Higher Engineering Mathematics

476 DIFFERENTIAL EQUATIONS

(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
problems of Section 50.3 may each be verified by

substituting expressions fory,

dy

dx

and

d^2 y

dx^2

into the

original equation.

50.3 Worked problems on differential

equations of the form

a

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 parti-

cular solution given that whenx=0,y=4 and

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: (2m−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=6 andB=−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

whent=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=0 where D≡

d

dt

(b) Substitutingmfor D gives the auxiliary equation

9 m^2 − 24 m+ 16 =0.

Factorising gives: (3m−4)(3m−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.

(d) Whent=0,y=3 hence 3=(0+B)e^0 , i.e.B=3.

Sincey=(At+B)e

4

3 t

then

dy

dt

=(At+B)

(

4

3

e

4

3 t

)

+ Ae

4

3 t, by the

product rule.