Game Development Reference

In-Depth Information

2

a

′′

()

′

()

()

xt axt

+

+

xt

=

0

.

(13.23)

4

It is a simple task to verify that the function

()

(

)

xt

=

te
−

at

(13.24)

is a solution to Equation (13.23), so the general solution to Equation (13.12)

when
1

rr

=

is given by

2

()

x t

=

e

rt

+

te

rt

,

(13.25)

where we have set

rr r

==

.

1

2

, then the roots of the auxiliary equation are complex. The solu-

tion given by Equation (13.19) is still correct, but it requires the use of complex

arithmetic. We can express the solution entirely in terms of real-valued functions

by using the formula

If

ab

2

−<

4

0

αβi

+

α

(

)

e

=

e

cos

β i β

+

sin

(13.26)

(see Appendix A, Section A.4). Assuming that
a
and
b
are real numbers, the roots

r
and
r
of the auxiliary equation are complex conjugates, so we may write

r α i

r α i

=+

=−

1

,

(13.27)

2

where

a

α

=−

2

1

β

=

4

ba

−

2

.

(13.28)

2

The solution given by Equation (13.19) can now be written as

(

)

(

)

()

αβit

+

αβit

−

xt

=

Ae

+

Be

αt

(

)

αt

(

)

=

Ae

cos

βti βt

+

sin

+

e βti βt

cos

−

sin

αt

[

(

)

(

)

]

=

eAB βt

+

cos

+

ABi βt

−

sin

.

(13.29)

This solution can be expressed using two real constants
C
and
C
by setting

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