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.
- 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
]
- 4
d^2 y
dt^2
− 5
dy
dt
+y=0; whent=0,y=1and
dy
dt
=−2.
[
y=4e
1
4 t−3et
]
- (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)]
- (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.