One solution to the initial-value problem is y(x) = 1. Determine another solution. Does...

One solution to the initial-value problem is y(x) = 1. Determine another solution. Does this contradict the existence and uniqueness theorem (Theorem 1.3.2)? Explain.

Reference : Theorem 1.3.2

(Existence and Uniqueness Theorem)

Let f (x, y) be a function that is continuous on the rectangle R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}.

Suppose further that ∂f/∂y is continuous in R. Then for any interior point (x₀, y₀) in the rectangle R, there exists an interval I containing x₀ such that the initial-value problem (1.3.2) has a unique solution for x in I .