Final answer:
The Existence and Uniqueness Theorem implies that the initial value problem y²dy/dx = x⁶, y(5)=0, has a unique solution, as the functions involved are continuous and meet the theorem's criteria.
Step-by-step explanation:
The initial value problem given is y² dy/dx = x⁶, with the condition y(5)=0. To determine whether the Existence and Uniqueness Theorem ensures a unique solution, we need to check if the function and its partial derivative with respect to y are continuous around x = 5. The function on the right side, x⁶, and the function that would be y² after dividing both sides by y², are polynomials, and therefore continuous everywhere. The only concern is the point where y = 0, which would cause division by zero. However, since we are not dividing by y², the original function x⁶ remains continuous, and so does its derivative with respect to y, which is zero.
Given this, we can say that both functions are continuous around the initial value, meaning the theorem applies, implying that there is a unique solution to the initial value problem near x = 5.