Even simple-looking differential equations can have complicated solution curves. In this...

Even simple-looking differential equations can have complicated solution curves. In this problem, we study the solution curves of the differential equation

y ' = −2xy². (1.3.8)

(a) Verify that the hypotheses of the existence and uniqueness theorem (Theorem 1.3.2) are satisfied for the initial-value problem

y ' = −2xy², y(x₀ ) = y₀

for every (x₀, y₀). This establishes that the initialvalue problem always has a unique solution on some interval containing x₀.

(b) Verify that for all values of the constant c, y(x) = 1/(x² + c) is a solution to (1.3.8).

(c) Use the solution to (1.3.8) given in (b) to solve the following initial-value problems. For each case, sketch the corresponding solution curve, and state the maximum interval on which your solution is valid.

(d) What is the unique solution to the following initial-value problem?

y ' = −2xy², y(0) = 0.

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 .