Higher Engineering Mathematics, Sixth Edition

482 Higher Engineering Mathematics

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.

Thus 4=A+B (1)

Velocity,

dx

dt

=− 2 Ae−^2 t− 4 Be−^4 t

dx

dt

=8cm/swhent= 0 ,

thus 8=− 2 A− 4 B (2)

From equations (1) and (2),

A=12 andB=− 8

Hence the particular solution is

x=12e−^2 t−8e−^4 t

i.e.displacement,x=4(3e−^2 t−2e−^4 t)cm

Now try the following exercise

Exercise 188 Further problems on second

order differential equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy= 0

1. The charge,q, on a capacitor in a certain
   electrical circuit satisfies the differential equa-
   tion

d^2 q

dt^2

+ 4

dq

dt

+ 5 q=0. Initially (i.e. when

t=0), q=Q and

dq

dt

=0. Show that the

charge in the circuit can be expressed as:

q=

√

5 Qe−^2 tsin(t+ 0. 464 ).

2. A body moves in a straight line so that its
   distancesmetres from the origin after time
   tseconds is given by

d^2 s

dt^2

+a^2 s=0, where a

is a constant. Solve the equation forsgiven

thats=cand

ds

dt

=0whent=

2 π

a

.

[s=ccos at]

3. The motion of the pointer of a galvanometer
   aboutits positionofequilibriumis represented
   by the equation

I

d^2 θ

dt^2

+K

dθ

dt

+Fθ= 0.

IfI, the moment of inertia of the pointer about

its pivot, is 5× 10 −^3 ,K, the resistance due to

friction at unit angular velocity, is 2× 10 −^2

andF, the force on the spring necessary to

produce unit displacement, is 0.20, solve the

equation forθin terms oftgiven that when

t=0,θ= 0 .3and

dθ

dt

=0.

[θ=e−^2 t( 0 .3cos6t+ 0 .1sin6t)]

4. Determineanexpressionforxfor adifferential

equation

d^2 x

dt^2

+ 2 n

dx

dt

+n^2 x=0 which repre-

sents a critically damped oscillator, given that

at timet=0,x=sand

dx

dt

=u.

[x={s+(u+ns)t}e−nt]

5. L

d^2 i

dt^2

+R

di

dt

+

1

C

i=0isanequationrepre-

senting currenti in an electric circuit. If

inductanceL is 0.25 henry, capacitanceC

is 29. 76 × 10 −^6 farads and R is 250ohms,

solve the equation forigiven the boundary

conditions that whent=0,i=0and

di

dt

=34.

[

i=

1

20

(

e−^160 t−e−^840 t

)

]

6. The displacementsof a body in a damped
   mechanical system, with no external forces,
   satisfies the following differential equation:

2

d^2 s

dt^2

+ 6

ds

dt

+ 4. 5 s= 0

where t represents time. If initially, when

t= 0 ,s=0and

ds

dt

=4, solve the differential

equation forsin terms oft.[s= 4 te−

3

2 t]