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

19.6 Constant coefficient equations 569
Solutions
1 (a) y=Ae x +Be 2x
(b)y=Ae −x +Be −6x
(c) x =Ae −2t +Be −3t
(d)y=Ae −t +Bte −t
(e) y=Ae 2x +Bxe 2x
(f) y = e −0.5t (Acos2.78t +Bsin2.78t)
(g)y=Ae x +Bxe x
(h) y = e −0.5t (Acos2.18t +Bsin2.18t)
(i) y=Ae −2x +Be x
(j) y=Acos3x+Bsin3x
(k)y=A+Be 2x
(l) x=Ae 4t +Be −4t
2 Lk 2 +Rk+ 1 C =0 i(t) =Aek 1 t +Be k 2 t
where √
R 2 C −4L
−R ±
C
k 1 ,k 2 =
2L
√ √ )
3x 3x
3 e
(Acos −x/2 +Bsin
2 2
19.6.2 Findingaparticularintegral
We stated in Property 2 (Section 19.5.2) that the general solution of an inhomogeneous
equation is the sum of the complementary function and a particular integral. We have
seenhowtofindthecomplementaryfunctioninthecaseofaconstantcoefficientequation.
We shall now deal with the problem of finding a particular integral. Recall that
theparticularintegralisanysolutionoftheinhomogeneous equation.Thereareanumber
of advanced techniques available for finding such solutions but these are beyond
the scope of this book. We shall adopt a simpler strategy. Since any solution will do
we shall try to find such a solution by a combination of educated guesswork and trial
and error.
Example19.26 Findthe generalsolutionofthe equation
d 2 y
dx − dy
2 dx −6y=e2x (19.14)
Solution ThecomplementaryfunctionforthisequationhasalreadybeenshowninExample19.18
tobe
y H
=Ae 3x +Be −2x
We shall attempt to find a solution of the inhomogeneous problem by trying a function
of the same form as that on the r.h.s. In particular, let us tryy P
(x) = αe 2x , where α is a
constant that we shall now determine. Ify P
(x) = αe 2x then dy P
dx = 2αe2x and d2 y P
=
dx 2
4αe 2x . Substitution inEquation (19.14) gives
thatis,
4αe 2x −2αe 2x −6αe 2x = e 2x
−4αe 2x = e 2x