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

19.6 Constant coefficient equations 567
WritingA =C +DandB =Cj −Dj,wefind
y =e αx (Acosβx+Bsinβx)
This isthe required solution.
Iftheauxiliaryequationhascomplexroots α +βjand α −βj,thenthecomplementaryfunction
is
y =e αx (Acosβx+Bsinβx)
Note that Example 19.21 isaspecial case of Example 19.22 with α = 0 and β = 2.
Example19.23 Find the general solution ofy ′′ +2y ′ +4y = 0.
Solution The auxiliary equation isk 2 +2k +4 = 0.This equation has complex roots given by
k = −2 ± √ 4−16
2
= −2 ± √ 12j
2
=−1± √ 3j
Using the result of Example 19.22 with α = −1 and β = √ 3 we find the general
solution is
y = e −x (Acos √ 3x+Bsin √ 3x)
Example19.24 Theauxiliaryequationofay ′′ +by ′ +cy = 0isak 2 +bk +c = 0.Supposethisequation
has equal rootsk =k 1
. Verifythaty =xe k 1 x isasolution ofthe differential equation.
Solution We have
y =xe k 1 x y ′ =e k 1 x (1+k 1
x) y ′′ =e k 1 x (k 2 1 x +2k 1 )
Substitutioninto the l.h.s. ofthe differential equation yields
e k x{ 1 a ( k 2 1 x +2k )
1 +b(1+k1 x)+cx } =e k x{( 1 ak 2 1 +bk 1 +c) x +2ak 1
+b }
Butak1 2 +bk 1 +c = 0 sincek 1
satisfies the auxiliary equation. Also,
k 1
= −b ± √ b 2 −4ac
2a
butsincetherootsareequal,thenb 2 −4ac = 0andhencek 1
= − b
2a .So2ak 1 +b=0.
We conclude thaty = xe k 1 x is a solution ofay ′′ +by ′ +cy = 0 when the roots of the
auxiliary equation areequal.