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

21.10 Solving linear constant coefficient differential equations 653
We also notethat
L(−t 2 ) = − 2 s 3
The Laplace transform of the differential equation is taken. This yields
s 2 Y(s)−2s−Y(s)=− 2 s 3
The equation isrearranged forY(s):
(s 2 −1)Y(s) = − 2 s 3 +2s
= 2(s4 −1)
s 3
= 2(s2 −1)(s 2 +1)
s 3
Bydividing the equation by (s 2 −1) weobtain
Y(s)= 2(s2 −1)(s 2 +1)
s 3 (s 2 −1)
= 2(s2 +1)
s 3
= 2s2
s 3 + 2 s 3 = 2 s + 2 s 3
Taking the inverse Laplace transform gives
y(t)=2+t 2
Example21.24 Solve
y ′′ +y ′ −2y=−2 y(0)=2,y ′ (0)=1
Solution Let L(y) =Y(s). Then
L(y ′ ) =sY(s) −y(0) =sY(s) −2
L(y ′′ ) =s 2 Y(s) −sy(0) −y ′ (0) =s 2 Y(s) −2s −1
We also notethat
L(−2) = − 2 s
We can now take the Laplace transform of the differential equation toget
and so
s 2 Y(s)−2s−1+sY(s)−2−2Y(s)=− 2 s
(s 2 +s−2)Y(s)=− 2 s +2s+3
(s −1)(s +2)Y(s) = 2s2 +3s−2 (2s −1)(s +2)
=
s s
Dividing the equation by (s −1)(s +2) gives
Y(s)=
(2s −1)(s +2)
s(s −1)(s +2) = 2s−1
s(s −1)