Higher Engineering Mathematics, Sixth Edition

Second order differential equations of the forma

d^2 y

dx^2 +b

dy

dx+cy=f(x)^491

(v) Substitutingvinto

(D^2 +D− 6 )v= 12 x−50sinxgives:

(D^2 +D− 6 )(ax+b+csinx+dcosx)

= 12 x−50sinx

D(ax+b+csinx+dcosx)

=a+ccosx−dsinx

D^2 (ax+b+csinx+dcosx)

=−csinx−dcosx

Hence (D^2 +D− 6 )(v)

=(−csinx−dcosx)+(a+ccosx

−dsinx)− 6 (ax+b+csinx+dcosx)

= 12 x−50sinx

Equating constant terms gives:

a− 6 b=0(1)

Equating coefficients ofxgives:− 6 a=12, from

which,a=−2.

Hence, from (1),b=−^13

Equating the coefficients of cosxgives:

−d+c− 6 d= 0

i.e. c− 7 d= 0

(2)

Equating the coefficients of sinxgives:

−c−d− 6 c=− 50

i.e. − 7 c−d=− 50

(3)

Solving equations (2) and (3) gives:c=7and

d=1.

Hence the P.I.,

υ=− 2 x−^13 +7sinx+cosx

(vi) The general solution,y=u+v,

i.e. y=Ae^2 x+Be−^3 x− 2 x

−^13 +7sinx+cosx

Problem 10. Solve the differential equation

d^2 y

dx^2

− 2

dy

dx

+ 2 y=3excos2x, given that when

x=0,y=2and

dy

dx

=3.

Using the procedure of Section 51.2:

(i)

d^2 y

dx^2

− 2

dy

dx

+ 2 y=3excos2x in D-operator

form is

(D^2 −2D+ 2 )y=3excos2x

(ii) The auxiliary equation ism^2 − 2 m+ 2 = 0

Using the quadratic formula,

m=

2 ±

√

[4− 4 ( 1 )( 2 )]

2

=

2 ±

√

− 4

2

=

2 ±j 2

2

i.e.m= 1 ±j 1.

(iii) Since the roots are complex, the C.F.,

u=ex(Acosx+Bsinx).

(iv) Since the right hand side of the given dif-

ferential equation is a product of an expo-

nential and a cosine function, let the P.I.,

v=ex(Csin2x+Dcos 2x)(see Table 51.1(f) —

again, constantsCandDare used sinceAandB

have already been used for the C.F.).

(v) Substitutingvinto(D^2 −2D+ 2 )v=3excos2x

gives:

(D^2 −2D+ 2 )[ex(Csin2x+Dcos 2x)]

=3excos2x

D(v)=ex( 2 Ccos2x− 2 Dsin 2x)

+ex(Csin2x+Dcos2x)

(≡ex{( 2 C+D)cos 2x

+(C− 2 D)sin2x})

D^2 (v)=ex(− 4 Csin2x− 4 Dcos2x)

+ex( 2 Ccos2x− 2 Dsin2x)

+ex( 2 Ccos2x− 2 Dsin2x)

+ex(Csin2x+Dcos 2x)

≡ex{(− 3 C− 4 D)sin2x

+( 4 C− 3 D)cos2x}

Hence (D^2 −2D+ 2 )v

=ex{(− 3 C− 4 D)sin2x

+( 4 C− 3 D)cos2x}

−2ex{( 2 C+D)cos2x

+(C− 2 D)sin2x}

+2ex(Csin2x+Dcos 2x)

=3excos2x