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562 Chapter 19 Ordinary differential equations I
Solution As suggested we try a solution of the formy = e kx . Differentiating we find dy
dx =kekx
and d2 y
dx 2 =k2 e kx . Substitution into the given equation yields
that is,
k 2 e kx −ke kx −6e kx =0
(k 2 −k−6)e kx =0
The only way this equation can be satisfied forall values ofxisif
that is,
k 2 −k−6=0 (19.13)
(k−3)(k+2)=0
so thatk = 3 ork = −2. That is to say, ify = e kx is to be a solution of the differential
equationkmustbe either 3 or −2. We therefore have found two solutions
y 1
(x) = e 3x and y 2
(x) = e −2x
These two functions are linearly independent and we can therefore apply Property 1 to
give the general solution:
y H
(x) =Ae 3x +Be −2x
Equation (19.13) fordeterminingkiscalled theauxiliary equation.
Example19.19 Findthe auxiliary equation ofthe differential equation
a d2 y
dx 2 +bdy dx +cy=0
Solution We try a solution of the formy = e kx so that dy
dx =kekx and d2 y
dx = 2 k2 e kx . Substitution
into the given differential equation yields
that is,
ak 2 e kx +bke kx +ce kx =0
(ak 2 +bk+c)e kx =0
Since thisequation istobe satisfied for all values ofx, then
ak 2 +bk+c=0
isthe required auxiliary equation.