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Differential Equations
By Harry W. Reddick
DIFFERENTIAL EQUATIONS
245 pages. 13 illustrations. 5% by 8~. Cloth.
.
1/1
By Harry W. Reddick and Frederic H. MiII~r
ADVANCED MATHEMATICS FOR ENGINEERS
473 pages. 130 illustrations. 6 by 9. Cloth.
/
. ,
..
DI FFERENTI A I.
Harry
W. Reddick
PROFESSOR OF MATHEMATICS NEW YORK UNIVERSITY (UNIVERSITY HEIGHTS)
NEW YORK: JOHN WILEY & SONS, INC. LONDON:
CH~,,>MAN
AND HALL. LIMITED
CoI'YRIGRT.l94a BY HARRY
W. REDDICK
All Rights &Served This book 01' any par! thereoj mu.! not be rejW9lJucea in. e;n,y . "Ie n with9Ut rJw wrilten permiuion ojfAe publisher.
SEVENTH PRINTING, MARCH,
1948
PRINTED IN THE UNITED STATES OF AMERICA
PREFACE
This book deals with methods of solving ordinary differential equations and with problems in applied mathematics involving ordinary differential equations; it includes no treatment of partial differential equations. It has been the aim of the author, in both the theory and the hundred illustrative examples, to achieve a clarity of explanation which will enable any student who desires to understand actually to do so. In the physical applications the · importance of setting forth clearly the physical units involved is stressed, so that numerical results with proper units attached may be obtained. There are more than 600 problems, It should be a source of llatisfaction to the student that answers are given to all of 'hem. The book is suitable for courses of various lengths in both Engineering and Liberal Arts schools. For a short course of two hours per week for one semester in an Engineering school, the following selection has been found appropriate: Chapters I, 3, 4, 5, and 6, omitting Articles 21, 25, 26, 41, 47, 48, and 49. HARRY W. REDDICK New York, N. Y. January, 1943
v
.
COI o. A symbol, such as C in equation (1), which is constant for any one curve, but which can take on an arbitrary number of different values, is called a parameter. A family of curves, such as (1), which contains one parameter, is called a oneparameter family, or a singly infinite family, or a family of 00 1 curves, the exponent on the 00 indicating the number of parameters in the equation of the family. Thus we may say that in a plane there are 00 1 concentric circles with fixed center. How many circles are there in the xyplane with centers on the xaxis? The answer is 00 2, since the equation of such a family is (x  a)2 + y2 = r2, where a and r are arbitrary constants or parameters. The differential equation of the fanrlly is of second order, since two arbitrary constants will have to be eliminated to produce it. On the other hand there are only 00 1 circles of radius 1 with centers on the zaxis: the equation of this family is (x  a)2+ y2 = 1, containing only one parameter, and the differential equation will be of first order. The phrase, "in the xyplane," is always understood in such an example, since we are dealing with only two variables. In general, an nparameter family of curves will be represented by a differential equation of nth order. To obtain the differential equation of the family, we write, by analytic geometry, the equation of the family containing the proper number of essential parameters, then eliminate the parameters by one of the methods explained in the preceding articles.
24
Chapter 2
1. Find the differential equation of the family of circles concentric about the origin. • The equation of the family of circles is EXAMPLE
x2
Differentiating, we have
x dx
+ y2 = c. + y dy = 0,
or
dy dx
+ x = 0, Y
which is the differential equation required. 2. Find the differential equation of all circles with centers on the zaxis. The equation of the family is EXAMPLE
(x  a)2
+ y2 =
r 2•
Differentiating twice, we have
x  a + yy' = 0, then
1 + yy"
+ s" =
0,
which is the required differential equation. 3. Find the differential equation of (a) all vertical parabolas, i.e., parabolas with vertical axes; (b) all vertical parabolas with vertices on the line y = z, There are 00 3 vertical parabolas. The equation of the family may be written in either of two ways: EXAMPLE
y  k = a(x  h)2,
y
= Ax2 + Bx + C.
Using the second form, and differentiating three times, we have
= 2Ax + B, y" = 2A, y'" = 0. y'
The last equation is the answer to part (a).
Article 8
25
For part (b) we use the first form, (h, k) being the vertex of the parabola. The condition requires that k = h; hence the equation is y  h
h)2,
= a(x 
which represents a twoparameter family. Differentiating twice, we have y' = 2a(x  h), y"
= 2a,
from which y' xh=y" ,
y" 2 '
a=
y  h
=
y 
x
y' +. y"
Substituting these values in the original equation, we have Y
X
+
y' y"
=
y,2 2y" ,
or 2(y  x)y"
= y,2  2y',
which is the answer to part (b). PROBLEMS Eliminate the arbitrary constants from the following equations, encountered in Chapter 1, and thus obtain the corresponding differential equations.
+ + + + + + + +
1. y = C1 cos (x C2) . 2. y = A sin x B cos x. 4 2 3. y = x C1X C.g; Cs. 4. y = x 0(1 xy). 5. X cos" a y = cx2  x sin2 a. 6. (x  a)2 (y  b)2 = r 2. '1. (ax + b) (ay + b) = c (c arbitrary). 8. (ax + b) (ay + b) = c (a, b, c arbitrary). 9. y = Ae x Be:", 10. y = a sinh (x b).
+
+
+
Eq. (9), Art. Eq. (10), Art. Prob. l(b), Art. Probe 2, Art. PrQb. 3(b), Art. Probe 3(c), Art. Probe 4, Art.
2. 2. 3. 3. 3. 3. 3.
Probe 7(a), Art. 3. Probe 7(c), Art. 3.
Find the differential equations of the following families of curves. 11. Circles with centers on the yaxis. 12. Circles of fixed radius r, with centers on the zexis. 13. Circles with centers on the line y =:= X.
26 14. 16. 16. 17. 18. 19. 20.
Chapter 2 Circles with centers on the line y = x, and passing through the origin. Circles tangent to the yaxis at the origin. Parabolas with axis on the xaxis. Parabolas with axis on the xaxis and focus at the origin. Horizontal parabolas tangent to the yaxis. Vertical parabolas passing through the point (a, b). The probability curves, y = (h/Y!;)e h2:r?.
Find the differential equations from the following general solutions. 21. (x 2 + a2) y 2 = a2x2• 22. y = Ae~+" + Be~+". 23. y = Ae~ + Be2~ + 24. (x 5y + 9)4 = C(x + 2y + 3).
ce.
+
26.
tanh(~+~) = V3tan(~x+c} +
26. y = Aellv':i BelI,;/i. 27. y = A V' 1 + x 2 Bz,
sa. y =
+
A(l +x+ ;) + Be".
+ x sin x) + B(sin x  x cos z), 30. y = A sinh x + B cosh x. x x SL y = A cos 2x cosh (2x + a) + B sin 2x sinh (2x + (j), where A, B, a, 29. y = A (cos x
and {j are arbitrary constants. S2. Find the differential equation whose general solution is
(a) if A, B are arbitrary; (b) if A, B, n are arbitrary.
33. Find the differential equation of (a) the doubly infinite family of catenaries, y = a cosh [(x  b)/a]; (b) the triply infinite family of catenaries, y = a cosh [(x  b)/a] + c. . 34. Find the differential equation of all parabolas tangent to the yaxis at the origin. 36. Write (a) in rectangular coordinates, (b) in polar coordinates, the equation of the family of all circles passing through the origin. Find the differential equation in each case and transform one differential equation into the other. 36. Find the differential equation of the family of hyperbolas having (a) asymptotes parallel to the coordinate axes; (b) asymptotes parallel to the
Article 8
27
coordinate axes and centers on the line y = x. Show, from the two differential equations, without using their general solutions, that every solution of the second differential equation is a solution of the first. 37. In Prob. 3(a), Art. 3, we found from the general equation of a. conic that the differential equation of all conics is of the fifth order. This differential equation may be found by dividing out one of the constants in the general equation (or, what amounts to the same thing, putting one constant equal to unity), then differentiating five times, and eliminating the five essential constants from the six equations by means of a determinant of the sixth order. The result comes out surprisingly simple. There is a shorter method, however, due to Halphen. (See GoursatHedrickDunkel, Mathematical Analysis, Vol. II, Part II, p. 5.) Find the differential equation of the conics by one of these two methods. 38. Find the differential equation of all parabolas. Show, from this differential equation and that of the conics, Probe 37, that every solution of the former is a solution of the latter.
elu.zptta 3 DIFFERENTIAL EQUATIONS OF FIRST ORDER 9. Introduction. In this chapter we consider various types ()f differential equations of first order and the methods of solving them, together with their physical and geometric applications. The equations will involve only two variables, say z and y, and will be of first degree in the derivative, dy/dx. Such an equation can be written, in differential form,
M dx + N dy = 0,
(1)
where M and N are, in general, functions of z and y. 10. Separable equations•. If, in equation (1), Art. 9, M is a function of x only (or a constant) and N is a function of y only (or a constant), the equation has its variables separated and can be integrated at once. A differential equation which can be written in the form, M(x) dx
+ N(y) dy
= 0,
is called separable. EXAMPLE
1. Solve dy dx
x
=
u'
Separating the variables, we have x dx  y dy
Integration gives
or
= O.
(1)
29
Article 10
which is the general solution. The constant of integration can be written as any function of 0; here we chose 0/2. 2. Solve the following differential equation, obtaining the general solution in algebraic form, and prove that the solution is correct: (2) EXAMPLE
We first write the equation in the form
dx
dy
1 + x 2 + 1 + y2 =
o.
Integration then gives arc tan x
+ arc tan y
= arc tan O.
Here we choose the constant of integration in the form arc tan 0, since in the next step we are going to take the tangent of both sides in order to express the result in algebraic form. In doing this we apply the formula for the tangent of the sum of two angles, namely, tan (ex with ex
+ (j)
tan ex
+ tan P
= 1  tan ex tan (j '
= arc tan x, (j = arc tan y, and obtain x+y = 0 1  xy ,
or x
+ y = 0(1
(3)
 xy),
(4)
which is the required solution. The solution may be proved correct either by eliminating 0 from the solution (4) and producing the differential equation (2), or by substituting in (2) the values of y and dy/dx obtained from (4), thus producing an identity in z. By the first method, writing (4) in the form (3) and differentiating, we have y dx) = 0, (1  xy)(dx + dy) + (z + y)(x dy (1 + y2) dx + (1 + x 2) dy = 0,
+
which is equation (2).
30
Chapter 3
By the second method we have, from equation (4), dy 1+C2 = dx (1 + Cx)2'
Cx y = 1 Cx'
+ 2)(1 1 + 11 = (1 + CX)2 + (C  X)2 = (1 + 0 + x2). (1 + CX)2 (1 + CX)2 Substitution of these values for 1 + y2 and dy/ dx in equation (2) produces an identity.
In the next example and frequently thereafter we shall have occasion to take antilogarithms of both sides of an equation in order to reduce it to an equation free of logarithms. Suppose, for example, that we wish to remove the logarithms from the equation (5) 2 log x  3 log y = log C + sin a, by taking antilogarithms of both sides. This means that we must write an equation such that, if we should take logarithms of both sides of it, we would have equation (5). Let us see whence each term of equation (5) would come, by the process of taking logarithms: 2 log x from x 2 , 3 log y from 11, the difference_2 log x  3 log y from the fraction x2 /1I; log C from C, si{x from esin z (the logarithm being to the base e), the sum log C + sin x from the product z. We can there= fore remove the 10 arithms from an equation b t' ro rly chan in coefficients to exponents, erences to fractions, sums to pro~u~s, and ter e of 10 aritluiis to ex onentials. Takiug antilogarithms, equation (5) becomes
ce:
x2 3 = Cesi n z • y
(6)
In applying this process to an equation having one member zero, it must be remembered that antilog 0 = 1. Thus, if equation (5) is written in the form . 2 log x  3 log y  sin x  log C = 0,
Article 11
31
we have, on taking antilogarithms, x2esin z
C1/
= 1,
which is the same as equation (6). EXAMPLE
3. Solve 2y cos y dy
= y sin y dx + sin y dy.
Transposing the last term to the left side and dividing by y sin y, \
~h~
( 2 cot Y 
~) dy =
dx.
Integration yields 2 log sin y  log y
=x
+ log C.
Then, taking antilogarithms, we obtain the general solution in the form sin2 y = Oyez.
11. Particular solutions and particular values. It happens frequently that we are more interested in a particular solution of a differential equation, which satisfies a prescribed condition, than in the general solution; or our chief interest may be in finding a particular value of one of the variables corresponding to an assigned value of the other. 1. Suppose that we wish to find the solution of the differential equation dx =1+2x (1) dt ' EXAMPLE
which satisfies the condition x = ~ when t = 0, and that we also wish to find the value of x when t = 1. We may proceed as follows: Separating the variables, we have
dx 1 + 2x = dt.
(2)
, Integration then gives ~ log (1
+ 2x) = t + C.
(3)
32
Chapter 3
The condition, x = ~, t = 0, determines C: ~ log 2
Equation (3) then becomes 1
log
2
1
1
= C.
+2 2x =
+ 2x = e2 t
X 
t,
2e2 t, 1:. 2'

(4)
the required particular solution. We then find the value of x corresponding to t = 1: 2 X]t=l = e  ~ = 6.89. Instead of finding the general solution (3) and determining the value of the arbitrary constant, we could have integrated equation (2) between limits. Integrating between the known pair of values ( ~, 0) and the general pair (z, t), we have
1
dx
:1:
=
~1+2x
[
dt 0'
~ log (1 + 2x)]~ = t]~, 1 2
log
1
X 
+2 2x = t,
e2 t

1 2'
X]'=l = 6.89. If, however, we are interested only in the value of x when t = 1, it is not necessary to find the particular solution (4). We can integrate equation (2) between the limits (~, 0) and (z, 1), thus:
1
:1:
dx
I»
+ 2x  0 1 1 + 2x ~ log 2 =
~ 1
x
= e2 
~
'
1,
= 6.89.
33
Article 11
2. Suppose that the derivative dx/dt is proportional to x, that x = 5 when t = 0, and that x = 10 when t = 5. What is the value of x when t = 12? The differential equation is EXAMPLE
I
dx dt
= kx
'
where k is an unknown constant of proportionality which, together with the constant of integration, makes two constants to be determined from the two given conditions. We shall solve the problem by two methods. In the first method, we determine both k and the constant of integration; in the second method we neither determine the value of k nor use a constant of integration. First method. Separating the variables, we have dx
 = kdt. x
Integrating, log x = kt
(5)
+ C.
(6)
The condition, x = 5, t = 0, gives log 5 = C, so that (6) becomes x
log 5 = kt.
(7)
The condition, x = 10, t = 5, now determines k = becomes
t
log 2, and (7)
or Then
X]'=12
= 5.2
2 4 .
= 26.4.
Second method. Integrating equation (5) between (5,0) and (10, 5), then between (5, 0) and (x, 12), we have
r: r:
=1c
5
x
15
dt,
0
 = k 112dt.
5
x
0
34
Chapter 3
Dividing the second result by the first, we find x
log 5" log 2
x
12
= 5 = 2.4,
= 5.22 •4 = 26.4. PROBLEMS
1. Solve the following differential equations and prove that the solutions are correct. (a) dy = 3y. dx x
(b) 2 sin x cos y dy (c)
VI 
= cos x sin y dx.
y2 dx + VI

x 2 dy = O.
(Algebraic solution.)
(d) dy = eZ+u, dx (e) y(x 2 dy  y2 dx) = x 2 dYe
2. Find the particular solutions of the following differential equations, satisfying the prescribed conditions. (a) dy dx
+ x = ~. y
(y = 0, x = 2.)
(b) a dx = b
+ ex.
(c) xC08Ydx
+ x'sinydy 
dt
(d) dx = k(I8  9x dt
(dx = dt
+x
2
).
1, t 0.) =
Express x in terms of t.
a·sinydy. (x (x = 0, t = 0.)
=~, y = ~) (x = 2,
t
= 10.)
3. If ~ dy + 1 = 3y2 and y = 1 when x = 2, find the positive value of y ydx . corresponding to x = i. 4. Given dx = k(6  X)2, with x = 0 when t dt t = 10. Find the value of x when t = 15.
= 0 and
x = 2 when
35
Article 12 du
6. If dt
+ ku = 0, U = Uo when t = 0, U =
uo/2 when t
= 16, U = nUo
when t = 2; find the value of n. 6. Find a curve through the origin whose slope is 'given by (a) dy t:b;
= _ y2 + 2y + 4 .
(b) dy = _ t:b;
+ 2x + 4 ' y + 2.
x2
x+2
Why are the answers to (a) and (b) the same? 7. Find a curve satisfying the following differential equation in polar coordinates and passing through the point (3, 1r/3) :

