Higher Engineering Mathematics

(Greg DeLong) #1

480 DIFFERENTIAL EQUATIONS


Thus 4=A+B (1)

Velocity,

dx
dt

=− 2 Ae−^2 t− 4 Be−^4 t

dx
dt

=8 cm/s whent=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 189 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 elec-
    trical circuit satisfies the differential equa-


tion

d^2 q
dt^2

+ 4

dq
dt

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

t=0), q=Qand

dq
dt

=0. Show that the
charge in the circuit can be expressed as:
q=


5 Qe−^2 tsin (t+ 0 .464)


  1. 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

=0 whent=

2 π
a
[s=ccos at]


  1. The motion of the pointer of a galvanometer
    about its position of equilibrium is repre-
    sented by the equation


I

d^2 θ
dt^2

+K


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 .3 and


dt

=0.

[θ=e−^2 t(0.3 cos 6t+ 0 .1 sin 6t)]


  1. Determine an expression forxfor a differ-


ential equation

d^2 x
dt^2

+ 2 n

dx
dt

+n^2 x=0 which
represents 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=0 is an equation repre-
senting current iin an electric circuit. If
inductanceLis 0.25 henry, capacitanceC
is 29. 76 × 10 −^6 farads andR is 250 ohms,
solve the equation forigiven the boundary

conditions that whent=0,i=0 and

di
dt

=34.
[
i=

1
20

(
e−^160 t−e−^840 t

)

]


  1. 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

wheretrepresents time. If initially, when

t=0,s=0 and

ds
dt

=4, solve the differential

equation forsin terms oft.[s= 4 te−

3
2 t]
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