Final answer:
The requested differential equation is one like dy/dx = 1/(x + 1), which has a unique solution around (3, -2) but not around (-1, 4) due to the division by zero at x = -1.
Step-by-step explanation:
The question asks for an example of a first order differential equation which has a unique solution around the point (3, -2) but not around the point (-1, 4). To satisfy these conditions, we can look at the theorem of the existence and uniqueness for first order differential equations, which states that if a function f(x, y) and its partial derivative ∂f/∂y are continuous in a region around the point, then the initial value problem has a unique solution at that point.
A typical example of a differential equation with these characteristics could be one that involves dividing by an expression that becomes zero. For instance, we might consider the differential equation:
dy/dx = 1/(x + 1)
This equation implies division by zero when x = -1, which would be problematic around the point (-1, 4). However, at the point (3, -2), there is no such issue as x + 1 is clearly not zero, allowing a unique solution in its vicinity. Thus, this differential equation has a unique solution around (3, -2) but not around (-1, 4).