082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

654 Chapter 21 The Laplace transform
The expression,
Y(s)= 1 s + 1
s −1
2s−1
, iswritten as itspartialfractions:
s(s −1)
Taking the inverse Laplace transform yields
y(t)=1+e t
Example21.25 Solve
x ′′ +2x ′ +2x =e −t x(0) =x ′ (0) =0
usingLaplace transforms.
Solution Taking the Laplace transform ofboth sides:
s 2 X(s) −sx(0) −x ′ (0) +2(sX(s) −x(0)) +2X(s) = 1
s +1
Therefore,
(s 2 +2s +2)X(s) = 1 sincex(0) =x ′ (0) =0
s +1
1
X(s)=
(s +1)(s 2 +2s +2) = 1
(s +1)(s −a)(s −b)
wherea = −1 +j,b = −1 −j.Usingpartialfractions gives
X(s)= 1
s +1 − 1 ( 1
2 s −a + 1 )
s −b
Taking the inverse Laplace transform
x(t) =e −t − 1 2 (eat +e bt )
=e −t − 1 2 (e(−1+j)t +e (−1−j)t )
=e −t − e−t
2 (ejt +e −jt )
=e −t − e−t
2 (cost+jsint+cost−jsint)
= e −t −e −t cost
Example21.26 Solve
x ′′ −5x ′ +6x=6t−4 x(0)=1,x ′ (0)=2
Solution TheLaplace transformofboth sides ofthe equation isfound.Let L{x} =X(s).
s 2 X(s) −sx(0) −x ′ (0) −5(sX(s) −x(0)) +6X(s) = 6 s 2 − 4 s
(s 2 −5s +6)X(s) = 6 s 2 − 4 s +s−3=s3 −3s 2 −4s+6
s 2