The Chemistry Maths Book, Second Edition

338 Chapter 12Second-order differential equations. Constant coefficients

EXAMPLE 12.1Show thaty

1

1 = 1 e

2 x

andy

2

1 = 1 e

3 x

are two solutions of the equation

We have

Therefore

Similarly,

and

0 Exercises 1–3

This example demonstrates the general result that it is always possible to find two

particular solutionsy

1

(x)andy

2

(x)of a homogeneous linear equation. If either of these

solutions is multiplied by a constant the new function is also a solution. For example,

lety

1

(x)be a solution of the general homogeneous equation (12.2),

and lety

3

(x) 1 = 1 cy

1

(x), where cis a constant. Then

and

dy

dx

px

dy

dx

qxy c

dy

dx

px

dy

dx

2

3

2

3

3

2

1

2

1

++=+() () () ++=

























qxy()

1

0

dy

dx

dcy

dx

c

dy

dx

dy

dx

c

dy

dx

311

2

3

2

2

1

2

==, =

()

dy

dx

px

dy

dx

qxy

2

1

2

1

1

++=() () 0

dy

dx

dy

dx

ye e e

xxx

2

2

2

2

2

333

−+=−+= 5691560

ye

dy

dx

e

dy

dx

e

xx x

2

3

2

3

2

2

2

3

=, = , 39 =

dy

dx

dy

dx

ye e ee

xxxx

2

1

2

1

1

2222

−+=−×+= −564526 41(006 0+=)

ye

dy

dx

e

dy

dx

e

xx x

1

2

1

2

2

1

2

2

=, = , 24 =

dy

dx

dy

dx

y

2

2

−+= 560