1550078481-Ordinary_Differential_Equations__Roberts_

166 Ordinary Differential Equations

on any interval on which the functions an(x), an-i(x), ... , a 1 (x), ao(x) are all

continuous and an(x)-=/-0. We will need to use this fact later.

Verify that the functions y 1 (x) = x^2 and y 2 (x) = x^3 are two distinct solu-

tions on ( -oo, oo) of the initial value problem

(3) x^2 y" - 4xy' + 6y = O; y(O) = 0, y' (0) = 0.

The fact that the IVP (3) has two distinct solutions does not violate the ex-

istence and uniqueness theorem, since the IVP (3) does not satisfy all of the

hypotheses of the existence and uniqueness theorem. Notice that the initial

conditions are specified at x 0 = 0 and when x = 0 the coefficient of y", the

leading coefficient of the differential equation, is a2(x) = x^2 = 0. That is,

the hypothesis of the theorem a 2 (0) -=/- 0 is not satisfied on any interval I

which contains the point x = 0- which is where the initial conditions are

specified. This example illustrates that if a single hypothesis of the existence
and uniqueness theorem fails to be satisfied, then nothing can be concluded
about the existence of the solution of an initial value problem or about the
uniqueness of a solution when there is one. In general, when a single hypoth-
esis of the existence and uniqueness theorem is not satisfied, then the initial
value problem may have no solution, it may have multiple solutions, or it may
have a single, unique so lution.

EXAMPLE 1 Analysis of a Differential Equation and

Verification of Its Solution

Analyze the differential equation

(4) y" - 3y' + 2y = 4x^2

and verify that

(5) y(x) = 2ex - 3e^2 x + 2x^2 + 6x + 7

is the unique solution on ( -oo, oo) of the initial value problem

(6) y" - 3y' + 2y = 4x^2 ; y(O) = 6, y'(O) = 2.

SOLUTION

The DE ( 4) is linear and nonhomogeneous. The functions a 2 ( x) = 1,

a1(x) = -3, ao(x) = 2, and b(x) = 4x^2 are all defined and continuous
on (-oo, oo) and a2(x) -=/- 0 on (-oo, oo ). Therefore, by the existence and
uniqueness theorem there exists a unique solution to the IVP (6) on the in-
terval (-oo, oo) which satisfies y(x 0 ) = c 1 and y' (x 0 ) = c 2 for any choice of
the constants x 0 , c 1 , and c2.