Higher Engineering Mathematics

SECOND ORDER DIFFERENTIAL EQUATIONS (NON-HOMOGENEOUS) 483

I

(ii) Substituting m for D gives the auxil-

iary equationm^2 − 3 m=0. Factorising gives:

m(m−3)=0, from which,m=0orm=3.

(iii) Since the roots are real and different, the C.F.,

u=Ae^0 +Be^3 x, i.e.u=A+Be^3 x.

(iv) Since the C.F. contains a constant (i.e.A) then

let the P.I.,v=kx(see Table 51.1(a)).

(v) Substitutingv=kxinto (D^2 −3D)v=9gives

(D^2 −3D)kx=9.

D(kx)=kand D^2 (kx)=0.

Hence (D^2 −3D)kx= 0 − 3 k=9, from which,

k=−3.

Hence the P.I.,v=− 3 x.

(vi) The general solution is given byy=u+v, i.e.

y=A+Be^3 x− 3 x.

(vii) Whenx=0,y=0, thus 0=A+Be^0 −0, i.e.
0 =A+B (1)
dy
dx

= 3 Be^3 x−3;

dy

dx

=0 when x=0, thus

0 = 3 Be^0 −3 from which,B=1. From equa-

tion (1),A=−1.

Hence the particular solution is

y=− 1 +1e^3 x− 3 x,

i.e. y=e^3 x− 3 x− 1

Problem 3. Solve the differential equation

2

d^2 y

dx^2

− 11

dy

dx

+ 12 y= 3 x−2.

Using the procedure of Section 51.2:

(i) 2

d^2 y

dx^2

− 11

dy

dx

+ 12 y= 3 x−2 in D-operator

form is

(2D^2 −11D+12)y= 3 x− 2.

(ii) Substitutingmfor D gives the auxiliary equa-

tion 2m^2 − 11 m+ 12 =0. Factorising gives:

(2m−3)(m−4)=0, from which, m=^32 or

m=4.

(iii) Since the roots are real and different, the C.F.,

u=Ae

3

2 x+Be^4 x

(iv) Sincef(x)= 3 x−2 is a polynomial, let the P.I.,

v=ax+b(see Table 51.1(b)).

(v) Substitutingv=ax+binto

(2D^2 −11D+12)v= 3 x−2 gives:

(2D^2 −11D+12)(ax+b)= 3 x−2,

i.e.2D^2 (ax+b)−11D(ax+b)

+12(ax+b)= 3 x− 2

i.e. 0 − 11 a+ 12 ax+ 12 b= 3 x− 2

Equating the coefficients ofxgives: 12a=3,

from which,a=^14.

Equating the constant terms gives:

− 11 a+ 12 b=−2.

i.e.− 11

( 1

4

)

+ 12 b=−2 from which,

12 b=− 2 +

11

4

=

3

4

i.e.b=

1

16

Hence the P.I.,v=ax+b=

1

4

x+

1

16

(vi) The general solution is given byy=u+v, i.e.

y=Ae

3

2 x+Be^4 x+

1

4

x+

1

16

Now try the following exercise.

Exercise 190 Further problems on differen-

tial equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x) wheref(x) is a

constant or polynomial.

In Problems 1 and 2, find the general solutions

of the given differential equations.

1. 2

d^2 y

dx^2

+ 5

dy

dx

− 3 y= 6

[

y=Ae

1

2 x+Be−^3 x− 2

]

2. 6

d^2 y

dx^2

+ 4

dy

dx

− 2 y= 3 x− 2

[

y=Ae

1

3 x+Be−x− 2 −^32 x

]

In Problems 3 and 4 find the particular solutions

of the given differential equations.

3. 3

d^2 y

dx^2

+

dy

dx

− 4 y=8; whenx=0,y=0 and

dy

dx

=0.

[

y=^27 (3e−

4

3 x+4ex)− 2

]