Sec.
1.6]
differential equations of the first order
Then integrating,
19
we have
_
qy2
2
vl
3
Since y = 0 when x = 0, it is evident that
the path of the boat is
x
ix
c
+
c
must be zero.
 (8ay' 
Henc
46
ADVANCED ENGINEERING
MATHEMATICS
[SEC. 2.5
but also these terms multiplied by x, and is therefore
ax
ci cos 2x + cj sin 2x + Cjx cos 1x +
To find a particular integral, we try Y
into the differenti
ADVANCED ENGINEERING
44
6. Using
MATHEMATICS
[SEC. 2.5
the method of variation of parameters, find a particular integral of the
y" + w*y = sin kx. Discuss the limiting case when k*u, and show
equation
linear differential equations
Sec. 2.5]
The characteristic
equation in this case is
m
+
5m
+ 9m +

5
45
0
By inspection*
m =
is seen
to be a root.
1
Hence
must be one term in the complementary funct
Solving Eqs.
we obtain
v
y'lt
v
a
found by
v? and
43
and R(x) are all known. Hence u and can
With u and known, the particular
single integration.
The functions yi, yi, y[,
be
for
(4) and (5)
v',
linea
42
ADVANCED ENGINEERING
MATHEMATICS
[SEC. 2.4
generality is usually the inconvenience of integrals which cannot be
evaluated in terms of familiar functions.
The fundamental idea behind the process is
LINEAR DIFFERENTIAL EQUATIONS
Sec. 2.4]
y" +
S.
3y'

2 sin x
+
V" + y' = 1 + 2x
y" + t/ = sin
e* + 2
y"
V
i
 
4.
5
V ~
y" =
y" +
cos 3a:
i
7.

13.
16.
Find


a particular
 2y' + y
y"
10.
(Z>
+
40
ADVANCED ENGINEERING
MATHEMATICS
Solving these for C and D, we obtain C = i,
_
Y
and finally
y =
e
_./
D
cos x + sin
j
J.
[SEC.
2.3
Hence
4
j
cos z + sin x
i r>
\
*(A cos x + B sin x)
^
Example 6
W
linear differential equations
Sec. 2.3]
39
The roots of the characteristic equation,
+
m*
are mi
2, m, 3.
+
5m
G
=0
Hence the complementary
function is
cie, + cut"
For
the trial solution correspondi
38
ADVANCED ENGINEERING
MATHEMATICS
[SEC. 2.3
Substituting this and its derivatives into the given equation, we find
2(Be*) + 3(Be') + (A + Be')
4
e*
  e'
or
A
6Be* +
 
4
This equation will be ide
ADVANCED ENGINEERING
36
24.
MATHEMATICS
[SEC. 2.3
If the roots of its characteristic equation are real, show that no solution of the
homogeneous differential equation can have more than one real zero.
linear differential equations
Sec. 2.2]
The characteristic
35
equation in this case is
and its roots are mi

+ m
3m
1 and mt
.
y =
2 = 0
The general solution is therefore
+
Cie~*
Cae1*"
There is no
linear differential equations
Sec. 2.2]
33
This com
= ma.
xemiZ is also a solution when wii
remains to verify that yi = em,x and j/2 = xemiX have a nonvanishing
This value of mi clearly reduces the la
ADVANCED ENGINEERING
34
The characteristic
equation of the differential

m*
Its roots are mi =
+
4m
[SEC. 2.2
equation is
= 0
4
m = 2; hence the general solution is
+
Cie,x
y
By differentiation
MATH
ADVANCED ENGINEERING
32
we have immediately
y

[SEC. 2.2
MATHEMATICS
(ci + cs)e* =
Cie'"
This cannot be the general solution of the given equation because it contains only one
arbitrary constant, wh
linear differential equations
Sec. 2.2]
31
devise a more convenient form for the general solution of
Eq. (2) in the
in which mi and m2 are conjugate complex quantities.
To do this, let us suppose that
LINEAR DIFFERENTIAL EQUATIONS
Sec. 2.6]
ence level.
49
Its potential energy is therefore
(<*+a + jf)!
(4)
Substituting from (1), (2), (3), and
(4) into the relation
 constant
K.E. + P.E.
we have
;
f
LINEAR DIFFERENTIAL EQUATIONS
Sec. 2.6]
the given conditions into Eqs. (1), (2), (3), and (4), we find
Substituting
0

a
+
0 =
0 =
0
Solving these simultaneously

b
b

+
6
+2
+2
c
+ 2d
2c
b +3c 
Sec. 3.3]
simultaneous linear differential equations
65
the equations of the given system (2) are all linear, sums of solu
tions will also be solutions.
Hence we can combine the three particular
solut
ADVANCED ENGINEERING
64
MATHEMATICS
Thus
[Sec.
3.3
Mi
2
and thus the first of the three particular solutions is
xi
Vi
For nij
= 1, we have
=
i
similarly, from Eq. (3),
9.42
A2

6B2
 3C2
= 0
+ 2Bi +
simultaneous linear differential equations
Sec. 3.3]
63
Now let us attempt to find solutions of this system of the form
y = Be",
x = Aeml,
z = Cemt
Substituting these values into the equations in (2)
ADVANCED ENGINEERING
62
MATHEMATICS
[Sec. 3.3
Thus four relations must exist among the original eight arbitrary con
stants.
Solving these equations, we find (among equivalent possibilities
E
=
A,
F
=
Sec. 3.2]
simultaneous linear differential equations
A particular integral is obviously
Y
= 3; hence the general solution is
y = Ae~' + Be*' + C cos
(6)
61
21
+ D sin
21

3
To find x, we can substit
CHAPTER
3
SIMULTANEOUS LINEAR DIFFERENTIAL EQUATIONS
In many applied problems there are not one, but
Introduction.
several dependent variables, each a function of a single independent
The formulation
LINEAR DIFFERENTIAL EQUATIONS
Sec. 2.6]
57
be neglected,
determine the natural frequency of the oscillations that occur when
the system is slightly disturbed.
(Hint: Use the energy method to obtain th
ADVANCED ENGINEERING
58
Determine
each one.
Am.
the possible deflection
If P
=
BI,
n =
in a deflected position given by x
curves and the loads required
1, 2, 3
MATHEMATICS
A sin
[Sec.
2.6
to produce
t
Substituting
equation
[SEC. 2.6
MATHEMATICS
ADVANCED ENGINEERING
56
these moments into the basic equation (12), we obtain the differential
Ely"
=
cos 8)y
(F

(F sin
$)x
or
 B sinh cfw_y^P
*)
and sub
Sec.
linear differential equations
2.6]
and the moment at x + Ax is
M(x +
Ax) =
(ii
 x + Ax)Fi +
+
(xn
55
 x + Ax)Fn
Subtracting these, we find
M(x +
Ax)
 Mix)
=
AM
Ax(Fi
=
+
+Fn)
=
VAx
Hence