Higher Engineering Mathematics, Sixth Edition

480 Higher Engineering Mathematics

3.

d^2 y

dx^2

+ 2

dy

dx

+ 5 y= 0

[y=e−x(Acos 2x+Bsin2x)]

InProblems4to9,findtheparticularsolutionofthe

givendifferential equationsfor the stated boundary

conditions.

4. 6

d^2 y

dx^2

+ 5

dy

dx

− 6 y=0; whenx=0,y=5and

dy

dx

=−1.

[

y=3e

2

3 x+2e−

3

2 x

]

5. 4

d^2 y

dt^2

− 5

dy

dt

+y=0; whent=0,y=1and

dy

dt

=−2.

[

y=4e

1

4 t−3et

]

6. (9D^2 +30D+ 25 )y=0, where D≡

d

dx

;when

x=0,y=0and

dy

dx

=2.

[

y= 2 xe−

5

3 x

]

7.

d^2 x

dt^2

− 6

dx

dt

+ 9 x=0; whent=0,x=2and

dx

dt

=0. [x= 2 ( 1 − 3 t)e^3 t]

8.

d^2 y

dx^2

+ 6

dy

dx

+ 13 y=0; whenx=0,y=4and

dy

dx

=0. [y=2e−^3 x(2cos2x+3sin2x)]

9. (4D^2 +20D+ 125 )θ=0, where D≡

d

dt

;when

t=0,θ=3and

dθ

dt

= 2 .5.

[θ=e−^2.^5 t(3cos5t+2sin5t)]

50.4 Further worked problems on

practical differential equations

of the forma

d^2 y

dx^2

+b

dy

dx

+cy= 0

Problem 4. The equation of motion of a body

oscillating on the end of a spring is

d^2 x

dt^2

+ 100 x= 0 ,

wherexis the displacement in metres of thebody

from its equilibrium position after timetseconds.

Determinexin terms oftgiven that at timet=0,

x= 2 mand

dx

dt

=0.

An equation of the form

d^2 x

dt^2

+m^2 x=0isadiffer-

ential equation representing simple harmonic motion

(S.H.M.). Using the procedure of Section 50.2:

(a)

d^2 x

dt^2

+ 100 x=0 in D-operator form is

(D^2 + 100 )x=0.

(b) The auxiliary equation is m^2 + 100 =0, i.e.

m^2 =−100 andm=

√

(− 100 ),i.e.m=±j10.

(c) Since the roots are complex, the general solution

isx=e^0 (Acos 10t+Bsin10t),

i.e.x=(Acos 10t+Bsin10t)metres

(d) Whent=0,x=2, thus 2=A

dx

dt

=− 10 Asin10t+ 10 Bcos10t

Whent=0,

dx

dt

= 0

thus 0=− 10 Asin0+ 10 Bcos0, i.e.B= 0

Hence the particular solution is

x=2cos10tmetres

Problem 5. Given the differential equation

d^2 V

dt^2

=ω^2 V,whereωis a constant, show that its

solution may be expressed as:

V=7coshωt+3sinhωt

given the boundary conditions that when

t=0,V=7and

dV

dt

= 3 ω.

Using the procedure of Section 50.2:

(a)

d^2 V

dt^2

=ω^2 V,i.e.

d^2 V

dt^2

−ω^2 V=0 in D-operator

form is(D^2 −ω^2 )v=0, where D≡

d

dx

.

(b) The auxiliary equation is m^2 −ω^2 =0, from

which,m^2 =ω^2 andm=±ω.