dp
p d8
=
p2  1 p2
+ 1 tanS.
8. Solve dy
x  xy2
=~
8y
t:b;
+ 2x2y
9. Solve
eZ dy dx
10. Solve
(1  xy) (dx
= 1
tan x
+ tan! x •
+ dy) + (x r: y) (dx 
dy) = O.
11. Solve
'12. Solve 2 cosh
Z:
= 1 + y2,
and show that the particular solution which satisfies the condition (x . y = 0) is x y = tanh· 2
= 0,
12. Dynamics. We consider now some physical problems which require the solution of separable differential equations. Newton's, second law of motion states that the rate of change of momentum of a particle is proportional to the force acting on it and is in the same direction as the force. Thus, if
36
Chapter 3
a particle of constant mass m moves with varying velocity v under the action of a force F, we have d dv F = k dt (mv) = km dt '
(1)
where dv I dt is the acceleration of the particle and k is a constant of proportionality depending on the system of units employed. When cgs (centimetergramsecond) units are used, so that mass is measured in grams and acceleration in centimeters per second per second, we may take F in dynes, so that k = 1, or we may take F in grams, so that k = Ilg, where g is the gravitational constant; approximately equal to 980.5 em/sec". When fps (footpoundsecond) units are used, so that mass is measured in pounds and acceleration in feet per second per second, F may be measured in poundals, making k = 1, or in pounds, making k = Ilg, where now g = 32.17 It/sec", approximately. If we consider a moving mass as concentrated at its center of gravity, we may set up the differential equation of its motion by equating the expression for force, given in (1), to the resultant force in the line of motion. I
1. A sled weighing 100 lb is being pushed in a straight line against the wind by a force of 10 lb. Suppose that friction is negligible but that there is an air resistance to motion whose magnitude in pounds is equal to twice the velocity of the sled in feet per second. If the sled starts from rest, find the velocity and the distance traveled at the end of 1 sec. Let x (ft), positive in the direction of motion, be the displacement from the initial position (x = 0), and let v (It/sec) be the velocity, at time t (sec). We equate the expression for force, given by (1), to the resultant force, which is positive and equal in magnitude to 10  2v, thus forming the differential equation of the motion, EXAMPLE
\
\
100 dv = 10 _ 2v g dt '
(2)
Article 12
37
where g = 32.17 ft/sec 2 • Separating the variables, we get dv g 5  v = 50 dt.
(3)
In order to obtain v in terms of t, we may integrate (3), adding a constant of integration, and then determine this constant from the fact that when t = 0 we must have v = O. Or we may use definite integrals, thus avoiding a constant of integration, and integrate from the known pair of values, v = 0, t = 0, to the general pair of values v, t. Using the latter method, we have
u:
f
dV =o 5 v 50
dt
0
'
g ~t log (5  v)~ 1Io =  50  t 0' 5log 5
f)
gt
=  50'
5  v _ gt/50 5  e ,
v = 5(1  eg t / 50 ) ft/sec. (4) The relation (4) gives the velocity at any time t; for t = 1 sec this becomes V]t=l = 5(1  e32.17/50) = 5(1  eo· 6434)
= 5(1  0.5255) = 2.37 ft/sec. We may notice from (4) that v does not increase indefinitely with t, but approaches a limiting velocity of 5 ft/sec. This agrees with the value of v found by making the acceleration dv/ dt equal to ~ in equation (2). In order to obtain x in terms of t, we replace v by dx/dt in equation (4) and integrate between (0, 0) and (z, t), since x = 0 when t = o. We thus obtain [ x
dx =
5[ 0 the exponential in (19) is greater than 1; also for Xo > 0, Vo  T2Xo > Vo  TIXO. Therefore if Vo > T2Xo the inequality (19) is satisfied and the particle remains to the right of x = 0; for Vo > 0 the particle starts at x = Xo, moves toward. the right, then reverses its direction of motion, and approaches x = 0; for T2XO < Vo < 0 the particle starts toward the left and approaches x = o. If Vo < T2XO the inequality (19) is reversed for sufficiently large t; the particle starts toward the left at x = xo, passes through x = 0, reverses its direction of motion, and approaches x = o.
Article 30
127
4. A particle starts from rest at time t = 0 (sec) with a displacement x = 5 (ft) to the right of the origin, and moves along the xaxis according to the law EXAMPLE
d 2x dt 2
dx
+ dt + 1.25x =
O.
Find (a) the time required for the damping factor to decrease 50 per cent, (b) the percentage decrease in the damping factor after one period, (c) the location of the particle after one period. The differential equation is given in standard form [equation (13)], with a = ~, b2 = 1.25, a < b. (a) The d.f, is eat = et / 2 • Its original value, for t = 0, is 1; it reduces to ~ when
e t / 2


!.
2'
t
  = log 2, 2
t
= 2 log 2 = 1.39 sec.
(b) The period is 21r/Vb 2

a2
= 21r sec. At this time the d.f.
has the value er = 0.0432. Hence the d.f. has decreased 95.7 per cent. (c) Only in part (c) do we need to use the initial conditions and solve the differential equation
+ D + 1.25)x = O. The roots of the quadratic D 2 + D + 1.25 = 0 are (D 2
(20)  ~ ± i; hence
the general solution of (20) is
x = et / 2 (C1 sin t + C2 cos t).
(21)
Differentiation of (21) gives
I
=
et / 2
[ (C1 
~2) cos t (~1 + C2 )
sin
t].
(22)
Substituting the initial conditions x = 5, v = 0, t = 0 in (21) and (22), we obtain
128
Chapter 4
Hence the equation of motion is
x = et /2 (g sin t + 5 cos t). Then
x] t=2r = e
11"
(5)
= 0.216 ft,
and the particle is located 0.216 ft to the right of the origin. EXAMPLE
5. A particle moves along the zaxis in accordance with
the law
d2x dt2
dx
+ 10 dt + 16x = O.
From a point 1 ft to the right of the origin the particle at time t = 0 (sec) is projected toward the left at the rate of 9 ft/sec. Find (a) the time when the particle passes through the origin; (b) the numerically greatest negative displacement; (c) the maximum (positive) velocity. The differential equation is in standard form (13), with a = 5, b = 4, a > b, Tl =  2, r2 =  8. Since Xo = 1 and Vo =  9, the condition Vo < T2XO is satisfied and we have the type of motion in which the particle passes toward the left through the origin, reverses its direction of motion, and approaches x = o. For x and v [equations (16) and (17)] we have 2t
+ C2e 8t,
x =
C1e 
v =
2C1e
2t 
8C2e 8t.
Substitution of the initial conditions x = 1, v = 9, t = 0, gives 1 9
= C1 + C2 } = 2C1  8C2
Hence
+ ~e8t,
(23)
2t !.e2 8 e8t 3 3 .
(24)
x =  ~e2t V 
7 C2  6·
(a) When x = 0, equation (23) gives
e6t = 7,
t = ~ log 7 = 0.324 sec.
(b) The numerically greatest negative displacement occurs when v = 0, that is, from equation (24), when e6t = 28, from which e2t =
Article 30 28~, e8t
129
= 28%.
Substituting these values in equation (23),
we obtain x = ~( 28
+ 7)28% =
3.5(28)% = 0.0412 ft.
(e) When the velocity is a maximum,
dv
2 3
 =   e2t dt
+ 224 e8t = 3
0
.
Then e6t = 112, from which e2t = 112~\ e8t = 112%. stituting these values in equation (24), we find Vmax
= ~ (112  28)112%
= 28(112)%
Sub
= 0.0519 It/see.
EXAMPLE 6. A spring, fixed at its upper end, supports a weight of 10 lb at its lower end, which stretches the spring 6 in. If the weight is drawn down 3 in. below its equilibrium position and released, find the period of vibration and the equation of motion of the weight, assuming a resistance in pounds numerically equal to ~ the speed in feet per second. (Problem 8, Art. 30(b), where the resistance is negligible, is a special case of this problem.) We first determine the spring constant. Since a force of 10 lb stretches the spring 6 in. or ~ ft, we have 10 = e( ~), and the spring constant is e = 20 lb/ft. In solving a problem of this type, it is important to specify and keep clearly in mind the direction chosen as positive, and the origin from which the displacement is measured. We shall take the downward direction as positive and the origin 0 at the equilibrium position of the weight, that is, the position of the weight when it is hanging at rest. Let x (ft) be the displacement of the weight from its equilibrium position at any time t (sec); then the elastic force tending to restore the weight to its equilibrium position is  20x lb, the minus sign denoting that this force and the displacement are opposite in sign. We also have a resisting force whose sign is opposite that of the velocity dx/dt, namely,  ~ dx/dt lb, so that the resultant force tending to restore the weight to its equilibrium position is  20x ~ dx/dt lb. The force acting on the moving 1QIb weight is also (10/g)(d2x/dt2 ) lb. Equating these two expressions for pounds force, we obtain the differential equation 10 d2x
 2=
g dt
1 dx 20x
2 dt '
130
Chapter 4
or, in the standard form of equation (13), (D 2
+ 2aD + b2)x = 0,
where a (j
g
= 40 = 0.804,
= Vb 2

a2 = V64.34  .65 = 7.98.
The period is therefore 211"/(3 = 0.787 sec. The general solution is x
=
«r«, sin (3t + C2 cos (jt).
(25)
Differentiating (25), we get
v = eat[(pC 1

aC2 ) cos (3t  (aC.
+ (3C2 ) sin (3t].
(26)
Substitution of the initial conditions x = ~, v = 0 t = 0, in (25) and (26) gives
C2
1
a C1 = 4(3
=
4'
=
0.0252.
Hence the equation of motion is x = eO.804t(0.0252 sin 7.98t
+ 0.250 cos 7.98t).
PROBLEMS 1. For the motion represented by equation (13), Art. 30(c) , where a < b, show that if v is the velocity at any time, the velocity one period later is eaTv, where T is the period. 2. Discuss the motion of the particle in Art. 30(c) for the case where a = b, with initial conditions x = xo > 0, v = Vo, when t = O. 3. For each of the two cases in Art. 30(c) where a > b and the particle reverses its direction of motion, obtain a formula for the time elapsed until the reversal takes place. 4. The motion of a particle is represented by equation (13), Art. 30(c) , with a > b and initial conditions x = Xo > 0, v = 0, when t = 0; at what time will the particle be moving fastest? 5. Discuss the motion of a particle along the xaxis under the action of a repulsive force proportional to displacement and a resistance proportional to velocity, .assuming as intial conditions x = Xo > 0, v = Vo when t = O.
Article 30
131
6. A particle executes damped vibrations of period 1.35 sec. If the damping factor decreases by 25 per cent in 16.3 sec, find the differential equation, with numerical coefficients, representing the motion. 7. A weight hung on a spring and vibrating in air with negligible damping has a period of 1 sec. It is set vibrating with a practically weightless damping vane attached to it, causing a resistance proportional to the speed, and the period is found to be 1.5 sec. Find the damping factor and write down the differential equation, with numerical coefficients, corresponding to the damped vibrations. 8. A weight suspended from a spring executes damped vibrations of period 2 sec. If the damping factor decreases by 90 per cent in 10 sec, find the acceleration of the weight when it is 3 in. below its equilibrium position and is moving upward with a speed of 2 It/sec. 9. A weight of 4 lb is hung on a spring causing an elongation of 2 in. It is set vibrating and its period is 1r /6 sec. Find the time required for the damping factor to decrease 75 per cent if resistance is proportional to velocity. 10. In Ex. 3, Art. 3O(b), suppose that the water offers a resistance (lb) numerically equal to K times the velocity (ft/sec), thus increasing the period of vibration by 0.04 sec. Find the value of K and the time required for the damping factor to decrease 50 per cent.. 11. A particle moves along the zaxis in accordance with the law d 2x
dt2
dx
+ 10 dt + 9x =
O.
From a point 2 ft to the right of the origin the particle at time t = 0 (sec) is projected toward the left at the rate of 20 ft/sec. Find (a) the time when the particle reaches its leftmost position; (b) the distance traveled and the velocity at the end of 1 sec. 12. If the particle of Prob. 11 is projected toward the right at the rate of 20 ft/sec, find (a) the time when it reaches its rightmost position; (b) the distance traveled and the velocity at the end of 1 sec. 13. If the particle of Prob. 11 is projected toward the left at the rate of 10 ft/sec, find the distance traveled and the velocity at the end of 1 sec. 14. A particle moves along the zaxis in accordance with the law d2x
dx + 6dt2 dt 16x =
0
.
From a point 2 ft to the right of the origin the particle at time t = 0 (sec) is projected toward the left at the rate of 10 ft/sec. Find (a) the time when the particle reaches its leftmost position; (b) the distance traveled and the velocity at the end of 1 sec.
132
Chapter 4
16. A lolb weight is suspended by a spring which is stretched 2 in. by the weight. Assume a resistance whose magnitude (lb) is 40/v'Y times the speed (ft/sec) at any instant. If the weight is drawn down 3 in. below its equilibrium position and released, find (a) the displacement of the weight from its equilibrium position after 1/(2v'Y)sec; (b) the time required to reach the equilibrium position. 16. A lolb weight hanging on a spring stretches it 2 ft. If the weight is drawn down 6 in. below its equilibrium position and released, find (a) the highest point reached by the weight; (b) the location of the weight after 1/v'Y sec. Assume a resistance whose magnitude (lb) is 10/0 times the speed (ft/sec). 1'1. A 10lb weight stretches a certain spring 1 ft. The weight is drawn down 6 in. below its equilibrium position and then released. Assuming a resistance whose magnitude (lb) is equal to the speed (ft/sec) at any point, find (a) the period; (b) the percentage decrease in the damping factor when the weight is at its highest point; (c) the distance (in.) of the weight from its starting position when t = 0.6 sec. 18. A weight of 16lb hanging on a spring, which it stretches 6 in., is given a downward velocity of 10 ft/sec. If the resistance (lb) of the medium is numerically equal to the velocity (ft/sec), is the weight above or below the starting point at the end of 0.5 sec, and how much? 19. A simple pendulum makes small oscillations with a period of 2 sec in a resisting medium. Assume an angular retardation whose magnitude (rad/ sec2) is 1/25 the magnitude of the angular velocity (rad/seo). If the pendulum is released from rest at an angular displacement of 10 , find the displacement at the end of 10 periods. 20. A 6lb weight of specific gravity 3 stretches a spring 4 in. when immersed in water. If the period of vibration in water is g sec, by what percentage will the damping factor decrease in 1 sec? Assume resistance proportional to velocity. 21. A weight having specific gravity p is immersed in water and supported by a spring which it stretches a in. It is set vibrating and the resistance of the water to the motion of the weight is proportional to the velocity.' If at the end of r periods the damping factor is l/n its initial value, obtain a formula for the period T in terms of p, a, n, r, and g. Use the formula to compute the value of T if p = 3, a = 8 in., n = 6, r = 1, 9 = 32.17 ft/sec 2•
31. The linear equation with R ~ 0; simplest case. We consider now the simplest case of the differential equation f(D)y = (ao D n + a1Dnl +...+ an1D + an)y = R, with R ~ 0, namely, the case in which an and all the other a's
133
Article 32
except one, sayan _ r , are equal to zero; the differential equation then has the form dry an  r dxr
= R,
where R is a function of z, or dry dxr
= F(x).
(1)
The solution of the differential equation is effected merely by integrating r times, adding each time a constant of integration. EXAMPLE
1. Solve
d3 y
dx 3 = 6x
+ 2 cos 2x.
Integrating three times, adding each time a constant of integration, we have
tPy dx 2 =
3
X
2
+ SIn• 2x + Ch
dy 3 cos 2x dx = x 2 x4
y="4
sin 2x 4
+ CIX + C2,
2 CIX +2+C2X+C3.
General solution: 4y
=
x4

sin 2x
+ C1X 2 + C2x + C3 •
32. Deflection of beams. As an application of equation (1), Art. 31, we consider some problems on the deflection of beams. Fig. 9 shows a piece of a bent beam. I t can be regarded as made up of fibers such as F'F, all originally of length 8 (ft). The surface PQ, containing fibers whose lengths are unaltered when the beam is bent, is called the neutral surface. The curve of one of these fibers B'B is called the elastic curve of the beam. Fibers below the neutral surface are stretched and those above are compressed when the beam is bent. Suppose that the
Chapter 4
134
fiber F'F at a distance z below the neutral surface has been stretched an amount e by the force S dA, where S (lb/It") is the stress per unit crosssection area and dA (ft 2 ) is the crosssection area of the fiber. Also let R (ft) be the length of BC, the
SdA
FIG.
9
radius of curvature at B of the fiber B'B, and let QQ' be perpendicular to B'B and BC. Now, by Hooke's law, the stress S per unit area is proportional to the stretch e/8 per unit length in the fiber F'F; that is e S = E
(1)
8 '
where the constant of proportionality E (lb/ft 2 ) is Young's modulus, or the" modulus of elasticity. Furthermore, from the figure we have the proportion (2)
Then, from (1) and (2),
Ez
S =. R The moment of the force S dA about the axis QQ' is
E zS dA =  Z2 dA. R
Article 32
135
Integrating this over the cross section of the beam we get the bending moment M:
But
jz2dA = I, where I (ft4) is the moment of inertia of the crosssection area of the beam with respect to the axis QQ'; hence
El M=·
R
(3)
Now take the zaxis horizontal through some point of the fiber B'B chosen as the origin and the yaxis positive upward. The formula for radius of curvature is R = (1 + y'2)3/2/y", but for small bending y' is small and v" may be neglected in comparison with unity, so that a close approximation to the radius of curvature is R = 1/y". By use of this approximation, (3) becomes (4) M = Ely", the expression for the bending moment at any section of the beam. Equating this expression to the algebraic sum of the moments, with respect to the axis QQ' of the section, of all forces on one side of the section which tend to bend the beam about this axis, we obtain the differential equation of the elastic curve of the beam. When the moments of all these forces are expressible in terms of x the differential equation is of the form (1), Art. 31, with r = 2. 1. Find the elastic curve and the maximum deflection of a beam L ft long resting on supports at the ends and slightly bent under a uniform load of q lb/ft, Take the origin 0 at the left support and the yaxis positive upward (Fig. 10). Then take a section of the beam through a point P at a distance x from 0 and equate the expression for the bending moment at P, Ely" [equation (4)], to the sum of the moments of all the forces EXAMPLE
136
Chapter 4
on one side of P, say to the left, which tend to bend the beam at P. Since y" is positive on a curve when the curve is concave toward the positive direction of the yaxis, we write a moment positive when it tends to produce concavity in the direction of the positive yaxis. Since v" is negative on a curve when the curve is concave toward the negative yaxis, we write a moment negative when it tends to produce concavity in the direction of the negative yaxis. The load qL is equally distributed on the supports, causing an upward force of qL/2 at the left support and a moment qLx/2 with respect to P. The load y
qL
qL
T
2'
k+:::JIIr,% p
FIG. 10
qx on OP acting downward at the middle point of OP produces a moment  qx2/2 with respect to P. The differential equation of the elastic curve of the beam is therefore
Ely" = qLx _ qx2. 2 2
(5)
In this equation the units of the various quantities involved are as follows: E (lbjft2 ) , I (ft 4) , y" (ft l ) , q (lb/ft), L (ft), x (ft). The differential equation (5) is solved merely by integrating twice and determining the two constants of integration from two conditions. Integrating once, we get
Ely' = The condition y'
q~x2  q~ + (JI'
(6)
= 0, x = L /2, gives
qL3 qL3 o = 16  48
+ 01 ;
01
qL3
= . 24
Article 32
137
Inserting the value of C1 in (6) and integrating again, we find Ely =
qLx3
12 
qx4
qL3x
24 
24 + C2 •
The condition y = 0, x = 0, gives C2 = of the beam is
o.
(7)
Hence the elastic curve (8)
There were just enough independent conditions to determine the two arbitrary constants. It may be noticed that a third known condition y = 0, x = L, is not independent; it satisfies (8) automatically, or, if substituted in (7), yields again C2 = 0. If we had not seen at the beginning that the reaction at each support is qL/2, we could have called one reaction R and the other qL  R; then all three conditions would have been needed to determine R, Cit and C2 • It may be noticed further that if we had taken moments to the right of P, instead of to the left, it would have been more complicated but the same differential equation would have been obtained; for in that case
EI "
= qL(L  x) _ qeL  x )2
Y
2
2'
which reduces to equation (5). The maximum deflection of the beam is the value of y when x = L/2, that is, from equation (8), ylz:L/2
q
(L
4
L
4
L
4
= 24EI 16  4" + 2
)
5qL
4
= 384EI ft.
EXAMPLE 2. Solve Example 1 if the beam, instead of being simply supported at its ends, is embedded horizontally in masonry at its ends. The differential equation is the same as in Example 1, except that there is an additional unknown moment at each end of the beam, exerted by the masonry, which keeps the beam horizontal at the ends (Fig. 11). Since this moment tends to produce concavity downward it is negative; we denote it by M. The differential equation is
Ely" =
q~
q:; 
M.
138
Chapter 4
Integrating once,
Ely' =
qLx2
4
qx3
 {\ 
Mx
+0
The condition y' = 0, x = 0, gives 0 1 = 0. y' = 0, x = L/2, gives
1•
Then the condition
qL3 qL3 ' ML qL2 M=· 0= 16  48  2; 12 Integrating again, qLx3 qx4 qL2x2 Ely = 12  24  24 + O2 • y
FIG.
11
The condition y = 0, x = 0, gives O2 = 0. Hence the equation of the elastic curve of the beam is
q
y
= 24EI (x4

