-step explanation: To determine a region in the xy-plane for which the given differential equation would have a unique solution passing through a point (x₀, y₀), we can solve the equation and analyze the behavior of its solutions. The given differential equation is (y-x)dy/dx = y+x. To solve this equation, we can first rewrite it as: dy/dx = (y+x)/(y-x). Next, we can separate the variables by multiplying both sides of the equation by (y-x): (y-x)dy = (y+x)dx. Now, we can integrate both sides of the equation: ∫(y-x)dy = ∫(y+x)dx. Integrating the left side gives us: (1/2)y² - xy = ∫(y+x)dx. Simplifying the right side of the equation: (1/2)y² - xy = xy + (1/2)x² + C. where C is the constant of integration. Now, let's analyze the behavior of the solutions: Since the differential equation is separable, we can see that the graph of the solutions is a family of curves defined by the equation (1/2)y² - xy = xy + (1/2)x² + C. For a given value of C, the solution curves will form a continuous curve in the xy-plane. However, if C is allowed to vary, we will have different solution curves corresponding to different values of C. To determine a region in the xy-plane for which the differential equation has a unique solution passing through the point (x₀, y₀), we need to find a specific value of C that satisfies the initial condition y(x₀) = y₀. By substituting x = x₀ and y = y₀ into the equation (1/2)y² - xy = xy + (1/2)x² + C, we can solve for C: (1/2)y₀² - x₀y₀ = x₀y₀ + (1/2)x₀² + C. Cancelling out the terms gives: (1/2)y₀² = (1/2)x₀² + C. Simplifying further: C = (1/2)y₀² - (1/2)x₀². Now, the region in the xy-plane for which the differential equation has a unique solution passing through the point (x₀, y₀) is the set of points (x, y) that satisfy the equation (1/2)y² - xy = xy + (1/2)x² + (1/2)y₀² - (1/2)x₀².