Higher Engineering Mathematics

SECOND ORDER DIFFERENTIAL EQUATIONS (HOMOGENEOUS) 479

I

When t=0,

dV

dt

= 3 ω,

thus 3 ω=Aω−Bω,

i.e. 3 =A−B (2)

From equations (1) and (2),A=5 andB= 2

Hencethe particular solution is

V=5eωt+2e−ωt

Since sinhωt=^12 (eωt−e−ωt)

and coshωt=^12 (eωt+e−ωt)

then sinhωt+coshωt=eωt

and coshωt−sinhωt=e−ωt from Chapter 5.

Hence the particular solution may also be

written as

V=5(sinhωt+coshωt)

+2(coshωt−sinhωt)

i.e.V=(5+2) coshωt+(5−2) sinhωt

i.e.V=7 coshωt+3 sinhωt

Problem 6. The equation

d^2 i

dt^2

+

R

L

di

dt

+

1

LC

i= 0

represents a current i flowing in an elec-

trical circuit containing resistance R, induc-

tance L and capacitance C connected in

series. If R=200 ohms,L= 0 .20 henry and

C= 20 × 10 −^6 farads, solve the equation fori

given the boundary conditions that whent=0,

i=0 and

di

dt

=100.

Using the procedure of Section 50.2:

(a)

d^2 i

dt^2

+

R

L

di

dt

+

1

LC

i=0 in D-operator form is

(

D^2 +

R

L

D+

1

LC

)

i=0 where D≡

d

dt

(b) The auxiliary equation ism^2 +

R

L

m+

1

LC

= 0

Hencem=

−

R

L

±

√

√

√

√

[(

R

L

) 2

−4(1)

(

1

LC

)]

2

WhenR=200, L= 0 .20 andC= 20 × 10 −^6 ,

then

m=

−

200

0. 20

±

√

√√

√

[(

200

0. 20

) 2

−

4

(0.20)(20× 10 −^6 )

]

2

=

− 1000 ±

√

0

2

=− 500

(c) Since the two roots are real and equal (i.e.− 500

twice, since for a second order differential equa-

tion there must be two solutions),the general

solution isi=(At+B)e−^500 t.

(d) Whent=0,i=0, henceB= 0

di

dt

=(At+B)(−500e−^500 t)+(e−^500 t)(A),

by the product rule

Whent=0,

di

dt

=100, thus 100=− 500 B+A

i.e.A=100, sinceB= 0

Hence the particular solution is

i= 100 te−^500 t

Problem 7. The oscillations of a heav-

ily damped pendulum satisfy the differential

equation

d^2 x

dt^2

+ 6

dx

dt

+ 8 x=0, where xcm is

the displacement of the bob at timetseconds.

The initial displacement is equal to+4 cm and

the initial velocity

(

i.e.

dx

dt

)

is 8 cm/s. Solve the

equation forx.

Using the procedure of Section 50.2:

(a)

d^2 x

dt^2

+ 6

dx

dt

+ 8 x=0 in D-operator form is

(D^2 +6D+8)x=0, where D≡

d

dt

(b) The auxiliary equation ism^2 + 6 m+ 8 =0.

Factorising gives: (m+2)(m+4)=0, from

which,m=−2orm=−4.

(c) Since the roots are real and different,the general

solution isx=Ae−^2 t+Be−^4 t.

(d) Initial displacement means that timet=0. At

this instant,x=4.