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' 
Hence the equation of
22/')
To find the distance downstream
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 differential equation gives
cos x
(.4
+ B sin x) + 8(
A
i + 9S si
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
Y
that it leads to
7. Show
=
1 eB
uX.
that the genera
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 function.
When the factor corresponding
to this root is divi
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',
linear differential equations
Sec. 2.4]
= uyi
completely det
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 this. Instead of using two
arbitrary constants, C\ and
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
What is the general solution of the equation
y"2y,+y~xe
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 corresponding to 3e~* we would normally try Ae'*. How~* is
already
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 identically true if and only if the coefficients of like t
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.
Find the solution
conditions.
3/
of each of the follow
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 need to differentiate in this problem because neither
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 last term in (5) to zero.
pletes the proof that y =
It

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
MATHEMATICS
ctxe1*
we find
y' = (2ci +
+
c2)e*
2c2ieto
Sub
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, whereas the general solution of a secondorder equation m
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
case
mi = p + iq
so
mt = p iq
and
that the general sol
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 (*)'
+
sf @y + a 
^
()'Ik.
+
 iA^ 
c + . + >ic
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 6d
for o,
12,
a =
and
47
and d gives
6, c,
6 = 12,
c =
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
solutions into the general solution
Since
Xi +
x2
x3 =
Ai +
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 + 3C2 =
4At
If
A
 3^!
+
0
2C2 = 0
we solve for B2 and C2
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) and dividing out
the common factor emt leads to the set
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
=
G =
2(C + D),
H
=
2(C
 D)
It is tedious but perfectl
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 substitute for y in either of the original differential
equati
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 of such problems in mathemati
variable, usually time.
c
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 the
Fio.
Fig.
2.7.
2.8.
of very small motions, determine
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
the column will be in equilibrium
y)
*
A sin
For all
oth
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 substituting,
.
cosh
V(F
cos
9)
Ly
"\
V^cos
.1
.
(tan
9/A
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, dividing by Ax and letting Ax > 0, we have
V
dcfw_EI