2Lx3
+ L 2x2),
and the maximum deflection is 4
qL Y]x=L/2 = 24El
(116  14 + 41)
4
qL = 384El ft.
Two other conditions y = 0, x = Land y' = 0, x = L are not independent of the ones used and are satisfied automatically. 3. In Example 2, if there is an additional concentrated load of F lb at the midpoint of the beam, find the elastic curve of the left half of the beam and the maximum deflection. EXAMPLE
Article 32
139
The effect of the load F at the midpoint is to increase the reaction at each support to (qL F) /2. But there is another important difference between this example and the preceding one. In Example 2 the point P can be taken anywhere along the beam and the moments to the left of P are given by the same expression; the differential equation obtained holds for the whole beam. In this example, however, if P is moved toward the right across the midpoint of the beam, an additional moment to the left of P is introduced, due to the concentrated load, so that the differential equation of the right half of the elastic curve is different from that of the left half; the two halves of the elastic curve have different equations. This may seem strange, since the elastic curve "looks like" a curve with a single equation, but if the student will draw for illustration the curve y =  x3 from 1 to 0 and the curve y = x 3 from 0 to 1, the result will be a curve whose two halves have different equations. Taking P to the left of the midpoint of the beam and taking moments to the left of P, the differential equation representing the left half of the elastic curve is
+
Ely" = (qL ~ F)x _ q;2 _ M. Following exactly the same steps and using the same conditions as in Example 2 (the student should do this), the elastic curve of the left half of the beam is found to be y
=
q 4 (x 24EI

2Lx3
F + L 2x2) + 4 (4x3  3Lx2 ) 8EI·
Notice that the two conditions y = 0, x = Land y' = 0, x = L, which were satisfied automatically in Example 2, are not satisfied here; these conditions are true only for the right half of the beam. For the maximum deflection we have 3
y]Z==L/2
(1 3)
4
+
3
qL4 FL qL 2FL = 384EI  48EI 2  4 = 384E' ft. PROBLEMS
1. A beam of length 20 ft is simply supported at the ends and carries a weight of 240 lb at its midpoint. Taking the origin at the left end and neglectingthe weight of the beam, find (a) the equation of the elastic curve of
140
Chapter 4

the left half of the beam; (b) the equation of the elastic curve of the right half of the beam; (c) the maximum deflection. 2. If the beam of Prob. 1 carries in addition a uniform load of 30 lb/ft, find the maximum deflection. 3. A cantilever beam (one end free and the other fixed horizontal) of length L ft weighs q lb/ft and carries a load of W lb at its free end. (a) Taking the origin at the free end, find the equation of the elastic curve of the beam. (b) What load, distributed uniformly along the beam, would produce the same maximum deflection as the load W at the free end? 4. A cantilever beam of length L ft weighs q lb/ft and carries a load of P lb at its midpoint. Find the maximum deflection of the beam. 6. A beam L ft long, carrying a uniform load of w lbjft, is fixed horizontally at one end and is simply supported at the other. Find (a) the deflection of the midpoint of the beam; (b) the maximum deflection of the beam. 6. A beam 10 ft long, fixed at one end and simply supported at the other, carries a uniform load. Find the distance from the supported end to the point where the maximum deflection occurs. 7. A beam 6 ft long is simply supported at its ends. Two concentrated loads, each equal to P lb, are supported at the points of trisection of the beam. Neglecting the weight of the beam, find the points at which the deflection has half its maximum value. 8. A cantilever beam of length L ft, vertical dimension h ft, and modulus of elasticity E Ib/ft2 carries a load of material whose density is w Ib/ft3 and whose depth at distance x ft from the free end is kx2 ft. Neglecting the weight of the beam, find its maximum deflection. 9. A beam of length 2L ft carrying a uniform load of q lb/ft is supported at its ends and at its middle point. (a) Taking the origin at the middle point, find the elastic curve of the right half of the beam. (b) Show that the maximum deflection is qL4 (39 + 55y'33) 216EI ' and that this is 42 per cent of the value it would have if the beam were cut in two at its middle point. 10. A simply supported beam L ft long carries a concentrated load of W lb at a distance of eft from the left end. (a) Taking the origin at the left end, find the equation of the elastic curve of the portion of the beam to the left of Wand of the portion of the beam to the right of W. (b) If c < L/2, show that the distance from the left end to the point of maximum deflection is greater than c and less than L/2, and find the maximum deflection.
Article 33
141
33. The linear equation with R ~ o. We now rewrite equations (1) and (2) of Art. 29: f(D)y = R, (1) f(D)y = 0, (2) where f(D) = .aoDn + a1Dnl + ... + aniD + an. Equation (2) has been solved in Art. 29; we now seek the general solution of equation (1), in which R is a function of x. The general solution of equation (2) is a value of y containing n essential arbitrary constants, n being the order of the differential equation; we call this the complementary function (c.f.) of equation (1) and denote it by Yc. Thenf(D) operating on Yc produces 0: (3) f(D)yc = o. Now suppose that in some way we are able to find a particular value of y, free of arbitrary constants, which satisfies equation (1); we call this a particular integral (p.i.) of equation (1) and denote it by yp. Then f(D) operating on YP produces R: f(D)yp = R.
(4)
We have seen in Arts. 2 and 11 that a particular integral of a differential equation can be obtained from the general solution by assigning particular values to the arbitrary constants which enter into the general solution; but here we are going to find a particular integral before the general solution is knownthe finding of a particular integral is now to be a step in the process of finding the general solution. Adding equations (3) and (4), we find
+ yp) = = Yc + YP
f(D)(yc
hence y
R; (5)
satisfies the differential equation (1), and since it contains n essential arbitrary constants it is the required general solution.
142
Chapter 4
Thus the general solution of (1) consists of two parts, the e.f. and a p.i, The operator j(D) , operating on the c.f., produces nothingthe c.f. merely acts as a carrier of the required arbitrary constants. The p.i., however, when operated on by jeD), yields the right member R. EXAMPLE
1. Solve
+ 2D + l)y = 4ex • by solving (D2 + 2D + l)y = 0, or (D 2
The c.f. is found = 0; it is
Yc = (0 1
(D
+ 1)2y
+ 02 x)e x.
A p.i., found by inspection, is YP = e", since (D 2
+ 2D + l)e = e" + 2ez + e = 4ex• Z
X
Therefore the general solution of the given equation is y
=
(0 1
+ 02 x)e + e', X
This example illustrates the plan of solving an equation of form (1). We have but to find the c.f. by the method of Art. 29, then find a p.i., and add them together. However, we cannot rely on inspection to reveal a p.i. except in very simple cases. We shall explain in the next two articles two methods for finding a p.i. The first method, undetermined coejficients, will apply when R is of certain special form; the second, variation oj parameters, is general and can be used theoretically for any form of R, although it is impracticable to carry it through for complicated forms of R. 34. Undetermined coefficients. We shall find it convenient in this article to call the part of a term which is multiplied by a constant coefficient the variable part (v.p.) of the term. Thus the v.p, of 7xe2z is xe2z , the v.p, of 6 cos 2x is cos 2x, and the v.p. of 3x' is x. If a term is a constant, say 5, it can be thought of as 5xo, where 5 is the constant coefficient and Xo or 1 is the v.p. Now a function may be of such kind that successive differentiation of it beyond a certain point ceases to yield terms with new variable parts.
143
Article 34
For example, take the function x 2e3z and its two successive derivatives:
x 2e3z ,
3x 2 e3z
+ 2xe
3z
,
9x2e3z
+ 12xe + 2e3z. 3z
Further differentiation would yield only terms with variable parts, x 2e3z , xe3z , e3z • As another example, take the function 3x2 + sin 2x and its two successive derivatives:
3x2
+ sin 2x,
6x + 2 cos 2x,
6  4 sin 2x.
These and further derivatives contain only terms with v.p.'s x 2 , X, 1, sin 2x, cos 2x. As an example of a function which does not possess this property, take tan z and its successive derivatives: tan z, sec2 x, 2 sec2 x tan x, Further differentiation continues to produce functions with new variable parts. I t can be shown * that functions possessing the above property are expressible as sums of terms of the form Cxpeaz cos ~x or CxPelXZ sin ~x, where p is a positive integrer or zero, a and {3 are any real constants (including zero), and C is any constant. Thus the function x 2e 3z in the first example above is obtained from the first form by putting C = 1, p = 2, a = 3, P = 0; and the function 3x2 + sin 2x in the second example is obtained by summing the first form with C = 3, p = 2, a = 0, {3 = 0, and the second form with C = 1, p = 0, a = 0, ~ = 2. Suppose now that R is a function possessing the above property; that is, R is a function such that successive differentiation of it beyond a certain point ceases to yield terms with new variable parts. Otherwise stated, R is a function expressible as sums of terms of the form CxPe lXZ cos ~x or CxPe lXZ sin Bx, where p is a positive integer or zero, a and {3 are any real eon* See Reddick and Miller, Advanced Mathematics for Engineers, Art. 11, for proof of this and the rule following.
Chapter 4
144
stants (including zero), and C is any constant; then it is shown (loc. cit.) that the following rule will yield a particular integral of equation (1), Art. 33. Write the variable parts of the terms in the right member Rand the variable parts of any other terms obtainable by succes8ively differentiating R. Arrange the v.p.'s so found in groups such that all v.p.'s obtainable from a single term of R appear in only one group. Any group consisting of v.p.'s none of which is a v.p, of a term of the complementary function Ye is left intact, but if any member of a group is a v.p. of a term of Yo all the members of this group are to be multiplied by the lowest positive integral power of x that will make them all different from the o.p, of any term in Ye. Now multiply each member of all the groups by a general constant (undetermined coefficient) and take the sum of the expressions 80 obtained as Yp. Finally, operate on YP with feD) and equate the complete coejJicients of the various v.p.'s to the coefficients in the corresponding terms of R in order to evaluate the undetermined coefficients in Yp. 1. Solve
EXAMPLE
(D2  D)y = 3x2

4x
+ 5 + 2~ + sin z.
(1)
We give all the work necessary, followed by an explanation of the steps taken in finding yp. Roots of f(D) = (D 2 .
yp=Ax3+ Bx2+ 1)
D2 yp=
3A=3
A=l
Ye = C1 + C2,ez • Cx +Eze"
0, 1.
+Fsin x+G cos x (2)
6Ax+2B+E~e:i:+2Ee:i:FsinxGcosx (4)
6A2B=412Ba=5 B=l 0=7
YP
= 0:'
D) = D(D  1)
3Ax2+2Bx+ C+Exe:i:+ Ee:i:;G sin x+F cos x (3)
Dyp= 1)

=  x3

,;2 
7x
2EE=2 E=2
+ 2x~ 
F+G=l, FG=O F=!, G=!
~ sin x
+ YP' is x 3 + (C2 + 2x)e
+ ~ cos x,
The general solution, y = Ye
y = C1

7x  x 2

Z

~ sin x
+ ~ cos z.
Article 34
145
The v.p.'s of 3x 2 and its two successive derivatives are x 2 , a, 1. Corresponding to the term  4x we would get x, 1, and corresponding to the term 5 merely 1, but these are already included in the group r, x, 1, so that this group of v.p.'s corresponds to the terms 3x2 4x + 5 of R. Now one of these v.p.'s, namely 1, is the v.p. of 0 1, a term of the c.f.; hence we multiply each member of the group x 2, x, 1, by x, obtaining x 3 , x 2 , x, none of which is a v.p. of a term of Ye. By multiplying by A, B, 0 respectively we get the first three terms of YP' namely, Ax3 + Bx2 + Ox. The only v.p. obtainable from 2eZ is ~, which itself forms the secondgroup. Since f? is the v.p, of the term 02ez in Ye, we multiply it by x and prefix a general constant E to obtain the fourth term, Exez , of Yp. From sin x the only new v.p. obtainable by successive differentiation is cos z, so that the third group consists of sin z, cos z. Since neither of these is a v.p. of a term of Ye, we multiply by F and G respectively, obtaining, for the last two terms of YP' F sin x G cos a; Now write down YP' and under it write Dyp and D2yp, arranging terms with the same v.p.'s in the same column. Since YP satisfies equation (1), the combination D 2yp  Dyp must be identical with the right member of (1). To produce this combination we multiply equation (4) by 1, equation (3) by 1, and add, placing the multipliers 1 and 1 at the left of the respective equations. We do not use a; multiplier (other than 0) for equation (2) since y, unaffected by a D, does not appear in the left member of (1). By equating the resulting coefficients of the various v.p.'s column by column to the coefficients of the corresponding terms in the right member of (1), we obtain just enough equations to solve for the values of the undetermined coefficients; any additional equations, although superfluous, must not be contradictory. In this example there are six equations, obtained from the 2nd, 3rd, 4th, 6th, 7th, and 8th columns, for determining the six undetermined coefficients, A, B, 0, E, F, G. Note that the first column yields nothing, and the fifth gives E  E = 0, or merely 0 = 0; if the fifth column had yielded a contradictory re ' sult, a mistake would have been indicated.
+
I
A short cut. In the case where R is the product of an exponential function e" and some other function of x, it is usually shorter to multiply the equation through by eaz and then
146
Chapter 4
make use of the reverse exponential shift [equation (6), Art. 28] in finding the p.i. As an illustration we shall work the following example first by the regular method just explained, then by use of the short cut. : :, ~__ EXAMPLE
1 (D l'if))
2. Solve
First solution. The roots of the quadratic D 2 (D  1)2 = 0 are 1, 1; hence Yc
=
(C1

2D +/1 =
+ C2x)e
te •
Writing the v.p.'s of 2xete and of the terms obtainable by differentiation, we have the group ze", e". Multiply the members of this group by x 2 in order to obtain the group x 3ete, x 2eX , both of whose members are different from the v.p, of any term in Yc. Now multiply these members by A, B, respectively, to obtain yp. 1)
2)
Dyp = 1)
Ax3e~
D2yp = Ax3e x
+ (3A + B)x2e~ + (6A + B)x2e~ + (6A + 4B)xe~ + 2Be~ 6A+4B4B=2 A  .!3
2B = 0 B=O
(Note that the first two columns give 0 = 0.) General. solution: Second solution. The c.f., as before, is
Multiplying the differential equation through byeX, we have eX(D  1)21/ = 2x.
Article 34
147
Applying the reverse exponential shift, D 2e Z y = 2x.
A p.i. of this equation is obtained by integrating twice, omitting the constants of integration:
General solution: y
= (C1 + C2x
+ ~x3)ez.
If R contains in addition other terms not of the above type, it is often convenient to divide R into parts, then find a p.i, corresponding to a part of the above type by means of the short cut, using the regular method to obtain a p.i. corresponding to the other parts. The sum of the separate p.i.'s will be the complete Yp. For, suppose that the differential equation is
and we find a p.i. Ypl corresponding to R 1 and a p.i. spending to R 2 , that is f(D)YPl
Yp2
corre
= Rl,
It follows that so that is a p.i. of the original differential equation. As an illustration, take Example 2 with the terms 3x 2 added to the right member: EXAMPLE
3. Solve
(D2
As before,

