1550078481-Ordinary_Differential_Equations__Roberts_

N-th Order Linear Differential Equations 167

The function ( 5) y( x) = 2ex - 3e^2 x + 2x^2 + 6x + 7 is defined and continuous

on ( -oo, oo). Differentiating twice, we find

and

y'(x) = 2ex - 6e^2 x + 4x + 6

y"(x) = 2ex - 12e^2 x + 4.

Substituting these expressions for y, y' and y" into the DE (4), we see that

y" - 3y' + 2y

= (2ex - 12e^2 x + 4) - 3(2ex - 6e^2 x + 4x + 6) + 2(2ex - 3e^2 x + 2x^2 + 6x + 7)

= (2 - 6 + 4)ex + (-12 + 18 - 6)e^2 x + ( 4 - 1 8 + 14) + (-12 + 12)x + 4x^2

= 4x^2.

Thus, since y' and y", as well as y , are all defined on (-00,00), y(x) is a

solution of the DE ( 4) on (-oo, oo ). Evaluating y and y' at x = 0, we find

y(O) = 2 - 3 + 7 = 6 and y'(O) = 2 - 6 + 6 = 2.

Thus, the initial conditions specified in the IVP (6) are satisfied. Hence, y(x)
is t he unique solution of the IVP (6) on the interval (-oo, oo).

EXAMPLE 2 Analysis of a Differential Equation and

Verification of Its Solution

Analyze the differential equation

(7) x^2 y" + xy' - 4y =^0

and verify that

(8)

1

y(x) = 2x^2 + 2

x

is the unique solution on (0, oo) of the initial value problem

(9)

SOLUTION

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

The DE (7) is linear and homogeneous. The functions a2(x) = x^2 , ai(x) =

x, a 0 (x) = - 4, and b(x) = 0 are all defined and continuous on (-oo, oo) and

a 2 (x) =/= 0 for x =/= 0. Since the initial conditions of t he IVP (9) are specified

at x 0 = 1 E (0, oo), by the existence and uniqueness theorem there exists

a unique solution to the IVP (9) in the interval (0, oo). Differentiating (8)

1. y(x) = 2x^2 + 2 twice, we get
   x

y '( x) = 4x - -^2

x3