Question

State where in the ty-plane the hypotheses of Theorem 2.4.2 are satisfied. y' = (1 – t² – y²)¹/²

Answer 1

The hypotheses of Theorem 2.4.2 are satisfied in the region where t² ≤ 1 - y²

Step-by-step explanation:

The question asks about the hypotheses of Theorem 2.4.2 in the context of the given differential equation y' = (1 – t² – y²)¹/². The hypotheses of Theorem 2.4.2 state that the equation must have a continuous solution in a neighborhood of the initial point (t₀, y₀) and the partial derivative ∂f/∂y must be continuous in some rectangle R around (t₀, y₀).

To determine where these hypotheses are satisfied, we need to analyze the given differential equation. The equation involves a square root, which is not continuous for negative values. So, the hypotheses of Theorem 2.4.2 will be satisfied only in the regions of the ty-plane where the expression inside the square root, i.e., (1 – t² – y²), is non-negative.

In other words, the hypotheses are satisfied in the region where t² ≤ 1 - y². This region can be represented as a shaded area in the ty-plane, such that t lies between -1 and 1 (inclusive) and for each t, y lies between -√(1 - t²) and √(1 - t²) (inclusive).