Higher Engineering Mathematics, Sixth Edition

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.