2D
+ l)y =
2xeZ
+ 3x 2 + 4.
+4
148
Chapter 4
Using the short cut to obtain the p.i. Ypl corresponding to 2xez , we have (second solution of Example 2) _
1 3
Z
Ypl  aX e .
To obtain Yp2, that is, a p.i. of the equation (D 2

2D
+ I)y =
3x2
+ 4,
we have I) yp2
2)
+
= Ax2
Bx
+C
2Ax+B
DYp2 =
I) D2Y p2 =
2A
A=3 Yp2
2A2B+C=4 6 24+C=4
4A +B=O B = 12
= 3x2 + 12x + 22
C == 22
General solution: Y = (C1 EXAMPLE
+ C2x + ~x3)ez + 3x2 + 12x + 22.
4. Solve (D 2
The roots of D 2
2D

Yc
2D

+ 2)y = e
+2 =
Z
cos z,
0 are 1 ± i; hence the c.f. is
= eZ(Cl sin X
+ C2 cos z).
.
Multiplying the differential equation by eZ gives
eZ[(D  1)2 + I]y == cos z, or, applying the reverse exponential shift, (D 2
+ I)eZy =
cos z.
We now find a p.i. of this equation, that is, a particular value of eZy, by the method of undetermined coefficients. The group of v.p.'s obtainable from cos x is cos x, sin z. But, if we were solving this equation for e:l:y, the c.f. would be C1 sin x + C2 cos z. Hence
Article 34
149
the members of the above group must be multiplied by x in order to make them different from the v.p.'s of terms in this c.f. We have 1) eZyp
=
+ Bz sin x
Ax cos x
D2e  zyp = Ax cos x e ZYp YP
+ A cos x + B sin x Bx sin x + 2B cos x  2A sin x
Bx cos x  Ax sin x
De zYP = 1)
= Ix sin x = Ixez sin x
2B
=1
B=I
1
2A. = 0
A=O
General solution: y
= (01 sin x
+ O2 COS X + ~x sin x)f?
If this example had been done by solving the original equation directly by the method of undetermined coefficients, the work would have been considerably longer, since a YP of the form, Axe~ cos x Bae" sin x, would have had to be differentiated twice.
+
PROBLEMS
Solvethe following differential equations. 1. (D2 + 2D + l)y = e2z • 2. (D2  l)y = x2 2x. 3. (D2  5D  6)y = 2e3z  ae2:1:. 4. (D2  5D 6)y = 2e3z  3e2z • 6. (D3 + D2)y = x2  X + 2. 6. (2 + D + D2  4D3)y = 1 + 2x2• 7. (D2  3D 2)y = x 2  2ez • 8. (D2  D  12)y = e9z  2ez + 1. 9. (D3  3D  2)y = 3eZ eZ  4. 10. (D2  2D  l)y = 3 sin x. 11. (D2 + 4)y = 2 sin x + cos 2x. 12. (D3  D)y = 2 sin x 4 cos x 6x.
+
+
+
+
+
+
13. (2D2+5D3)y=W+l+V;=; 14. 2 illy
dx2
+ dy + 8x 
16. d2x2 + x dt
dx =
4t sin t.
4 = e:e/ 2 + 17 cos 2x.
150
16.
Chapter 4 d2
dx~
= 4y
+ 12x2e 2z •
1'1. (DS + D2  D  l)y = xeZ
+ e
sin x.
Z
18. d4y _ d8y = 120(x3 + eSZ). dx4 dx 3
+ +
19. d2y2  2 dy 2y = 10 cos x cosh x. dx dx 20. (3D4 8D3 6D2)y = (XS  6x 2
+
+ 12x 
24)e~c.
d2y dy . 21.  2  2   8y = 5e t SIn 2t. dt dt 22. (DS D 2)y = eZ cos x. 23. (DS + 3D2  4)y = 9xe2z + 4x.
+ + (D3  D2 + 4D 
(V + +
24. 4)y = 4 sin2 3x/ 2). 26. (D2 4D 404)y = eZ(sin 20x 40 cos 2Ox). z 26. (DS  3D 2)y = (1& 6)e ge 2z • 2'1. (4D2  l)y = ez/ 2[x  cos (x/2)]
+
+ +
+
28. Determine x so that it will vanish when t = 0 and when t = log 2 and satisfy the differential equation
d2x dt2

dx 3 dt
+ 2x = e3t •
Show that x will vanish for no other value of t. 29. Find a solution of the differential equation
d2y dy 4 dx2 = 4  x dx
o.
if y = 1/4 and dy/dx = 0 when x = 30. Find a solution of the differential equation
d8y _ 2 d2y dx
3
dx
2
+ dy =
2(eZ
+ x)
dx
if yand its first two derivatives vanish when x =
o.
31. Solve
Q d22 + 106Q = 100' sin SOOt dt if Q = 0, dQ/dt = 0 when t = 0; find the value of dQ/ dt when t = 1/100 (i.e., when 500t = 5 rad).
Article 35
151
35. Variation of parameters. There was developed in the preceding article the method of undetermined coefficients which enables us to find a p.i. of the differential equationj(D)y = R, where R is of a certain special form. It happens that R is of this special form in most of the linear differential equations which arise in practice, but it is well to have a method for finding a p.i. in case an equation is encountered in which the method of undetermined coefficients does not apply. Such is the method of variation of parameters, due to Lagrange. The method is general and can be used theoretically for any form of R, and even for a linear differential equation with variable coefficients * provided the complementary function is known, although it is impracticable to carry it through except for fairly simple forms of R. The difficulty in carrying _it through increases not only with the complexity of R, but also with the complexity of the complementary function. Even for Problem 22, Art. 34, where the complementary function is rather simple, the method is hardly practicable; in Problem 27 of the same article, however, although R is more complicated, the e.f. is simpler, and, as we shall see in Example 2, this problem can be solved by the method of variation of parameters with little more work than by the method of undetermined coefficients. The chief value of the method from the practical point of view is in solving an equation like that of the following Example 1, where R is a function whose successive derivatives continue to yield terms with different variable parts, so that the method of undetermined coefficients would involve a YP having an infinite number of terms. The method of variation of parameters is powerful for use in theoretical considerations and is justly described by mathematicians as "elegant." We shall explain the theory for the differential equation of second order with constant coefficients, fCD)y = CaoD 2 + aiD + a2)y = R, (1) then indicate how it applies to equations of higher order. * For example, Prob. 12 at the end of this article.
Chapter 4
152
First find the c.f. by the familiar method of Art. 29. We use, for convenience, A and B instead of the arbitrary constants C, and O2 , and write the c.f. in the form
Yc = Au + Bo, where u and v are certain known functions of x. Our problem is now to find a p.i. of equation (1), that is, a function such thatf(D) operating on it produces R. We know that feD) operating on the function Au + Bv produces 0 if A and B are constants; it is reasonable to surmise that if A and B were functions of x (which can be called parameters), it might be possible to determine them (vary the parameters, so to speak) so that j'(D) operating on Au + Bv would produce R instead of o. Accordingly we write y
= Au + Bv
(2)
where now A and B are undetermined functions of z. If we can determine A and B (free of arbitrary constants) so that (2) satisfies (1), the corresponding value of y will be the YP that we are seeking. We need two conditions on A and B in order to determine them. We may choose the first condition in some way that will simplify the problem. But the second condition is forced on us; it is that y = Au + Bv must satisfy equation (1) after the first condition has been imposed. The form in which these two conditions appear is made clear by the following compact arrangement of the problem. Subscripts denote differentiation with respect to z.
= Au + Bv Dy = Au~ + BVt + A1u D2 y = AU2 + BV2 + Atut y
+ B1v = 0 + BtVI = R/ao
Arrange the four terms obtained by differentiating Au + Bv so that the first group of two terms contains A and B, and the
Article 35
153
second group contains Al and B 1 • Now draw a vertical line separating the two groups and set the second group equal to zero. This is the first condition, mentioned above, which we arbitrarily impose on the two functions A and B, namely, that the sum of their derivatives multiplied by U and v respectively shall vanish. Next differentiate the part of Dy which remains on the left of the vertical line, obtaining four more terms for D2 y arranged as before. If now we imagine that the values of y, Dy and D2 y are substituted in (1), the parts to the left of the vertical line yield nothing, since these would be the values of y and its derivatives if A and B were constant. There remains onlyao(A1UI + B1VI) which must equal R if the second condition mentioned above is satisfied. To say that y = Au + Bv must satisfy (1) when A1u + B1v = 0 is the same as to say that A1UI + B1VI = R/ao, which is the second condition that the functions A and B must satisfy. In order to determine A and B, the equations
+ B1v = 0, A1uI + B1VI = R/ao, A1u
must be solved for Al and B 1 ; these expressions are then integrated (omitting constants of integration) to determine A and B. Substitution of these values of A and B in (2) yields a p.i. YP of equation (1). The general solution of (1) is then
y
=
Yc
+ Yp·
If, when Al and B 1 were integrated to obtain A and B, con.. stants of integration had been added and the resulting values of A and B substituted in (2), the general solution of (1) would have been obtained. Since Yc must be found at the beginning, however, it is expedient to use the method of variation of parameters merely to find yp, then add Yc and Yp to obtain the general solution.
154
Chapter 4
EXAMPLE
1. Solve (D 2
The c.f. is Yc = Ct sin x
+ l)y = tan »:
+ C2 cos a:
Asinx+Bcosx
y=
AcosxBsinx+
Dy=
D 2 y ,= AsinxBcosx+ A
B
=
cosx
= sin x
At sin x
At cos x  B1 sin x = tan x At
 log (sec x
+ tan x)
+ B I cos X = 0
= sin x
sin2 x Bt = cos x
= cos x
 sec x
+ tan x) log (sec x + tan x)] cos z
Yp = cos x log (sec x
Y = CI sin x
+ [C2 
In solving the two equations to the right of the vertical line for Al and B, we multiplied the first of these equations by sin x, the second by cos z, then added to obtain At; we then multiplied the first equation by cos x, the second by sin z, and subtracted to obtain B t • The expressions for Al and B I were then integrated (omitting constants of integration) to find A and B. When these values of A and B were substituted in the equation y = A sin x + B cos x, the terms cos x sin x and sin x cos x cancelled and the expression for 'UP was obtained. Finally the general solution was written down by adding 'Ie andyI'.
In studying this method the student may have been curious about the statement that we may choose a condition on A and B that will simplify the problem. In the above example we arbitrarily imposed on A and B the condition that Al sin x + B, cos X = O. Was it necessary to impose this particular condition? Suppose we had written some other constant or a function of x instead of 0, would the result have been the same? The answer is that this particular condition was not necessary; it was convenientit simplified the problem. If the student has sufficient curiosity, he should try working this example by
Article 35
155
imposing the condition A 1 sin x + B 1 cos X = »; the final result will be the same but the work will be longer. We shall now solve Problem 27, Art. 34, by the method of variation of parameters, omitting explanations. EXAMPLE
2. Solve (4D 2
l)y = e.,/2 (x  cos ;)

The c.f. is Yc = C1ez / 2
+ C2e
Z
/
2
•
y= AeZ/ 2+ B ez / 2 Dy=
A 1ez / 2 + B 1e z / 2 =0
~AeZ/2 ~Bez/2+
D2y= lAez/2+ lBeZ / 2+ ~Alez/2 ~Blez/2 = lez/ 2[xcos (x/2)]
z /2
y=C1e
+
x 1 x) e" C   x+2 . +2
('
2
X
4
8
5
SIn
2 5
COS 
Z/2'
2
The method of variation of parameters is applicable theoretically to differential equations of any order, but practically it is seldom used for equations of higher order than the second. The amount of work necessary to carry it through increases rapidly with the order of the equation, owing to the necessity of solving a system of simultaneous equations for At, BIt ... , the number of equations in the system being equal to the order of the differential equation.
156
Chapter 4
In order to illustrate the method for a third order equation, we give the following example, although it could be solved much more readily by the method of undetermined coefficients. Note that here we need three conditions for determining At, BI, CI, so we arbitrarily set the tails of Dy and D2y equal to zero for the first two conditions. ExAMPLE
j(D)
=
3. Solve
(D  l)(D  2)(D  3)
+ Ce3a: Dy = A~ + 2Be2a: + 3Ce3z + D2y = A~ + 4Be2a: + 9Ce3z + D 3y = A~ + 8Be2z + 27Ce3a: + y
= Aea: +
Be2a:
+ B le2z + Cle3a: = 0 AI~ + 2B 1e2z + 3Cle3a: = 0 Al~ + 4B1e2z + 9Cle3z = ez Alez
Solving the system of three simultaneous equations to the right of the vertical line for At, Bt, 0 11 we obtain I.
A 
4: e2z
  e3z B1 
B =
~e3a:
4 a: C1  .le2
C 
le4a: 8
Al
=
~e2a:
4: YP  I.ea:
+ I.e:l: 3
I.ea:  _ 8
Y  C1ea: + C/2 e2z + C'3 e3Z 
~a: :Il4~
Lea: 24 •
PROBLEMS Solve the following differential equations, working 5, 9, and 10 by method of Art. 34 and also by the method of Art. 35. 1. (D2 l)y = esc z. 2. (D2 4)y = cot 2z. 3. (9D2 l)y = sec (zj3). 4:. (D2 l)y = tan2 z. 6. (D2  l)y = sin 2 x.
+ + + +
th~
Article 36 6. 7. 8. 9. 10.
157
(4D2 + 4D + 5)y = 4e z / 2 sec x + 4ez / 2• (D2  l)y = x2e z2 / 2• (D2  2D + 2)y = 3x + eZ tan x. (D2 + l)y = x cos x. (D3  7D  6)y = 26e2z cos x.
11. (a) Solve by the method of variation of parameters:
(D2 + 3D
+ 2)y =
sin ete•
(b) Show that; although successive differentiation of the function sin ~ continues to yield terms with different variable parts, nevertheless the differ
ential equation may be solved without using the method of variation of para
meters. 12. Solve by the method of variation of parameters: (D
+ P)y = Q,
where P and Q are functions of x, obtaining the solution (8), Art. 19.
36. Forced vibrations. In Example 6, Art. 30. we saw that the differential equation 10 d2x
  2= g dt
1 dx
20x2 dt
(1)
represents the vibrations of a 10lb weight, hung on a spring with spring constant 20, in a medium offering resistance numerically equal to half the velocity. If resistance is negligible, the last term is absent and the differential equation 10 d 2x
2 =
g dt
(2)
20x
represents the special case of simple harmonic motion in a nonresisting medium [Problem 8, Art. 30(b)]. These differential equations, of standard form. (D2
+ 2aD + b2)x = 0, (D2 + b2 )x = 0,
(a
< b),
158
Chapter 4
respectively, represent free or natural vibrations, that is, vibrations due to the inherent forces in the system. The system involved in (1) consists of the spring and the weight, together with the resisting medium in which they are immersed; in (2) the system is merely that of the spring and the weight. Suppose now that there is applied to the weight, vibrating in accordance with equation (1), a periodic force A sin wt, external to the system, of amplitude A and period 211"/w. The differential equation of the motion will then be 10 d 2x
g dt2
1 dx
=  20x  2
dt
.
+ A SIn wt,
of standard form (D2
+ 2aD + b
2)x
=
C sin wt,
(a
< b).
(3)
Neglecting the resistance due to the medium, the standard form of the ditierential equation is (D2
+b
2)x
= C sin wt.
(4)
Equations (3) and (4) represent forced vibrations respectively with and without resistance proportional to velocity and due to the medium. The solutions of these equations consist of a complementary function Xc plus a particular integral Xp • The form of the solution of equation (4) will depend on whether w = b or ClJ ¢ b, that is, whether the period 211"/w of the impressed external force is or is not equal to the period 2r/b of the natural vibrations. If w = b, we have the case of resonance, where the vibrations get larger and larger. This condition would cause dangerous stresses in some vibration problems; on the other hand a resonant condition is desirable in certain acoustical and radiocircuit problems. EXAMPLE
1. A 12lb weight hangs at rest on a spring which is
stretched 4 in. by the weight. The upper end of the spring is given the motion y = sin .y2gt, where y (ft) is the displacement, measured positive upward, of the upper end of the spring from its original posi
Article 36
159
tion at time t (sec). Find (a) the equation of motion of the weight, (b) the position of the weight 7r/ v'3Y sec after the motion starts. Since a force of 12 lb stretches the spring ~ ft, 12 = c ~, and the spring constant is c = 36 lb1ft. Take the origin at the equilibrium position of the weight and let x (ft), measured positive upward, represent the displacement of the weight from its equilibrium position at time t (sec). Now when the upper end of the spring is raised a distance y the weight will follow a (smaller) distance z. Thus there is an additional stretch of y  x in the spring, producing an upward force of 36(y'  x) lb. The differential equation is therefore 12 fPx
 2 = 36(y 
z).
9 dt
By substituting y = sin ~t, multiplying by 9112, and using operator notation, the differential equation takes the form (D 2
+ 3g)x = 3g sin v'2Yt.
This equation is of standard form (4) with w ~ b. The eomplementary function is Xc
= C1 sin v'3Yt + C2 cos v'3Yt.
For the particular integral we have 3g) Xp
1)
=
A sin V2!jt
+
B cos V2!jt
D2X p =  2gA sin V2!jt  2gB cos V2!jt x p = 3 sin V2!jt
gA = 3g gB A=3
=0
B=O
Hence the general solution is x = C1 sin
The condition x yields
v'3Yt + C2 cos V3gt + 3 sin ~t.
= 0, t = 0,
gives C2
= 0.
Then differentiation
160
Chapter 4
and the condition dx/dt = 0, t = 0, gives
The equation of motion is therefore x
When t
=  V6 sin ~t + 3 sin v'2Yt.
= 7r/~,
Z]'=r/V3U
we find
= V6 sin 11" + 3 sin ~ = 0 + 3 sin 146° 58' = 3(0.5451) = 1.64.
Hence, when t = 7r/v'3g sec the weight is 1.64 ft above the equilibrium. position.
37. Electric circuits. In Art. 23 electric circuits leading to linear differential equations of first order were discussed. Using the same notation as in that article, the differential equation for a circuit containing resistance, inductance, and capacitance, is * i = Dq.
(1)
Here D = djdt and, as in Art. 23, the capital letters L, R, and C represent constants of the 'circuit whereas the small letters are, in general, variables. 1. An inductance of 1 henry and a capacitance of 106 farad are connected in series with an emf e = 100 sin 500t volts. If the charge and current are both initially zero, find (a) the charge and current at time t sec, (b) the value of the current when t = 0.001 sec, (c) the maximum value of the current. Substituting L = 1, R = 0, C = 106 and e = 100 sin 500t in (1)1 we have the differential equation EXAMPLE
(D2
+ 106 )q =
100 sin 5OOt.
* For references to the derivation of this equation see footnote, Art. 23.
1)
28. Using the formulas of Probe 27, with equal k's, find the ratio of the times required to travel a units of distance starting from rest at x = a, iii the two cases of an attractive and of a repulsive force varying inversely as the square root of the distance from a point O. 29. Subject to an attractive force which varies inversely as the i power of the distance from a point 0, a particle starts from rest 16 ft from 0 with an acceleration numerically equal to 4 ft/sec. Find the time required for the particle to reach O. 30. A flexible chain L ft long is hung over the upper end of a smooth inclined plane which makes an angle 8 with the horizontal. The end of the chain which hangs down is a ft below the end which rests on the plane. Show that the time required for the chain to slide off is
t
=
J L cosh ~ (sec) "Jg(1 + sin 8) a' 1
where g = 32.17 ft/sec 2• Assume a space of at least L ft below the upper end of the plane. 31. Using 8 = 0° and 90° respectively in the formula of Prob. 30, find the time required for a chain 13 ft long, with a = 5 ft, to slide (a) off a smooth , horizontal table; (b) off a smooth peg. 32. If "upper" is replaced by "lower" in Probe 30, obtain a formula for the time required for the chain to slide off the plane. 33. If the lower end of the chain in Probe 32 is initially at the lower end of the plane, show that the time required for the chain to slide off is
t=
"JJg(1  L.SIn 8) cosh"! esc 8j
also show that, as 8 approaches 1r /2, t approaches the time required for a freely falling body to fall L ft. Assume that the plane is at least L ft long. 34. A particle Q starts from the origin and moves uniformly along the xaxis. It is pursued by a particle P which starts at the same time from a point on the yaxis at a distance a from the origin. The velocity of Q is k times that of P. Find the curve of pur8uit, i.e., the path of P, (a) i(k;= 1; (b) if k = 1. Show that capture will occur if k < 1 after a time ak/(I  k2)r sec, where a is measured in feet and r (ft/sec) is the rate of Q.
e~6 SIMULTANEOUS EQUATIONS
44.. System of two first order equations. The simplest system of simultaneous differential equations contains two equations of first order in three variablesone independent variable and two dependent variables. Taking x as independent variable and y and z as dependent variables, we now consider a system of two first order equations which can be written in the form dy = M dx
'
(1)
dz = N dx
'
where M and N are functions of x, y, and z, or in the form (2)
where P, Q, and R are functions of x, y, and z, Any or all of the functions M, N, P, Q, and R may actually contain less than three variables; they may even all be constant. Solving such a system as (1) or (2) will be understood to mean finding two relations, free of derivatives, which together involve ,the three variables and two arbitrary constants, and which satisfy the equations. We take first an example of the simplest case where the capital letters are all constants, then an example in which M and N are not constant, but one at least of equations (1) contains only two variables. 185
186 EXAMPLE
Chapter 6
1. Solve dy _ 1
dx
,

