Higher Engineering Mathematics

488 DIFFERENTIAL EQUATIONS

5. A differential equation representing the

motion of a body is

d^2 y

dt^2

+n^2 y=ksinpt,

wherek,nandpare constants. Solve the

equation (givenn=0 andp^2 =n^2 ) given that

whent=0,y=

dy

dt

=0.

[

y=

k

n^2 −p^2

(

sinpt−

p

n

sinnt

)]

6. The motion of a vibrating mass is given by
   d^2 y
   dt^2

+ 8

dy

dt

+ 20 y=300 sin 4t. Show that the

general solution of the differential equation is

given by:

y=e−^4 t(Acos 2t+Bsin 2t)

+

15

13

( sin 4t−8 cos 4t)

7. L

d^2 q

dt^2

+R

dq

dt

+

1

C

q=V 0 sinωt represents

the variation of capacitor charge in an

electric circuit. Determine an expression

forqat timetseconds given thatR= 40 ,

L= 0 .02 H,C= 50 × 10 −^6 F,V 0 = 540 .8V

andω=200 rad/s and given the boundary

conditions that when t=0, q=0 and

dq

dt

= 4. 8

[

q=(10t+ 0 .01)e−^1000 t

+ 0 .024 sin 200t− 0 .010 cos 200t

]

51.6 Worked problems on differential

equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)where

f(x)is a sum or a product

Problem 9. Solve

d^2 y

dx^2

+

dy

dx

− 6 y= 12 x−50 sinx.

Using the procedure of Section 51.2:

(i)

d^2 y

dx^2

+

dy

dx

− 6 y= 12 x−50 sinxin D-operator

form is

(D^2 +D−6)y= 12 x−50 sinx

(ii) The auxiliary equation is (m^2 +m−6)=0,

from which,

(m−2)(m+3)=0,

i.e. m=2orm=− 3

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

u=Ae^2 x+Be−^3 x.

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

tial equation is the sum of a polynomial and a

sine function let the P.I.v=ax+b+csinx+

dcosx(see Table 51.1(e)).

(v) Substitutingvinto

(D^2 +D−6)v= 12 x−50 sinxgives:

(D^2 +D−6)(ax+b+csinx+dcosx)

= 12 x−50 sinx

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−50 sinx

Equating constant terms gives:

a− 6 b= 0 (1)

Equating coefficients of xgives:− 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=7 and

d=1.

Hence the P.I.,

υ=− 2 x−^13 +7 sinx+cosx