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

19.6 Constant coefficient equations 563
The auxiliary equation ofa d2 y
dx 2 +bdy dx +cy=0is
ak 2 +bk+c=0
Solution of this quadratic equation gives the values of k which we seek. Clearly the
nature of the roots will depend upon the values ofa,bandc. Ifb 2 > 4ac the roots will
be real and distinct. The two values ofkthus obtained,k 1
andk 2
, will allow us to write
down two independent solutions:
y 1
(x) =e k 1 x
y 2
(x) =e k 2 x
and so the general solution of the differential equation will be
y(x) =Ae k 1 x +Be k 2 x
Iftheauxiliaryequationhasreal,distinctrootsk 1
andk 2
,thecomplementaryfunction
will be
y(x) =Ae k 1 x +Be k 2 x
On the other hand, if b 2 = 4ac the two roots of the auxiliary equation will be equal
andthismethodwillthereforeonlyyieldoneindependentsolution.Inthiscase,special
treatmentisrequired.Ifb 2 <4acthetworootsoftheauxiliaryequationwillbecomplex,
that isk 1
andk 2
will be complex numbers. The procedure for dealing with such cases
will become apparentinthe following examples.
Example19.20 Findthe generalsolutionof
d 2 y
dx 2 +3dy dx −10y=0
Find the particular solution which satisfies the conditionsy(0) = 1 andy ′ (0) = 1.
Solution By lettingy = e kx , so that dy
dx =kekx and d2 y
dx = 2 k2 e kx , the auxiliary equation is found
tobe
k 2 +3k−10=0
Therefore,
(k−2)(k+5)=0
sothatk = 2andk = −5.Thusthereexisttwosolutions,y 1
= e 2x andy 2
= e −5x .From
Property 1 we can writethe general solution as
y=Ae 2x +Be −5x