dz = 2.
(3)
dx
Here each of the two equations may be integrated immediately, giving the solution y = x + Ch Z
EXAMPLE
= 2x
(4)
+ C2 •
2. Solve
dz
dy
 = 2x, dx
=
y
dx
+z
(5)
x
Integration of the first equation gives y = x2
+ C1•
(6)
Substituting the value of y from (6) into the second of equations (5) and rearranging the terms, we find dz dx


z x
 =
x
C1 +_. x
This is a linear equation which may be solved by the method of Art. 19. It may be solved also by use of an integrable combination, as follows: _x _dz_  _z_d_x _ 2 + 2 dx,
(1 0.)
X
X
(7)
Equations (6) and (7) constitute the required solution of the system (5).
Geometric interpretation. The term space will be used in this chapter to denote threedimensional space in which a point is z. determined by its three rectangular coordinates, z, y, A single equation (not a differential equation) in the three variables represents a surface. The surface is a plane if the
anti
Article 44
"187
equation is of first degree. If the equation contains only two variables it represents a cylindrical surface with elements parallel to the axis denoted by the missing variable; if the equation is linear in the two variables the surface is a plane parallel to the axis denoted by the missing variable. Two simultaneous equations in three variables represent the intersection of the two surfaces represented by the single equations, i.e., a curve in space; if the two surfaces are planes, the curve is a straight line. In Example 1, the first of equations (4), for a fixed C 1 , represents a plane parallel to the zaxis; the second, for a fixed C2 , represents a plane parallel to the yaxis; hence the two equations taken simultaneously, for fixed 0 1 and O2 , represent the straight line intersection of the two planes. As 0 1 and C2 independently take on an infinity of values we have 00 1 parallel planes intersecting another 00 1 parallel planes in a family of 00 2 parallel straight lines, one through each point in space. This doubly infinite family of straight lines is the geometric picture of the solution of the system of differential equations (3). In Example 2, for a fixed C1 , the parabolic cylinder (6), with elements parallel to the zaxis, intersects the singly infinite family of parabolic' cylinders (7), with elements parallel to the yaxis, in a family of 00 1 curves. Thus, for each value of C1 we have 00 1 curves of intersection; for C1 arbitrary, there are 00 2 curves of intersection forming the geometric picture of the solution of the system of differential equations (5). What does the system of differential equations tell us about the direction of the curves representing the solution? Let us look at the equations of the system written in the form
dx
P
dy
=Q=
dz
R·
(2)
The functions P (z, y, z), Q (z, y, z), and R (z, v, z) determine at any point (z, y, z) of space three numbers, and these three numbers, regarded as direction components, fix a direction at
Chapter 6
188
that point.' Thus if P = z + y, Q = 2y, R = x + y + z, then, at the point A (3, 2, 1), P = 5, Q = 4, R = 6. If from A we travel 5 units in the positive direction of the zaxis, then 4 units in the negative direction of the yaxi~, then 6 units in the positive direction of the zaxis, we arrive at a point B, and the line joining A to B fixes a direction at the point A; we denote this direction by the notation [5, 4, 6]. Furthermore, the three differentials dx, dy, and dz are, at the point (z, y, z), the direction components of the tangent line to the curve through this point and satisfying the differential equations. Relations (2) state that dx, dy, and dz are proportional to P, Q, and R, and hence that the directions determined by these two sets of direction components are the same. Therefore the system of simultaneous differential equations (2) defines a family of curves, one passing through each point of space in a direction [P, Q, R]. In Example 1 the differential equations can be written in the form
dx 1
dy
== 
1
dz
= . 2
The solution (4)
represents a family of parallel straight lines in the direction [1, 1, 2]. If y = 2 and z = 3 when x = 1, then C1 = 3, C2 = 1, and the particular solution through the point (1, 2,3) is
y
=x
 3,
z = 2x
+ 1.
(8)
This particular solution could be written down directly from the differential equations, since the equations of a line through (1,  2, 3) in the direction [1, 1, 2] are xI
1 the equivalent of (8).

y+2 z3 12'
(8')
Article 44
189
In Example 2 let us find the particular solution through the point (1, 2, 3) and its direction at this point. Substituting x = 1, y = 2, Z = 3 in equations (6) and (7),
01
= 1,
and the particular solution is
y = x2
+ 1,
Z
= x2
+ 3x 
1.
(9)
Writing the differential equations in the form
dx dy dz X=2x2 = y + z ' we have for the direction of the curve (9) at the point (1, 2, 3), [P, Q, R]
= [1, 2, 5].
Systems in which both equations (1) contain all three variables. Certain systems in which both equations (1) contain all three variables may be solved by proceeding as in the following example which we shall solve by two different methods. EXAMPLE
3. Solve dz dx = x
+ y,
dy dx = x
+ z,
(10)
Ftrst solution. By using a system of multipliers l,m,n, which may be either constant or variable, any of the fractions in equations (2) may be set equal to
1dx + m dy + n dz lP + mQ + nR·
w.· . ' ( ). riting equations 10 In
form (2), we have dx
 1
dy dz x+z x+y
.
Then dx
  dydz 1 zy
dy+dz+2dx y+z+2x+2
190
Chapter 6
Integrating the first two fractions, then the first and third, we obtain log (y  z) =  x
+ log 0h (11)
log (y
+ z + 2x + 2) = x + log O2 , Y + z + 2x + 2 = 02ez.
(12)
By adding and subtracting (11) and (12),
= Aez z = Aez
y
+ Be?" 
xI,
(13)
Bez  xI.
(14)
Equations (11) and (12), or equations (13) and (14), constitute the solution of the system (10). Second solution. Subtracting the first of equations (10) from the second, we have d
dx (y  z) = z  y,
(D
+ 1)(y 
z)
= 0, (15)
Substituting the value of z from equation (15) into the second of equations (10), we obtain dy dx  Y
=x

0 z
Ie ,
a linear equation, whose solution is
«» =
f
(xe"  C1e 2") ax + C2 ,
= e x Z
(
1)
+ c21 e2z + O2,
or (16) Equations (15) and (16) constitute the solution of the system (10); if the value of y from (16) is substituted in (15), the equivalent solution consisting of equations (13) and (14) is obtained.
191
Article 44
PROBLEMS 1. Find the equation of the cylinder formed by the straight lines which satisfy the equations dy dz = 3  = 1, dx ' dx and (a) pass through the circle y2 + Z2 = 1, x = 0; (b) pass through the parabola 3y2 = x, Z = o. 2. Find the particular solution of
dz
Y
=1 d» x '
through the point (1,2,3), and its direction at this point. 3. Find a curve through the point (1, 1, 1) satisfying the equations dy
 = 2xz
dx
'
dz
dx = 2xy.
4. Find the two most general functions of x such that the derivative of each one equals the other one. Solve the following systems of differential equations.
dy y  =0 dt x '
5.
dx dy =l+y l+x
6. 7.
dx t    = O. dt x
dy dx
+ z = sin. x,
dz
dx
+ y = cos x.
8. In a certain type of chemical reaction, if a substance A forms an intermediate substance B, which in turn changes into a third substance C, the respective concentrations, x, y, e, of the three substances obey the relations dx +ax=O, dt
x
+y +z =
c,
where a, b, and c are constants. Solve for x, y, and e, using the conditions y = z = 0 when t = 0, (a) if a ¢ b; (b) if a = b.
Chapter 6
192
9. There .are two tanks, each of 100 gal capacity, the first being full of brine holding 50lb of salt in solution and the second being full of water. If water runs into the first tank at the rate of 3 gal/min and the mixture, kept thoroughly stirred, passes out at the same rate through the second tank, when will the first tank contain twice as much salt as the second, and how much salt will have passed through the second tank at that time? 10. There are two tanks, each of 100 gal capacity, the first being full of brine holding 50 lb of salt in solution and the second being full of water. If the brine runs out of the first tank into the second at the rate of 3 gal/min and the mixture, kept thoroughly stirred, runs at the same rate out of the second tank into the first, when will the first tank contain twice as much salt as the second? 11. Tank A initially contains 100 gal of brine in which 20 lb of salt are dissolved. Two gallons of fresh water enter A per minute and the mixture, assumed uniform, passes at the same rate from A into a second tank B initially containing 50 gal of fresh water. The resulting mixture, also kept uniform, leaves B at the rate of 1 gal/min. Find the amount of salt in tank B at the end of 1 hr.
45. Systems of two linear equations. A differential equation is linear if it is of first degree in the dependent variables and their derivatives. All the differential equations encountered so far in this chapter, except those of Problems 5 and 6, are linear but of the simplest type, each being of first order and containing only one derivative. We now develop a method for solving a system of two linear equations of any order, with constant coefficients, involving one independent and two dependent variables. The method may be used also for systems of three or more equations, and for systems of equations with variable coefficients, of the type discussed in Art. 38, which can be reduced to linear equations with constant coefficients. Two examples follow which illustrate the method. EXAMPLE
1. Solve dy
dz
+=4y+l dx dx '
Article 45
193
Using operator notation, we place all terms containing the dependent variables y and z on the left, those containing y in the first column and those containing z in the second:
+ Da = 1, 3)y + z = x 2 •
(D  4)y
(1)
(D 
(2)
It can be shown * that for a system of differential equations written in this form the number of arbitrary constants appearing in the gen. eral solution is equal to the exponent of the highest power of D in the expansion of the determinant of the (operator) coefficients of the dependent variables. The highest power of D in the expansion of the determinant D4 D D  3 1
is D2 ; hence two arbitrary constants will appear in the solution of the system of equations (1) and (2). If more than two arbitrary constants appear in the process of solution, the extra ones must be evaluated, or expressed in terms of only two. Three methods of procedure are possible. We may eliminate z from equations (1) and (2), solve the resulting equation for y, then substitute this value of y in (2) to obtain z, If the value of y, containing two arbitrary constants, were substituted in (1) to obtain z, another constant of integration would appear and would have to be evaluated by substituting the values of y and z in (2); hence it is better to obtain z from equation (2). Another method, obviously more complicated in this case, is first to eliminate y from equations (1) and (2), solve the resulting equation for z, then substitute the value foundfor z in (1) or (2) to obtain y; this method introduces a third arbitrary constant which must be evaluated. A third method would be to eliminate z from equations (1) and (2), then solve the resulting equation, obtaining y in terms of x and two arbitrary constants, say C1 and C2 • Next eliminate y from equations (1) and (2), solve the resulting equation and obtain z in terms of x and two arbitrary constants, say C~ and C;. Finally C~ and C; would be evaluated or expressed in terms of C1 and C2 by substituting the values of y and z in (1) or (2). • See Forsythe's A Treatise on Differential Equations, fifth edition, Art. 171.
~apter 6
194
We proceed to solve Example 1 by each of the three methods. First solution. Multiplying equation (2) by D, (D 2
3D)y

+ Dz = 2x.
(3)
Subtracting (1) from (3), (D2  4D + 4)y = (D  2)2 y= 2x  1.
(4)
Solving (4) by the method of Art. 34, we find
=
Yc
(C1
+ C2x )e2x •
YP
= Ax + B
4) 4)
=
A
D2 yp =
o
Dyp 1)
4A = 2
A Y
=
 .l. 2
(C1
4B  4A = 1
B
_1 4
+ C2x )e2x + ~x + ~.
(V"/ ,.
(5)
Differentiating (5), Dy
= (2C 1 + C2
+ 2C2x)e2z + ~.
From (5) and (6), (D  3)y
"
= (C 1 + C2

.' _2
C2x)e x
Then. from (2),

3
Vx 
1 4
!
= (C1 
Z
(6)
C2
(7) /'
+ C2x )e2x + '~2 + ~x + ~. ,
(8)
Equations (5) and (8) comprise the general solution of the system of equations (1) and (2). Second solution. Multiplying equation (1) by D  3 and (2) by o  4, we have
+
(D  3)(D  4)y (D2  3D)z = 3, (D  3)(D  4)y + (D  4)z = 2x  4x 2•
(9) (10)
Subtracting (10) from (9), (D2

4D
+ 4)z =
CD  2)2Z
= 4x2 
2x  3.
(11)
Article 45
195
The solution of (11) proceeds as follows: Zc =
(a~
+ a~x)e2:ll.
4)
zp=Ax2+ Bx+ a
4) 1)
4A = 4
De;
=
D2Zp
=
4B  8A = 2
2Ax+ B 2A 40  4B
+ 2A =
B=~
A=1 z
=
(a~
3
0=1
+ a;x)e2:ll + x2 + ~x + 1.
(12)
. Substituting the value of z from (12) into (2), we obtain the linear equation (D  3)y =  (C~ + C;x)e211:  ~x  l, whose solution is found by the method of Art. 19:
e&y = 
f
[(0;.
+ C;x)e" + (~x + ~)e3Z] dx + 0;,
3 e311: X x .= a;e  C;e ( x  1)  2 9 (3x  1)
y = (C~ To evaluate
+ 12 + a;,
+ C; + a;x)e211: + ~x + l + c;e3x.
a; we multiply (12) by D and (13) by D Dz
=
e311:
(2~
(13) 4, giving
+ a; + 2a;x)e 211: + 2x + i,
(D  4)y = (2C~  a;  2a;x)e211:  2x  ~  a;e 311:,
which, substituted in (1), yield
1  a;e3x = l. Hence
a; = 0, and equation (13) becomes y = (~
+ a; + a;x)e2x + ~x + l.
(14)
Equations (14) and (12) comprise the general solution of (1) and (2). If = O2 and a~ = 0 1  O2 , equations (14) and (12) reduce to (5) and (8).
a;
196
Chapter 6
First eliminating z and then eliminating y from equations (1) and (2), and solving the resulting equations, as above, Third solution.
y
z=
= (C1 + C2x)f!'z +
+ l, (C~ + C;x)e2z + x 2 + ~x + l. ~x
(5)
(12)
In order to find the relations connecting the constants C1, C2 , O~, C;, substitute (5) and (12) in (2), making use of the value of (D  3)y given by (7): (C 1 + O2  02X)f!'z  ~x l + (C~ + C;x)e2Z+x2 + ~x + l = x 2, (0 1  O2 + 02x)e2z = (O~ + C;x)e2Z, O~
= 01 
C2 ,
C;
= O2 •
Hence (12) reduces to
z = (0 1

