1550078481-Ordinary_Differential_Equations__Roberts_

N-th Order Linear Differential Equations 221

SOLUTION

Since the differential equation of this problem is equation (10) of section 4.4,

we know the general solution is y(x) = Yc(x) + Yμ(x) where Yc(x) and Yμ(x)

are as calculated in example 4 of section 4.4. That is ,

-3x 4 3 4 2 8 2 -3x

y(x) = c 1 + c2e + - x - - x + - x + -xe.

9 9 27 3

Differentiating, we find

y I ( x ) = -3c2e -X +^4 - x^2 - - x^8 + - + -^8 2 ( -3x + 1 ) e -3X.

3 9 27 3

The constants c 1 and c 2 must be chosen so that the initial conditions y(O) = 1

and y' (0) = 0 are satisfied. Hence, c 1 and c2 must be chosen to satisfy the

following system of linear equations.

8 2

-3c2 + - + -= 0

27 3

(from the condition y(O) = 1)

(from the condition y'(O) = 0)

. 55 26.

Solving this system, we find c 1 = - and c2 = -. Therefore, the solut10n of

81 81

the IVP (5) is

55 26 _ 3 x 4 3 4 2 8 2 -3x

y(x) = - + - e + - x - - x + - x + -xe.

81 81 9 9 27 3

EXERCISE S 4 .5

Solve t h e fo llowing initial value problems.

1. y<^3 l - 3y<^2 l - 4yCll + 12y = O; y(O) = 1, yCll (0) = 5, y(^2 l (0) = - 1


2. y(4) _ 2y(3) + 2yCll -y = O; y(O) = 1, y<^1 l(O) = -1, y(^2 ) (0) = - 3,

y(^3 l(o) = 3

3. y(3) y(2) + y(l) y = 2ex; y(O) = 1, yCll (0) = 3, y(^2 l(O) = -3


4. y(^4 ) + 2y(^2 ) + y = 3x + 4; y(O) = 0, yCll(O) = o, y(^2 ) (0) = 1,

y<^3 l(o) = 1