In Example 3 , it was stated that the solution y=-√(25-x^2) is defined on the open interval (-5,

step 1 Welcome to this. We have new solutions for Y, that is 5 .1 x of square width of 25 minus squared

In Example 3 , it was stated that the solution y=-√(25-x^2)...

In Example 3 , it was stated that the solution $y=-\sqrt{25-x^{2}}$ is defined on the open interval $(-5,5) .$ Why would it be incorrect to say that the solution is defined on the closed interval [-5,5]$?$

Key Concepts

Differentiability in Differential Equations

In many differential equations, solutions are required to be not only continuous but also differentiable on their defined interval. Although the function y = -?(25 - x²) may be computed at x = -5 and x = 5, the derivative at these endpoints may fail to exist or be well-defined. This lack of differentiability disrupts the conditions needed for applying existence and uniqueness theorems, which is why the function is commonly considered on the open interval (-5, 5).

Endpoint Behavior and Problem Conditions

While a function might be defined at the endpoints in a purely algebraic sense, the behavior at these endpoints is critical in the context of initial value problems. If the endpoints do not support the necessary smoothness (continuous derivatives), then the solution on the closed interval would include points that do not satisfy the underlying differential equation’s requirements. Restricting the domain to an open interval excludes these problematic points and ensures that the solution is consistent with theoretical assumptions.

Domain of a Function

The domain of a function refers to the complete set of input values for which the function produces a valid output. Even if the formula may yield real numbers at the endpoints of an interval, in the context of differential equations the domain is often restricted to an open interval because additional conditions like differentiability or continuity of the derivative are required to fully satisfy the problem’s needs.

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