C2 + 02x)e2z
+ x2 + Ix + ~;
(8)
equations (5) and (8), as before, comprise the general solution of the given system of equations. EXAMPLE
2. Solve
d2x + 2 (dY _ a) = dt dt
0
2
'
~: 2(:  b) = O. First write the equations in the form
+ 2Dy = 2a,
(15)
2Dx  D 2 y = 2b,
(16)
D 2x
where D now denotes d/dt. The determinant of the coefficients of the dependent variables x and y yields D 4 as the highest power of D; hence the general solution will contain four arbitrary constants. Adding D X (15) and 2 X (16), we get (D3
Then Xc
+ 4D)x =
4b.
= 0 1 + O2 sin 2t + 0 3 cos 2t, X
=
C1
Xp
= bt,
+ C2 sin 2t + 0 3 cos 2t + bt.
(17)
197
Article 45

Substituting in (15) the value of D 2x = 4C2 sin 2t  4C3 cos 2t, we find Dy = 2C2 sin 2t + 2C3 cos 2t + a. Integration of (18) gives
(18)
cos 2t + C3 sin 2t + at + C4 • (19) Equations (17) and (19) contain the proper number of arbitrary constants and constitute the general solution of the given system. This solution may also be written in the form y
= C2
x y
where A, B, C, and
= A cos (2t + a) = A sin (2t + a)
a
+ bt + B, + at + C,
are arbitrary constants. PROBLEMS
Solve the following systems of differential equations. dy 1.   y  z = 2 cos 2x, dx 2. dy dx
+ dz
3. :
+ 2x = t 
dx
dz   3y dx
+ e = o.
= cos x  3y
dy = 11 sin x 'dx
y,
dx
4. dt  Y = cos 2t,
::
+: = 1 
+y 
z.
y.
dx dy . dt  dt  x  y = SIn 2t
+ cos 2t.
dy dy dz 6.  = e~  y  z, 2  +  = cos x  z. dx dx dx dy liz . dy 6.    + y + z = 4 SIn x, 2   y + 3z = 2 cos z; dx dx . dx dz . dy dz dy 7.  +  + y = 2 SIn 2x,  +  + 2y + 2z = COS 2x. dx dx dx dx
+
. t, 8• dx = y  x 3 SIn dt dy d 2y 9. dx 2  dx = 1  s, 2y
10. d _ dy dx 2 dx
+
dx dy = 2y  3x dt dt dy dz X dx + dx = 4e + y.
+ z = 10e2~
'
dy dx
+ dz
dx
_ y
+ 2et +· 6 SIn t 
= 1 + 7e 2x
.
2 cos t.
198
Chapter 6
dx 11. t dt
dy t dt + x
+ y = t + x,
d2y 12. t2 dt2
dy
+ t dt + x = 1 + log t,
= t + y. dx
t dt
dy
+ t dt
 x  y
= log t.
13. Show that the system of differential equations
+ (D + l)z = x, l)y + [(n 2  n + I)D + n]z = Dy
(D 
eZ ,
where D = djdx, has none, one, or two arbitrary constants in its general solution when n = 0, 1, and t respectively. Find the general solution in each of these cases. 14. Find a solution of the system
(7D2
+ 23)x 
8y
=
0,
(3D2  13)x + 2(D2 + 5)y = 0,
where D = djdt, subject to the conditions: x = 0, y = 0, Dx = 1, and Dy= 3, when t = 0. Find the values of x and y when t = 1. 16. Solve the system d2x dy 8+9x=at+h dt2 dt ' d2y 2
dx +8+9y=bt+k. dt dt
16. Solve the three simultaneous equations dx
 =x+y dt '
dy  =zx dt '
dz
=y+z. dt
46. Motion of a projectile. The theory of the motion of a projectile in the atmosphere is quite difficult, owing partly to complications connected with the law for variation of resistance with velocity. We shall treat here only the case in which resistance is assumed to be proportional to velocity, and the path is considered to be a plane curve traced out by the center of gravity P (z, y) of the projectile. If t is the time, d3;/dt and dy/dt are the components of velocity in the x and ydirections, and d2 x/ dt2 and d2y/dt 2 represent the corresponding components of acceleration. Let x and y be measured in feet and t in seconds; also let w (lb) be the weight of the projectile and g (ft/see"), as usual, the gravity
199
Article 46
constant. We form a system of two simultaneous differential equations by equating (w/ g) (d2 x/ dt2 ) and (w/ g) (d2y/ dt2 ) respectively to the components of force in the x and ydirections:
wd
2x
g dt
2
= k dx dt ' .
wd
2y
g dt
= w
2
k dy.
(1)
dt
Here each of the equations contains only two variables and hence can be solved separately. The solution will consist of two equations giving the parametric equations of the path, x and y in terms of t. Four arbitrary constants will appear in the course of solution, for whose determination four conditions are necessary. We shall solve equations (1) under the assumption that the projectile. starts at the origin with an initial velocity Vo (ft/sec) in a direction making an angle a with the horizontal; that is, the four conditions are x = 0, y = 0, dx/dt = Vo cos a, dy/dt = Vo sin a, when t = O. After writing equations (1) in operator notation, with D = d/ dt, the solution proceeds as follows:
(D + :OD)Y = 2
x
= C1 + C2e  (k
y
g /w ) t
Dz =  kg C2e(kg /w)t W
.
o = C1 + C2
= Cs + C4e
Dy
=

o=
Os
4
w
wt
(kg /w) t 
kg C e 

g
k
(k g /w) t 
+ C4
Vo cos a =  kg C2
· kg O4  w Voslna=
C1 = C2 =
C3 = C4 =
W
W
kg
x
Y=
vocosa
k
W
= w Vo cos a (1 kg
~ ( Vo sin a + ~) (1 
w k

w (voSina+ w) kg k
e(kg/w)t)
e(kg/
1D
lt) 
~ t.
(2)
200
Chapter 6
Equations (2) are the parametric equations of the path and locate the position of the projectile at any time t. The a, y, equation of the path could be obtained by eliminating t between the two equations. PROBLEMS
1. Assuming that there is no resistance, i.e., k == 0 in the example of Art. 46, find (a) the x, y equation of the path; (b) the maximum height of the projectile and the maximum range. 2. (a) Find the maximum height of the projectile in the example of Art. 46. (b) Show that the result found in (a) approaches the value found for the maximum height of the projectile in Prob. 1(b) as k . o. 3. The small oscillations of a certain system with two degrees of freedom are given by the equations
dh
d2x
+3x2y=0 lit2 '
d2y
+3x+5y=0. lit2 dt2
If x = 0, y = 0, £lx/dt = 1, liy/lit = 3, when t = 0, find (a) the values of x and y when t = 1; (b) the maximum and minimum values of x and y. ' 4. A particle of unit mass moves in accordance with the law
d2x lit2
=
li2y lit2
y,
 .'
= x.
If it starts from rest at the point ( 1, 0), find the parametric equations of its path. Plot the path up to the point where it crosses the xaxis, and find the coordinates of the maximum point on this portion of the path curve. 5. The primary of a transformer has inductance L1 henries and resistance R1ohms, the secondary has inductance L 2 henries and resistance R2 ohms, and the mutual inductance is M henries, where L1L2 > M2. The free oscillations in the two circuits are given by
L1lii1 + M lii 
2
lit
. lit
+ R1't1.  0
'
M liil lit
+ L2 lii + R21.2. = 0, lit 2
where il and i2 (amp) are the respective currents in the primary and secondary at time t (sec); show that i1 and ~ approach zero as t becomes infinite. Suppose that the values of the circuit constants are L1 = 3, L2 = 6, R1 = 7, R2 = 10, M = 4; also that il = 2, i 2 = 3, when t = O. (a) Show that ~"2 continually decreases but that it increases to a maximum and then decreases indefinitely as t becomes infinite. (b) Find the maximum value of il
Article 47
201
and the time required for it to reach the maximum. (c) Find the values of il and i 2 when t = 0.01 sec. 6. If a particle of unit mass at P (x, y) is attracted toward the origin 0 by a force F"the x and ycomponents of the force are F cos 0 and F sin fJ, where 0 is the vectorial angle of P. The differential equations of motion are then . Fy d2y
2=  r
dt
where r is the radius vector of P . ....(a) Multiply the second equation by x, the first by y, and subtract; then, making use of the formula for differential sector area, Art. 3, show that the areal velocity (time derivative of sector area) is constant. , (b) Multiply the first equation by~2 dx, the second by 2 dy, and add, then letting r = 1/R, show that
d2R d0
2
R_
+ 
L
h2R2'
'where h is twice the areal velocity. (c) Solve the differentialequation in (b) if F = kR2 (inverse square law), subject to the conditions R = l/ro, dR/dO = 0, when 8 = 0, and show that the path is a conic. 7. In Thomson's experimental determination of the ratio m/e of the mass to '9te charge of an elec.tron, in which the electrons were subjected to an electric field of intensity E and a magnetic field of intensity H, the equations
d2x dy m+He=Ee 2
'dt
dJ,
d 2y dt
dx dt
mHe=O 2
,
were employed." If x = y = dx/dt = dy/dt = 0 for t path is a cycloid whose parametric equations are'
x = Em(l _ cos He H2e m
t),
y
'
= 0,
show that the
= Em (He t _ si~ He H2e m
m
t).
47. The roots of unity. The n roots of the equation x", = 1,
(1)
where n is a positive integer, are called the nth roots of unity. De Moivte's theorem gives the n values of x which satisfy equation (1) in the form 21rk
x = cos'n * See Phil.
21rk + .•SInn ' 1,
(k
Mag., Vol. 48 (1899), p. 547.
= 0, 1'" 2 ... n  1),
(2)
202
Chapter 6
where i = v=I. These values of x are the nth roots of unity. They may be represented graphically by n points equally spaced around a unit circle centered at the origin, the first point being at x = 1, corresponding to k = O. If r denotes the next of these points in the counterclockwise direction, corresponding to k = 1, 21r .. 211" ( ) r=cos+'tsln3
n
n'
and we may call r the basic nth root of unity; all the nth roots of unity are obtained by raising r to the powers 0, 1, 2, .'., n  1. Thus instead of (2) we may write 'r" tIV
= 1
r r2
" "
r ...
,
rn
l • ,
(rn
=
1).
(4)
> 1),
(5)
The relation 1
+ r + r 2 +...+ r n  l
= 0,
(n
follows from the symmetrical arrangement of the points. Furthermore 1 +rk
+r2h
+.. ·+r(nl)k = 0,
and ",nk
(k = 1,2,3,·· ',n 1),
= 1.
(6) (7)
When n = 2, r = 1; the square roots of unity, 1, r, are 1, 1.
When n = 3, r = ~ + iy'3/2 and r 2 = ~  iy'3/2. It is customary to denote r by the Greek letter '" when n = 3. Thus", = ~ +iv'3/2, (JJ2 = ~  iv'3/2, and the cube roots of unity are 1, "', ",2. When n = 4, r = i; the fourth roots of unity, 1, r, r 2 , r 3 , are 1, i, 1, i. 48. Cyclic systems of differential equations. By a cyclic interchange of the letters Xl, X2, X3, " ' , Xnl, Xn, we mean a change of Xl to X2, X2 to X3, ••• , Xnl to X n, Xn to Xl' We call a system of differential equations cyclic if it is unchanged by a cyclic interchange of the dependent variables; for example, the
Article 48
203
following system in which the x's are the dependent variables and D = d/dt: DXl DX2
= =
X2,
Xg,
(1) DXn_l = X n , DXn
= Xl.
Certain cyclic systems of n equations in the dependent variables Xl) X2, ••• , X n , may be solved by changing to a new set of dependent variables, Ul, U2, " ' , Un, connected with the old set by means of. the substitutions
= Xl + rklx2 + (rkl )2Xg +...+ (rkl )nlxn, (2) where k = 1, 2, 3, ... , n, and r is the basic nth root of unity. Uk
The original dependent variables Xl, X2, •• " X n and the independent variable t are assumed to be real. Usually we wish to express the x's in a form involving t and n arbitrary constants but involving no fixed complex constants, that is, to obtain a solution in real form, although complex quantities are used to effect the solution. . . EXAMPLE
1. Solve the system Dx=ax+by+ez,
+ bz + ex, Dz = az + bx + cy,
Dy = ay
where D = d/dt, and a, b, e are real constants. In the substitutions (2) let n = 3, r = w, and replace Xt, x, y, z, and Uh U2, Ua by u, v, w. Then
(3)
X2, Xg
+ y + z, = z + wy + w2z,
by
u = x v
(4)
w=x+w2y+wz.
The last term of the third equation is w4z, but it is written tJJa = 1.
wZ
since
204
Chapter 6
Adding equations (4), we find
u+v+w=3x, since 1 + w + w2 = 0 [equation (5), Art. 47]. Multiplying equations (4) by 1, w2 , w respectively and adding, then by 1, w, w2 and adding, we find similarly
+ w2v + ww = 3y, u + wv + w2w = 3z, u
so that the solution of equations (4) for x, y, z gives
+ v + w), ~ (u + w2v + ww), ~ (u + wV + w 2w).
x = ~ (u y = Z
=
(5)
Now, adding equations (3), we obtain Du = (a Ir b + c)u, whose solution is
u
= 3CleCa+b+c),.
(6)
Multiplying equations (3) by 1, w, w2 respectively and adding, we find Dv = (a + bw2 + cw)v, whose solution is (7) Finally, multiplying equations (3) by 1, w2 , w respectively and adding, we have Dw = (a + bw + c (2 )w, 'W
= 3CseCa+oo,+cw2)t.
(8)
Substituting (6), (7), and (8) in (5) gives
x
= Cte(a+b+c),t + C2eCa+oo,2+cw)t + C se(a+bw+cw2)t,
+ w2C2eCa+bw2+cw)t + wCse(a+bw+cw2)t, z = CleCa+b+c)t + wC2e(a+bw2+cw)t + w2Cse(a+bw+cw!)t. y = CleCa+b+cH
(9)
Equations (9), where the C's are arbitrary constants, represent the general solution of the system (3), but the solution is not in real form
Article 48
205
since it involves the fixed complex constant w. This solution, however, may be transformed into the following form (see Problem 1 at the end of this chapter):
x
= Ge(a+b+c)t
+ KJ't sin (J.Lt + a),
(pt + a  2;), z = Ce(a+b+o)' + Kt!" sin (pt + a  ~), y
= Ce(a+b+o)I + KeA• sin
(10)
where b
c
2
2'
X=a
and C, K, and a are arbitrary constants. Equations (10) represent the general solution of the system (3) in real form.; for any real values of the arbitrary constants G, K, and a, the values of z, y, and z are real. EXAMPLE 2. Solve the system (1). This system may be solved by use of the substitutions (2), but here it is easy to eliminate all but one of the dependent variables, say xl, then solve the resulting equation for Xl and find the other x's by successive differentiation. Substituting the value of X2 from the first equation into the second, then the value of Xa from the second equation into the third, ... , then the value of X n from the (n  l)th equation into the nth, we have or (D"  I)XI = 0, from which we write down the value of Xl, then obtain DXl = X2, ••• , Xn from DXnl = Xn• The general solution of system (1) is therefore Xl
=
Gle
t
X2 = Glet •
X2
+ G 2ert + G aer2t + + c.s">, + rG2ert + r2Gaer2t + + rnlGnernlt, •
•
•
•
•
•
•
•
from
(11) •
•
•
•
206
Chapter 6
in which the derivative of each equation gives the following one and the derivative of the last gives the first. As in the case of equations (9), the equations (11) represent a solution in imaginary form for n > 2, but in any particular case the solution can be exhibited in real form, for example in the following special case of (1).
EXAMPLE,3. Write in real form the solution of the system
DX2 =
Xs, .
Dxs
X4,
The first of equations (11) for n Xl
=
(12)
= 4, r == i, is
= Glet + G2eit + Gset + C4e it•
By means of Euler's relation we may replace 02eit + C4eit by K sin (t ex) as in Art. 29(c). Also let 0 1 == A, 0 3 = B. We thus write Xl in real form. and obtain the other x's by successive differentiation:
+
+ Bet + K sin (t + a), t t X2 = Ae  Be + K cos (t + ex), t t Xs == Ae + Be  K sin (t + ex), t t X4 = Ae  Be  K cos (t + ex). Xl =
Aet
(13)
Equations (13), in which A, B, K, and a are arbitrary constants, form the required solution of the system (12).
49. A special case of Einstein's equations. Einstein's law of gravitation is expressed by a system of ten partial differential equations of second order. It has been shown by Kasner * * "Solutions of the Einstein equations involving functions of only one variable," by Edward Kasner, Transactions oj the American Mathematical Society, Vol. 27 (1925).
207
Article 49
that in a special case these equations reduce to the following system of three ordinary differential equations of first order:
Dz = yz  x2 , Dy = z:c  y2, Dz
= xy 
(1)
Z2,
where D = d/dt. We notice that this system of equations is cyclic in :1;, y, z, but not linear as in the case of the system in Example 1, Art. 48; however, the method used in that example will be followed in solving system (1). Letting
+ =x+
u =x 1)
W
we find uv
= x2 +
vw = wu
+ z, 6J'Y + CJ)2Z, y
= X + ClJ2y +
+ ClJ2,,2 x 2 + y2 + Z2 ClJy2
= 3:2 + (JJ2y2 +
C1JZ2 
(2)
ClJZ,
ClJ2:tJJJ 
yz 
~,
xy 
yz 
zx,
611:Y 
yz 
6)2ZX•
(3)
Now, adding equations (1), multiplying them by 1, "', 6)2 respectively and adding, then multiplying by 1, lA>2, W respectively and adding, we reduce system (1) to the system
Du =vw, Dv = uv, Dw
(4)
=wu.
From the last two of these equations dv/v = dw/w, hence v = k2w, where k 2 is an arbitrary constant written in this form for convenience, and the system (4) is equivalent to
Du
= k 2w2 ,
Dw = wu, v
= k 2w.
(5)
208
Chapter 6
Eliminating u from the first two of equations (5), we get k 2w2 = WD2W + (DW)2
w2
or
,
WD2W  (DW)2 = k 2w4.
Solving this equation by the method of Art. 40, letting Dw = p, D2W = p dp/dw, we have wp dp _ p2 = k2w4 dw '
then, as in Art. 20(a) , 2p dp _ 2 p2 = 2k2u1 dw w '
where the constant of integration is taken as c 2 ; if the + sign is used the resulting solution will involve hyperbolic instead of circular functions. Solving for p( = dw/dt), separating the variables, and integrating again, we find
1
kw
c
c
 sec" 
c
= ± (t
w = k sec c(t
+ b) '
+ b).
The last two of equations (5) then give v and u: v u
= ck sec c(t + b), = c tan c(t + b).
Article 49
209
The relations (5) of Art. 48 yield the values of x, y, z: k2
x =
=
y
+ 1 c sec c(t + b)
3k
CJ)2k 2
+ CJ) c sec c(t + b) 
3k
CJ)k 2
Z =
c
 3 tan c(t
+ CJ)2 csec c(t + b) 
c
3 tan c(t C
3k
+ b),
3 tanc(t
+ b),
(6)
+ b).
Equations (6), in which k, b, and c are arbitrary constants, represent the general solution of the system (1). It may be noticed that if we write k
al =
2
+1
CJ)
a2 =
3k'
2k2
+ CJ)
3k
,
'
then at, a2, and aa satisfy the relations
+ a2 + aa = 0, ala2 + a2aa + aaal = k +1 ala2aa = 27k3 ; al
~,
6
hence at, a2,
aa are the roots of the cubic a3 
+
k6 1 la = 0• 3 27k3
(7)
The solution (6) of the system (1) may therefore be written in the form z, y, z
=
aiC sec c(.t
+ b)
c
 3 tan c(t
+ b),
(i = 1,2,3), (8)
where the a's are the roots of the cubic (7). This is the solution obtained by another method in the paper by Kasner previously cited. An integral of a system of differential equations expressed as a relation free of derivatives, involving some or all of the dependent variables and an arbitrary constant, is of importance
210
Chapter 6
in the discussion of certain physical problems. From the third of equations (5) we see that
+ wy + Z = x + w y + wZ 2
x
W
2
C
(9)
is an integral of the system (1). PROBLEMS
1. Transform equations (9), Art. 48, into equations (10), making use of Euler's relation, et.fJ = cos 8 + i sin 8. 2. Obtain the solution of the system dx dt
= Y+~
dy dt
= z + x  y,
dz dt
=
in the real form
x

