1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

6.2 Partial Fractions and Convolutions 371

The rules of elementary algebra suggest thatUcan be written as a sum,

cs+d

(s−r 1 )(s−r 2 )=

A 1

s−r 1 +

A 2

s−r 2 , (1)

for some choice ofA 1 andA 2. Indeed, by finding the common denominator
form for the right-hand side and matching powers ofsin the numerator, we
obtain

cs+d

(s−r 1 )(s−r 2 )=

A 1 (s−r 2 )+A 2 (s−r 1 )

(s−r 1 )(s−r 2 ) ,

c=A 1 +A 2 , d=−A 1 r 2 −A 2 r 1.

WhenA 1 andA 2 are determined, the inverse transform of the right-hand side
of Eq. (1) is easily found:

L−^1

(

A 1

s−r 1 +

A 2

s−r 2

)

=A 1 exp(r 1 t)+A 2 exp(r 2 t).

For a specific example, suppose that

U(s)=

s+ 4

s^2 + 3 s+ 2.

The roots of the denominator arer 1 =−1andr 2 =−2. Thus

s+ 4

s^2 + 3 s+ 2

= A^1

s+ 1

+ A^2

s+ 2

=(A^1 +A^2 )s+(^2 A^1 +A^2 )

(s+ 1 )(s+ 2 )

.

We fi n dA 1 =3,A 2 =−2. Hence

L−^1

( s+ 4

s^2 + 3 s+ 2

)

=L−^1

( 3

s+ 1 −

2

s+ 2

)

= 3 e−t− 2 e−^2 t.

A little calculus takes us much further. Suppose thatUhas the form

U(s)=qp((ss)),

wherepandqare polynomials and the degree ofqis less than the degree ofp.
Assume thatphas distinct rootsr 1 ,...,rk:

p(s)=(s−r 1 )(s−r 2 )···(s−rk).

We t r y t o w r i t eUin the fraction form

U(s)=

A 1

s−r 1 +

A 2

s−r 2 +···+

Ak

s−rk=

q(s)

p(s).