1550078481-Ordinary_Differential_Equations__Roberts_

10 Ordinary Differential Equations

I EXAMPLE 1 Verification of a Solution

Verify that y = xe^2 x is a so lution of the linear second order differential
equ ation y" - 4y' + 4y = 0 on the interval ( -oo, oo).

SOLUTION

Differentiating y = xe^2 x twice, we find

y' = 2xe2x + e2x

y" = 2(2xe^2 x + e^2 x) + 2e^2 x = 4xe^2 x + 4e^2 x.

The three functions y , y', and y" are defined on ( -oo, oo) and

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

= (4 - 8 + 4)xe^2 x + (4 - 4)e^2 x = 0

for all x in (-oo, oo ). So y = xe^2 x is a so lution of y" - 4y' + 4y = 0 on
( -oo, CXl).

EXAMPLE 2 Being a Solution D epends on the Interval

a. Show that y = x + ~ is a not solution of x^2 y" + xy' - y = 0 on the

interval ( -oo, oo).

b. Verify that y = x + ~ is a solut ion of x^2 y" + x y' -y = 0 on the intervals

( -oo, 0) and (0, oo).

SOLUTION

a. The function y = x + ~ is not defined at x = 0, so it cannot be the

so lut ion of a ny differential equation on any interval which contains the

point x = 0.

b. Differentiating y = x + ~ = x + x-^1 twice, we find

x

y / =l-x -2 =1--^1

x2 and

2

y" = 2x-3 = 3·

x

The three functions y, y', and y" are defined on the intervals (-oo, 0)

and (0, oo). And for x ~ 0

x^2 y II + xy I - y = x 2( -2 ) + x ( 1 - -^1 ) - ( x + -1)

x^3 x^2 x

2 1 1

= - + x - - - x - - = o.

x x x