8/11/2019 Second Order Diff Equ

1/6

68

LINEAR

EQUATIONS

ANSWERS

[CHAP.

3

4. a 2+

e ,

1+ 2e ; the condition

of

the problem rules out the choice of

the

zero

function.

b CI +

C2e ; any initial conditions can be satisfied with a function

from

this family.

c

x

+

+

c2e

5. u x

=

1

U2 andu3

must be linearly dependent

8. a No, b No

9. a , b , and c

10.

E

12,

F 2e

6

G cos

xe

2

ll -sin x

4 econd order equations with constant coefficients One of the

simplest and most useful cases of the linear equation is the second order

equation

Y

+

PlY

+

P2Y

=

q x ,

1

where

PI and

P2 are constants. We will examine this equation in some

detail

and

apply our results in Section 3-5

to

some examples from me-

chanics, electricity, etc.

First , let us look at the reduced equation

y + PlY + P2Y =

o

2

The

problem is to find two linearly independent solutions. Since

two

functions are linearly dependent only

one is a constant multiple of

the

other Problem 1 , the check for independence can be made

by

inspection.

Substituting Y =

e

TX

in the left side of 2 , we get

Hence

e

TX

is a solution of 2 if

r

is a root of the algebraic equation

r

2

+ Plr +

P2

=

3

4

The

equation 4 is called

the

auxiliary

equation

for

1

or 2 .

4 has

.two real roots,

r l

and

r2,

then

e

T X

and

e

T X

are

the

required two solutions

of 2 .

4

has only one real root, ro,

then

r

2

+ Plr + P2 = (r - ro 2 = r

2

-

2ror

+

That is,

PI =

-2ro and P2

=

In this case

the

solutions of 2 are

e

ToX

and xeToX Problem 2 . 4 has no real roots, the roots are conjugate

8/11/2019 Second Order Diff Equ

2/6

3 4]

SECOND

ORDER EQUATIONS

CONSTANT

COEFFICIENTS

69

complex numbers, say a + ib, a - ib,

and

r

2

+ Plr +

P2

= [r - a + ib ][r - a - ib

r

2

- 2ar + a

2

+ b

2

.

other words,

5

PI

= a

and

6

One can verify by

substitution

Problem 3

that the

solutions of 2 m

this case are e

ax

cos bx and e

ax

sin bx.

To

summarize, the family of solutions of 2 is

Cle

T X

+ C

eT

cle

Tox

+ C

xeT

cle

ax

cos bx +

C2eax

sin bx,

i f

r

I

and r

2 are

distinct real roots of 4 ,

if

1 0

is the only real root

of

4 ,

if a ib are

the

roots of 4 .

EXAMPLE

1. A

y

+ y

- 2y

=

The auxiliary equation is r

2

+ r - 2

=

0, or r + 2 r - 1

=

The

roots are 1 and -2 , so the solutions of A are cle

x

+ C2e-2x.

B

y +

4y

+

4y =

The auxiliary equation is r

2

+ 4r + 4

=

0, or r + 2 2

=

The

single

root is

-2 ,

so the solutions of B are y

=

cle-

2x

+ C2xe-2x.

C

y - 2y

+

5y

=

The

auxiliary

equation

is

r

2

- 2r

+ 5

= O.

Since

the

discriminant is

_2 2 - 4 1 5

=

-16