x,
+ y ~,
+ Be2t , Ae t + 062t , Ae t  (B + O)e2t ,
x = Ae t y = Z =
(a) by putting a = 1, b = c
=
1 in equations (9), Art. 48; (b) by putting
a = 1, b = e = 1, hence ~ = 2, p, = 0 in equations (10), Art. 48. Solve in real form the following systems, where D = d/&:
3. Dx = ay, Dy
=
D~
= ax.
(lB,
4. Dx=Dy=Dz=x+y+~. 6. Dz + Dy = z, Dy+D~=x,
Ds + Dz = y. 6. Dx = I (y + z + t), Dy = ! (z + x + t),
+ y + t). 7. tDx = y + s, tDy = z + x, tDz = x + y. D~ =
i
(x
"
211
Article 49
8. t Dx1 = ax2, t DX2 = axa, tDxa = ax4, tDX4 = axl. 9. Show that the substitutions
u=3:+Y+s, v = xy + yz + zx,
w = X'gZ, reduce·system (1) of Art. 49 to the system
Du+u2 = 3v, De = 0,
Dw+3uw = v2• 10. Obtain the integrals
zy =C zx
and
xy+yz+zx=C
of system (1), Art. 49. 11. Prove that the three integrals of system (1), Art. 49, given in equation (9) and in Prob. 10, namely, J(x, y, s)
=
+ ",y2 + ",2s = C1, X + '" y + "'z
g(x, y, z)
=
zy = C2, zx
x
h(x, y, s) = xy + yz + zx = Ca, are not independent by showing that the Jacobian of J, g, h with respect to x, y, z vanishes. (Cj. Art. 21.) 12. Letting x/z = a, y/z = {j, the first two integrals in Prob. 11 may be written
0(0:, (j)
= 1{j = Ia
C2 •
Show that F and G are not independent and find the relation connecting them.
e~7 SERIES SOLUTIONS 60. Power series. There are many simplelooking differential equations, for example, d2 y/ dx 2 + xy = 0, for which the methods discussed up to this point are inadequate to yield a general solution. When such an equation is encountered it may be possible to solve it by use of an infinite series. We shall consider in this chapter only certain =Ie equations of the form (1)
where the f's are functions of x and primes denote differentiation with respect to ». Assume that a value of y, expressible in the form of a power series in x, satisfies equation (1), and write down the expressions for y and its first two derivatives: (2) y = ao + alX + a2 x2 + aax3 +...+ anxn +...,
+ 2a2x + 3aax2 +...+ (n + 1)an+lXn + , y" = 2a2 + 3 ·2aax + 4·3a4x2 + ... + (n + 2)(n + l)an+2xn +....
y' = al
(3)
(4)
Not all differential equations of form (1) can be solved by the methods of yf  y = 0 (cf. Probe 26, this chapter. For example, the equation 4x 3y" + Art. 8) has a general solution which is not expressible by series of type (2), Art. 50, or of type (1), Art. 51. In this case the assumption of a solution of type (2), Art. 50, leads merely to the trivial solution y = 0, and the assumption of a solution of type (1), Art. 51, leads to a contradiction ao = o. Furthermore, in other cases these assumptions might lead to series which do not converge except possibly for x = o. It is understood that a solution expressed in series form is valid only for values of x for which the series is convergent. A fuller discussionmay be found in Ince's Ordinary Differential Equations, Chapter VII, or in Cohen's· Differential Equations, Chapter IX. 212 III
6r
Article 50
213
Now substitute the values of y, y', and y" from (2), (3), and (4) in equation (1), then equate to zero the complete coefficient of each power of x in order that (1) shall be satisfied identically. It may happen that two of the a's thus remain arbitrary but that the others are determined or expressible in terms of the two arbitrary ones. In this case equation (2) expresses y as a power series in x involving the two a's as arbitrary constants. If the series converges, equation (2) furnishes the general solution of (1), valid for values of z for which the series is convergent. Some of the following examples illustrating the series method could be solved without the use of series; in such cases it is interesting to notice that the results obtained by the use of series are equivalent to those obtainable by previous methods. EXAMPLE
1. Solve (1
+ x 2)y" + xy' 
y
=
o.
(5)
Substitute the values of y, y', and y" from (2), (3), and (4) in (5) and arrange the result in tabular form. At the left of the table are the terms of equation (5), at the top the various powers of x which occur, with their coefficients in the body of the table.
y xy' y" x 2y"
con.
x
x2
Xs
· ..
ao
al al
a2
as
· ..
2a2
3as
3·2aa
4·3a4
5· 4a6
~
3·2as
~
·.. ·.. ·..
·..
xn
an nan
(n
+ 2) (n + 1)an+2 n(n  l)a n
· ..
· .. ·.. · ..
The sum of each column of coefficients must vanish. The first two columns give 2a2  ao = 0,
= 0,
ao arbitrary,
a3
=
0,
al
arbitrary.
214
Chapter 7
The next two columns give
4·3a4
+ 3a2 = 0,
5·4as
+ 8as =
0,
as
= 0,
and so we see that all a's with odd subscripts vanish, except at which is arbitrary, and that each a with even subscript can be expressed in terms of ao which is arbitrary. Perhaps the best way to compute the a's is to obtain a general formula by equating to zero the column of coefficients of x": (n
+ 2)(n + l)an + 2 + (n2 an + 2
Then, letting n
= 0,
=
l)an
= 0,
nl n
+ 2 an·
1, 2, 3, 4, "., we find
n=O, 11,
= 1,
as = 0, 1
n = 2,
a, 
n = 3,
as = faa = 0,
 .l.,., 4:Ul2 
 a0,

2·4
1·3
n=4,
a6
•
= la, = 2.4.6 ao,
•
Substituting these values of the a's in equation (2) and remember.. ing that at is arbitrary, we have
Y = ao ( 1 + 1 x 2   1 x4 + 1·3 x 6  ... ) 2 2·4 2·4·6
+ alX,
(6)
which, with ao and al arbitrary constants, is the general solution of equation (5). We might recognize the series in parentheses as the expansion of (1 X2)~ and write the solution (6) in the finite form
+
y = aoV
I + x2 + atx.
(7)
215'
Article 50
However, if the series in parentheses were not recognized we could obtain the finite form of solution as follows. Putting ao = 0, at = 1 in (6) yields the simple particular solution y = x. We now apply the method of Art. 41 to find the general solution of (5) when a particular solution is known. By substituting y
=
y' = xv' + v,
xv,
y"
= xv" + 2v',
in equation (5); it is reduced to
~ +G+l~X2)dx=O. Two integrations give log v'
+ 2 log x + ~ log (1 + x 2 ) = log Ct,
or
dv dx
=
C1 x 2V l
+ x2
,
then
hence, multiplying by x, we obtain y
= A VI + x2 + Bx,
where A and B are arbitrary constants, which is form (7). Notice that the series solution (6) is valid only for 1 < x < 1, whereas there is no such restriction on x in solution (7). The solution may be verified by eliminating the arbitrary constants from the solution in finite form and thus obtaining the original differential equation (Problem 27, Art. 8). EXAMPLE
2. Given the differential equations
x2 y"  2xy' 4x 2y"
+ 2y = 0,
+ 4:vy' 
y = 0,
(a) (b)
obtain the general solution of (a) by the power series method and show that this method is inadequate to yield the general solution of (b).
216
Chapter 7
Substitution of the values of y, v', and y" from equations (2), (3), and (4) in equation (a) results in the following table:
2y 2xy' x 2y"
con.
z
x2
2ao
2al  2al
2a2 2·2a2 2a2
XS
2aa 2·3aa 3·2as
·.. · ..
·..
·..
xn
·..
2an 2nan. n(n  l)a n
·.. ·.. · ..
Equating to zero the sum of each column of coefficients, we see that all the a's are zero except al and a2 which are arbitrary since the columns in which they occur vanish identically. This could also be seen by summing the coefficients of z": (n2

3n
+ 2)an = (n 
l)(n  2)an
= OJ
hence an = 0 unless n = 1 or 2, al and a2 being arbitrary. Equation (2) then reduces to
the general solution of equation (a) in finite form. Applying the same method to equation (b) we have the table:
y 4xy' 4x 2y"
con.
x
ao
al 4al
x2
x3
tl2
aa
4·2~
4·3aa 4·3·2aa
4·2~
·..
xn
· ..
·.. ·.. ·..
an 4nan 4n(n  l)an
·.. ·.. ·..
Equating to zero the sum of each column we find that all the a's are zero; equation (2) gives only the trivial solution y = O. There is no power series in x which satisfies equation (b). The power series method is inadequate for the solution of (b) but we shall obtain the solution by a more general series method in the next article. Both equations (a) and (b) can be solved by the method of Art. 38.
217
Article 50 EXAMPLE 3. Solve by the power series method
xy"  (x
+ 2)y' + 2y =
and express the solution in finite form. Substituting the values of y, y', and (8), we have the following table: con.
2y 2y' xy' xy"
2ao 2al
x2
x
0,
(8)
v" from
(2), (3), and (4) in
2~
2al 2· 2a2 al
2·3as
2~
3·2as
2as 2·4a4 3as 4·3a4
2~
xn
·..
2an 2(n l)an+l nan (n l)nan+l
·.. ·.. ·.. ·..
·..
x3
·.. ·.. ·.. ·..
+ +
Equating to zero the sum. of each column, the first column give~_ al = llo, with ao arbitrary; the next gives  2Cl2 + al = 0, a2 = al/2~ the x2 column vanishes identically, making aa arbitrary; the remaining columns determine a4, a5, ..• in terms of aa. The general formula, obtained by summing the coefficients in the x" column, is (n
+ 1) (n 
2)an +l
=
(n  2)an ,
or
an an+l = n
(n ~ 2),
+ 1'
from which we find
n = 0, n = 1, n = 2,
aa arbitrary,
n = 3, n
= 4,
•
•
•
•
•
•
Chapter 7
218
Substitution of these values of the a's in (2) gives y=ao (
1+x+2
X2)
4
5
X X ) +a3 x3+++ ... 4 4·5 ' (
(9)
the general solution of equation (8) in series form, ao and a3 being arbitrary constants. In order to express the solution in finite form we might use the method of Example 1, but it is simpler to notice that the second part of the solution may be written 3 ) .3!a3 (x3!+4!+5!+··· x4 x5
,
and that the series in parentheses is now the series for first three terms, 1 + x + x 2 /2. Hence Y = ao
(1 +
x
~
minus its
+ ~) + 3la3 [e'"  (1 + x + ~)],
or, letting ao  3 tas = A, 3 !a3 Y= A (1
= B,
+ x + ~) + s«,
(10)
where A and B are arbitrary constants. Equations (9) and (10) are equivalent, the latter being the solution of (8) in finite form. This example could have been solved without using series, by the method of Art. 41. Verification of solution (10) is found in Problem 28 of Art. 8. EXAMPLE 4. Solve
v" + xy = o.
Substituting the values of y, y', and (11), we have the following table: con.
xy y"
x2
x
ao 2112
3· 2a3
al 4·3lZ4
x3
tl2
5·4a6
(11)
v" from (2), (3), and (4) in
·.. ·.. ·..
·..
xn
(n
+ 2) (n +
anl 1)an+2
·.. · ..
Article 50
219
Equating to zero the sum. of each column, we see that a2 = 0; ao and al are arbitrary; aa, a6, ll9, .•. are expressible in terms of llo; a4, a7, alO, ... are expressible in terms of al; as, as, au, ... are expressible in terms of a2 and are all zero. The general formula is
an+~
=
(n
+ 1)(n + 2) •
Then, letting n = 1, 2, 3, ... , we have
ao aa =  2.3' aa
4ao
a6==5·6 6!' ll9=
1·4· 7ao 9!.' alO =
 8·9 = •
=
•
•
101
9·10
•
•
2X4 al ( x  4!
+
= 0,
as
= 0,
' au
= 0,
•
•
2· 5a l
6·7 7! ' 2·5· 8a l a7 =
xa 1.4x6 1.4.7x9 , ) y=ao ( 1 3 1+ 6! 91 +...
as
•
•
•
•
+
2.5x7 2.5.8x10 7! 10 !
)
+... ,
(12)
the general solution of equation (11), where ao and al represent arbitrary constants. Here we leave the solution expressed in terms of two infinite series which do not represent expansions of any elementary functions so far encountered. The solution may be expressed in terms of Bessel functions. * PROBLEMS Solve the following differential equations by the power series method; also, in case the result involves an infinite series, express the solution in finite form if possible. In the first six problems verify the result by obtaining the original differential equation from the general solution as in Chapter 2. 1. x(x 2)y"  2(x l)y' 2y = O. 2. x(x + l)y" (x  l)y'  y == O.
+
* See Reddick
+
+
+
and Miller's Advanced Mathematics for Engineers, Chapter VI. Following the theory of this chapter it may be shown that (12) is equivalent to y = Ax~J~(ix%) + BX~J~(ix%).
220 3. 4. 5. 6. 7. 8. 9. 10. 11.
Chapter 7
+ +
(1  X 2)y"  xy' 4y = O. (1  4x 2)y"  4xy' 4y = O. xy"  (2x2 3)y' + 4xy = o. xy"  y' 4x 3y = O. y"  xy'  y = O. (1  x 2)y"  2xy' + 2y = O. y"  xy' 2y = O. xy"  2y' + xy = O. (Cf. Prob. 29, Art. 8.) 2(x2 + 8)y" + 2xy' + (x 2)y = O.
+ +
+
+
12. (a) Show that the power series method gives only a particular integral of the differential equation xy" + 2y'  xy = OJ find the particular integral and express it in finite form. (b) By making use of the particular integral found in (a), obtain the general solution of the differential equation. (c) Eliminate the arbitrary constants from the general solution found in (b), thus obtaining the original differential equation. (Cf. Prob. 30, Art. 8.)
51. The series of Frobenius. In Art. 50 we found that the differential equation (b) of Example 2 could not be solved by the method of that article since it has no solution in the form of a power series in z. We shall solve this differential equation as the next illustrative Example 1 by employing the more general Frobenius series. The series of Frobenius is obtained from the power series (2) of Art. 50 by multiplying it by xc: y
=
XC(ao
+ alX + a2x2 +... + a",x'" +. ·.), ao ~ O.
(1)
Here c is a constant to be determined, as well as the a's, by substituting the series in the equation to be solved and equating coefficients as in Art. 50. For zero or positive integral values of c, (1) is merely a power series, but for negative or nonintegral positive values of c the series represents a new type. The Frobenius series (1) is therefore a generalization of a power series, including it as a special case. The assumption ao ~ 0 means that XC is the lowest power of x appearing in the series. The method of solving a differential equation by use of series (1) is similar to that of Art. 50. Assume that a value of y
Article 51
221
expressible in the form of series (1) satisfies the differential equation of form (I), Art. 50, and write down the expressions for y and its first two derivatives:
+ a1XC+1 + a2XC+2 + aax + + ..., = Cac¢l + (c + l)alx + (c + 2)a2XC+1 +. ·.,
y = aoXC
y'
C
C
3
(2) (3)
v" =c(cl)ao1f2+(c+ l)catxcl+ (c+2) (c+ 1)a2xc+. .•• (4) Substitute the values of v, y', and v" from (2), (3), and (4) in the differential equation, form a table like those in Art. 50, starting with the lowest power of x which occurs after the substitutions are made, then equate to zero the complete coefficient of each power of x. Suppose that two values of c are thus determined. If, for one value of e, we are able to determine all the a's in a manner involving only one of them which is arbitrary, we substitute these values in (2) and obtain a value of y, say Yh which is a particular solution involving one arbitrary constant. * If, for the other value of c, we can determine the set of a's, which we now call a"s, involving one of them which is arbitrary, we have another particular solution, say Y2. The sum
involving two arbitrary constants, is the general solution of the differential equation. It may happen that for one value of c the a's are determined in terms of two of them which are arbitrary; the corresponding value of Y, involving two arbitrary constants, then represents the general solution. Solutions in series are valid for values of x for which the series converge. There follow some illustrative examples, the first being the differential equation (b) of Example 2, Art. 50. * It is understood here as in Art. 50 that the series represents a solution only for values of x for which it converges.
222
Chapter 7
EXAMPLE
1. Solve 4x2y"
+ 4xy' 
y = O.
(5)
Substitute the values of y, y', and y" from (2), (3), and (4) in (5) and arrange the result in tabular form, starting with the lowest power of x present after the substitutions, namely XC:
y 4xy' 4ry"
x c+2
xc+1
XC
ao 4cao
a2 al 4(c+l)al 4(c+2)~ 4c(cl)ao 4(c+l)cal 4(c+2) (c+l)~
·..
x c+r
·..
·.. ·.. ·..
a r 4(c+r)ar 4(c+r) (c+rl)a r
· .. ·.. ·..
Equating to zero the sum. of the coefficients of xc, we find c = ~ or  ~,
(4c2  l)ao = 0,
ao arbitrary.
The set of a's corresponding to each of the two values of e will now be determined. We obtain a general formula by equating to zero the sum of the coefficients of xc+~, first for c = ~, then for c =  ~. For c = ~: [4{~
+ r)2 
I]«,
=
(4r
+ 4r)a, = 4r(r + l)a, = OJ
hence a, = 0 unless r = 0, with ao arbitrary. For c =  ~: [4(  ~
+ r)2 
l]a~
= (4r2

