Uniqueness Questions. In Chapter 1 we indicated that in applications most initial va...

Uniqueness Questions. In Chapter 1 we indicated that in applications most initial value problems will have a unique solution. In fact, the existence of unique solutions was so important that we stated an existence and uniqueness theorem, Theorem 1, page 11. The method for separable equations can give us a solution, but it may not give us all the solutions (also see Problem 30). To illustrate this, consider the equation

(a) Use the method of separation of variables to show that

is a solution.

(b) Show that the initial value problem dy/dx =

(c) Now show that the constant function y = 0 also satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution.

(d) Finally, show that the conditions of Theorem 1 on page 11 are not satisfied.

(The solution was lost because of the division by zero in the separation process.)