Higher Engineering Mathematics, Sixth Edition

Second order differential equations of the forma

d^2 y

dx^2 +b

dy

dx+cy=f(x)^485

(ii) Substitutingmfor D gives the auxiliary equa-

tionm^2 +m− 2 =0. Factorizing gives:(m− 1 )

(m+ 2 )=0, from whichm=1orm=−2.

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

u=Aex+Be−^2 x.

(iv) Since the term on the right hand side of the

given equation is a constant, i.e. f(x)=4, let

the P.I. also be a constant, say v=k (see

Table 51.1(a)).

(v) Substituting v=k into (D^2 +D− 2 )v= 4

gives (D^2 +D− 2 )k=4. Since D(k)=0and

D^2 (k)=0then− 2 k=4, from which,k=−2.

Hence the P.I.,v=− 2.

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

y=Aex+Be−^2 x− 2.

Problem 2. Determine the particular solution of

the equation

d^2 y

dx^2

− 3

dy

dx

=9, given the boundary

conditions that whenx=0,y=0and

dy

dx

=0.

Using the procedure of Section 51.2:

(i)

d^2 y

dx^2

− 3

dy

dx

=9 in D-operator form is

(D^2 −3D)y=9.

(ii) Substitutingmfor D gives the auxiliary equation

m^2 − 3 m=0. Factorizing 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)thenlet

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

= 3 Be^3 x−3;

dy

dx

=0whenx=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) Substituting m for D gives the auxiliary

equation 2m^2 − 11 m+ 12 =0. Factorizing gives:

( 2 m− 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−2gives:

(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 ofx gives: 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