478 Higher Engineering Mathematics
50.2 Procedure to solve differential
equations of the form
a
d^2 y
dx^2
+b
dy
dx
+cy= 0
(a) Rewrite the differential equation
a
d^2 y
dx^2
+b
dy
dx
+cy= 0
as (aD^2 +bD+c)y= 0
(b) Substitutemfor Dand solve the auxiliary equation
am^2 +bm+c=0form.
(c) If the roots of the auxiliary equation are:
(i) real and different,saym=αandm=β,
then the general solution is
y=Aeαx+Beβx
(ii) real and equal, saym=αtwice, then the
general solution is
y=(Ax+B)eαx
(iii) complex,saym=α±jβ, then the general
solution is
y=eαx{Acosβx+Bsinβx}
(d) Given boundary conditions, constantsAandB,
may be determined and theparticular solution
of the differential equation obtained.
The particular solutions obtained in the worked prob-
lems of Section 50.3 may each be verified by substi-
tuting expressions fory,
dy
dx
and
d^2 y
dx^2
into the original
equation.
50.3 Worked problemson
differential equations of
the forma
d^2 y
dx^2
+b
dy
dx
+cy= 0
Problem 1. Determine the general solution of
2
d^2 y
dx^2
+ 5
dy
dx
− 3 y=0. Find also the particular
solution given that whenx=0,y=4and
dy
dx
=9.
Using the above procedure:
(a) 2
d^2 y
dx^2
+ 5
dy
dx
− 3 y=0 in D-operator form is
(2D^2 +5D− 3 )y=0, where D≡
d
dx
(b) Substitutingmfor D gives the auxiliary equation
2 m^2 + 5 m− 3 = 0.
Factorising gives: ( 2 m− 1 )(m+ 3 )=0, from
which,m=^12 orm=−3.
(c) Since the roots are real and different thegeneral
solution isy=Ae
1
2 x+Be−^3 x.
(d) Whenx=0,y=4,
hence 4=A+B (1)
Since y=Ae
1
2 x+Be−^3 x
then
dy
dx
=
1
2
Ae
1
2 x− 3 Be−^3 x
When x= 0 ,
dy
dx
= 9
thus 9 =
1
2
A− 3 B (2)
Solving the simultaneous equations (1) and (2)
givesA=6andB=−2.
Hence the particular solution is
y= 6 e
1
2 x− 2 e−^3 x
Problem 2. Find the general solution of
9
d^2 y
dt^2
− 24
dy
dt
+ 16 y=0 and also the particular
solution given the boundary conditions that when
t=0,y=
dy
dt
=3.
Using the procedure of Section 50.2:
(a) 9
d^2 y
dt^2
− 24
dy
dt
+ 16 y=0 in D-operator form is
(9D^2 −24D+ 16 )y=0whereD≡
d
dt
(b) Substitutingmfor D gives the auxiliary equation
9 m^2 − 24 m+ 16 =0.
Factorizing gives: ( 3 m− 4 )( 3 m− 4 )=0, i.e.
m=^43 twice.
(c) Since the roots are real and equal,the general
solution isy=(At+B)e
4
3 t.