Higher Engineering Mathematics, Sixth Edition

Second order differential equations of the forma

d^2 y

dx^2 +b

dy

dx+cy=^0481

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

solution is

V=Aeωt+Be−ωt

(d) Whent=0,V=7 hence 7=A+B (1)

dV

dt

=Aωeωt−Bωe−ωt

When t= 0 ,

dV

dt

= 3 ω,

thus 3ω=Aω−Bω,

i.e. 3 =A−B (2)

From equations (1) and (2),A=5andB= 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=7coshωt+3sinhωt

Problem 6. The equation

d^2 i

dt^2

+

R

L

di

dt

+

1

LC

i= 0

represents a currentiflowing in an electrical circuit

containing resistanceR, inductanceLand

capacitanceCconnected in series. IfR=200ohms,

L= 0 .20 henry andC= 20 × 10 −^6 farads, solve the

equation forigiven the boundary conditions that

whent=0,i=0and

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=0whereD≡

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 heavily damped

pendulum satisfy the differential equation

d^2 x

dt^2

+ 6

dx

dt

+ 8 x=0, wherexcm is the

displacement of the bob at timetseconds.

The initial displacement is equal to+4cmandthe

initial velocity

(

i.e.

dx

dt

)

is 8cm/s. Solve the

equation forx.