4r)a:.
=
4r(r  1)~
= OJ
hence a:. = 0 unless r = 0, 1, with a~ and a~ arbitrary. Then equation (2) gives, for c = ~ and c =  ~ respectively, Yl
=
aox~,
Y2
=
~1/2
+ a~x~.
The first of these results is superfluous since the second contains two arbitrary constants and represents the general solution. However, we would get the same general solution by setting y = Yl + Y2 and replacing ao + a~ by a new arbitrary constant. The general solution of equation (5) is therefore
Y = Ax~
+ Bx~,
where' A and B are arbitrary constants. This solution could be found without the use of series, by the method of Art. 38.
Article 51
223
As a second example we solve the differential equation of Problem 12, Art. 50, by use of a Frobenius series. EXAMPLE
2. Solve xy"
+ 2y' 
xy = 0,
(6)
and express the result in finite form. Substituting the values of y, y', and y" from (2), (3), and (4) in (6), we have the following table: xCI
x c+ 1
XC
xy 2y' 2cao 2(c+l)at xy" c(cl)ao (C+l)cal
XC
ao 2(c+2)~
(c+2)
(c+l)~
...
+2
X
... ... ...
al 2(c+3)aa (c+3) (c+2)a,J
C
+T
arl 2(c+r+l)ar+l (c+r+l) (c+r)ar+l
·.. ·.. ·.. ·..
Adding the first column and equating the sum to zero, we have
c(c + l)ao = O. Hence c = 0 with ao arbitrary, or c = 1 with ao arbitrary (= ~). The second column gives (c + l)(c
+ 2)at = o.
Hence if c = 0, at = 0, and if c = 1, at is arbitrary (= a~). The general column gives FOR
FORC=O (r
+ 1) (r + 2)ar+t = ao ao a2 · !, 2 3
3
arI,
r(r
C
= 1
+ l)a~+l ,
=
a:.h ,
, ao ao a 2  1.2  2!'
,
,
, at al a3  ! ,  • 2 3
,
3
,
ao a4  3.4  4!' ,
a2
,
a3
,
a3
as
= 5.6 =
0,
ao a6  6.7  7!' a4
•
•
,
at
a 5 
4.5  51'
,
,
a4 ao a 6  5.6  61' ,
224
Chapter 7
It happens here that for o = 1 all the a's are determined in terms of two of them, ~ and a~, which are arbitrary; hence this value of e yields the general solution: 3
5
2
4
6
I x x x ) ( x x x ) . y=a~ (  +  +  +  + ... +a~ 1++++··· . (7) x 2! 4! 6! 3! 5! 7!
The second series of (7) would give the particular solution corresponding to c = o. T~ express (7) in finite form, we make use of the expansions: x3 x5 x1 sinhx=x+3t+51+71+ •.. , x2 x4 x6 coshx=I++++··· 21 4! 61 •
Then (7) reduces to ,coshx + ,sinhx y = ao al · x x
(8)
A trick method. If we notice that
D(xy) = xy' where D
+ y,
D 2 (xy)
= xy" + 2y',
= d/dx, then equation (6) can be written (D 2

l)xy
=
o.
Hence we have at once by the method of Art. 29, xy = Clf?
+ C2 e z ,
or
(9) which is the general solution of equation (6) and is equivalent to (8). EXAMPLE
3. Show that the differential equation xy"  xy'
+y =
0
(10)
has only one particular solution of type (2). Make use of this particular solution in finding the general solution.
Article 51
225
Substitute the values of y, and form the table: xc l
y', and v" from (2), (3), and (4) in (10)
·..
xc+l
XC
·..
xc+r
y
ao
al (c + l)al 2) (c 1)t12
xy' 000 xy" c(c  l)ao (c + l)cal (c +
+

·.. ar · .. ·.. (c +r)ar ·.. ·.. (c +r + 1) (c +r)ar+l ·..
The first column gives c(c  l)ao = 0, from which c = 0 or e = 1 with ao arbitrary. The second column gives (c + 1)001 = (c  l)ao which cannot be satisfied by c = 0, since ao ~ 0; but for c = 1, at = o. Furthermore all the successive a's are 0 when c = 1, as can be seen from the formula (r + 2)(r
+ l)ar+1 = rare
Therefore the only solution of type (2) is for c = 1, ao arbitrary, a1 = a2 = ... = 0, namely, the particular solution y
= aox.
(11)
To find the general solution we use the method of Art. 41 with y = x as a simple particular solution obtained from (11) by putting ao = 1. Substituting y = xv, y' = xv' + v, y" = xv" 2v', in equation (10), we obtain
+
+ 2xv'  x2v' = 0, dv' + 2  x dx = O.
x 2v" v'
Integrating,
x
log v' + 2 log x  x dv
Olf?
;k=7· Then
= log 0 1,
226
Chapter 7
Substituting for eZ its power series expansion,
v = 01
f 0 + ~ + ;1 + :1 + :~ + ...) ax +
= 01 (
I 
X
x
2
O2
3
x ) + log x + 2! + 2.3! + ~4 t + ... + O2 , x
and, since y = xv, the general solution of equation (10) is
x3 x4 ) y=OI ( 1+xlogx+2t+2.3t+3;4t+ ... X2
+02 X•
52. Bessel functions of zero or positive integral order. Bessel functions are named after the German mathematician and astronomer, Friedrich Wilhelm Bessel, who was director of the observatory at Konigsberg. He obtained them in solving a differential equation connected with a problem in planetary motion, but they occur in various other problems in applied mathematics. Bessel functions are particular solutions of the differential equation x 2 y"
+ xy' + (x 2 
n 2 )'11
= O.
(1)
The number n may have any real or complex value, but we limit this discussion * to the case where n is zero or a positive integer: n = 0, 1, 2, 3, . . .. With this restriction on n let us make use of a Frobenius series to find a particular solution of equation (1). Substituting in (1) the values of y, y', and y" from equations (2), (3), and (4) of Art. 51, we form the following table of coefficients: ZC
n2y x2y X'/I' x 271"
XC+ 2
zC+l
n2ao
 n2al
n2~
(c+l)al c(cl)ao (C+l)cal
(c+2)~
ao cao
* See footnote to Ex. Bessel functions.
(c+2)
(c+l)~
x C+ 3
·..
 n2aa al (c+3)aa (c+3) (c+2)aa
·.. ·.. ·.. ·..
4, Art. 50, for reference to
8.
ze+ r
n2ar ar2 (c+r)a,. (c+r) (c+rl)a,.
·..
·.. ·.. ·.....
more extensive discussion of
Article 52
227
Setting the sum of the first column equal to zero, we find (c  n 2 )ao = 0, which with ao arbitrary is satisfied if c = n. We now obtain the particular solution of equation (1) corresponding to c = n. The second column gives al = 0, then the fourth column gives aa = 0, ete.; all a's with odd subscripts vanish. We compute the a's with even subscripts from the formula obtained by equating to zero the sum of the coefficients in the x c +r column with c = n: [(n + r)2  n 2]a r + ar2 = 0, ar  2 . a  r r(2n + r) Hence 2
a2 =  2 (2n a2
4(2n •
a2k
•

+ 4)
ao
2·4(2n
•
=
+ 2) , + 2)(2n + 4) ,
•
(I)k
+
•
ao
2·4 ... 2k(2n 2)(2n + 4) ... (2n . 2 .4 . . . 2k = 2k (1 .2 .'. . k) = 2k k r
Now since
+ 2k) .
and (2n+2) (2n+4) ... (2n+2k) 
2k(n+l)(n+2) .•. (n+k),
we may write a2k
=
(
1
)
ao
k
22kk !(n
+ 1) (n + 2)
... (n
+ k) .
Multiplying both numerator and denominator by 2n n !, 2n n!ao (1) 2n +2k k !(n k)! ' k
a2k
=
+
or, setting 2nn !l2 = C. x(1 + y) = Cel/ 1J . Art. 16 '
1. 2.88ft/sec; 1.05ft. 2. 0.557. 3. (a) 146ft/sec; (b) 945 ft; (c) 3.73 sec; (d) 3.65sec. 4. (a) 6.83V sec; (b) 5.70V sec. 5. (a) 2.16 sec; (b) 67.0 ft. 6. t = [wV/g(w  E)] log [100/(100  p)] sec. 7. (a) 8.83 It/sec: (b) 1.01 sec. 8. 126 It/see; 3.10. 9. 147 It/sec; 2.67. 10. 16A/7. 231
232 11. 13. 17. 22.
{
Answers
(a) 29.3 per cent; (b) O.415n yr. 12. (a and b) 2.41 min; (c) 27.3°F. (a) 2.36; (b) 2.57. 14. 10.3. 16. 2a/(n 1). 16. 17,900 ft.
+
8.85Ib/in2• 18. 15,800 ft. 19. 344 X 107Ib/in3• 20. 1.12 in. 3.45 mi. 23. 64.9 lb/ft"; 680,000 Ib/ft2• 24. 3.11 X 108 cal/daYi 63.4°C. 26. 117 cal/sec; 69.1°C. 26. (a) U = UI  (UI  U2) log (r/n)/log (r2/rl); (b) 86.4°C. 27. (a) U = UI [(UI  U2)(r  rl)/(r2  rl)](r2/r); (b) 76°C. 28. 5.47 em. 29. klk~(Ul  u2)/(k2h kl~) cal/sec, 30. 2rL(Ui)  Un)/'I,~= 1 (Iller) log (Xr/Xrl) cal/sec. 32. 10.2 min. 33. 3.93 min. 34. 2.28t. 36. (a) 45.3 min; (b) 54.3 min. 36. (a) 19.2 min; (b) 1.31 ft. 38. t. 39. 56.9 hr. Art. 18
+
2. :c = y3 + Cy. 3. y = x(log x + C). 4. y = x tan (x + C). 6. y2 = xe'U + C. 6. x 2  y2 = Cx. 7. xe~/'U = C. 8. 3x2 + 2y2 = Cx4• 9. log (y/x) = 1 + Cz: 10. (x + y)e~+'U = C(x  y). 11. x 2  2xy  y2 • C. 12. 6X 2y2 + y4 = C. 13. x 2 + 2y = Cy2. 14. y3 = Cx2 + x 3• 16. x 2 = Cy3esJn 3:. ~~"y = C[l + log (x/y)]. 17. x 2 = C sin (y/x). 18. (x2 + y2)2 = Cxy. 19. 7x 2 + 6xy + 11y2 + lOx + 34y = C. 20. x3 + y3 = xy(x + y + 0). 21. :c + y = Ce~/1/. 22. y2 + yVx 2 + y2 = Cx. 23. xy +Vx2  1 V y2  1 = cosh (xy + 0). 24. V:c  y(V; = Cevu/("'iV"iJ). 26. y + V x 2 + y2 = Cx2e"1/:£2+112/11.
vY)
Art. 21 1. y = (x C)e~. 2. 15x3y = 3x 5 5x3 + C. 3. y = (2/x) cV;. 4. x 2y = 3 C~. 6. 3y = 1 x 2 C(l x2)l/2. 6. uv' 1  x 2 = sinl X C. 7. 7x 2y4 = .CvY. 8. xy = xI + Ce~. 9. 9y = 6x2 log :D  4x2 cV;. 10. y = Cx  x 2• 11. 2y2 = 2x  1 Ce2~. 12. xy3(x + 0) = 1. 13. y = e" C(x  1). 14. (y  2x 4)4 = C(y  3x 6)3. 16. x y 2 = Ce". 16. 2(x  y) log (2x 2y 1) = C. 17. (x  1)2 + y2 = Ce2 tan  1 11l/ (Z  l)). 18. i 3 = sin y cos Y..+ ce:». 19. 3y4 = X2(y6 C). 20. (x 3y  5)2 = C(x 2y  2). 2 21. y = x (1 Cfil/~). 22. l/y = sin x C cosx. 23. 2x y2 + 1 = Ce,}. 24. (x + y + 1)e(~+2) J(~+1/+1) = C. 26. y2 = Ce~/'U. 26. xel/~'U = C. 27. x 4 = Cy y2. 28. 3y2 = 1 CX2y2. 29. ye:rt/2,} = C. 30. xy = C(xy 1)e,}/2. 31. y2 = (x  2)2 fil(~/2). 32. 4. 33. y =  sin x. 34. y = 2 sin 2 x  2 sin x; 4, i.
+ +
+
+ + +
+
+ +
+
+
+ + +
+ + +
+
+
+
+
+
+ + + +
+
+
+
+
233
Answers
36. xyln = 3(1  n)/(2n  3) + CX S 2n (n ¢ 1, !); y = Cx2e S / z (n = 1); 2x = (C  3 log x)vY (n = i). 36. 3y = 5x  2 tan1 [! tan (2x + C)]. 37. 2(x + 4y + 1) = 3 tan (6x + C). 38. x log (yIx) = sin x  x cos x + C. 39. (a) x 2(x2y2 + 2) = Cy; (b) y = Cx~y2/4. 40. (a) x = 1 + Ce(4z4) f(Z4 y  2) ; (b) 6(y  x) + C = log (2x + 2y + 1)(x2 + 2xy + y2 + X + y + 1)4. 41. (a) y = x 2 tan [(x2/2) + C]; (b) y (sin x + C) = sec x + C tan x. 42. Cl = C2 + r/3. 43. Cl = C2  r/3.
Art. 24 1. 1.32 lb/gal. 2. (a) 45 lb; (b) 39/64 lb/gal. 3. (a) 124 lb; (b) 54 min. 4. (a) 4 hr 25 min; (b) 2 hr 27 min. 6. 2871b; 14.9 min. 6. crV(l  r)(lr)fr lb; V(l  r)(lr)fr gal. 7. (a) 0.416 amp; (b) 0.413 amp; (c) 0.432. 8. (a) 0.476 amp; (b) 0.299 amp. 9. (a) q = EC KE t / R C ; (b) q = [EC/(! w2R2C2)](sin wt  wRC cos wt) KE t/ R C • 10. i = [wECI(l w2R2C2)J(COS wt wRC sin wt) KE t/ RC • 11. (a) 1.35 amp, 0.0432 coulomb; (b) 8.50 amp, 0.0131 coulomb. 12. 1.35 amp; 0.00677 coulomb. 13. 0.181 amp. 16. L = [2aILI(1 1L2)J(1 f!''Ir). 16. (a) 21L = (1  1L 2) f!''Ir/ 2; (b) 0.732. 17. 2.01 lb. 18. 0.630 ft. 19. 8.38 ft. 20. 19.6 lb; 2.22. 21. 730 4 4 ' from the top of the cylinder. 22. 1  101' 1L2 = 5CIL  (1  1L2) ( 0/2)Je 2I£'Ir/ 3. 23. 41 0 4 4 ' . 24. 5.85 ft; 0.13 ft. 26. 13,900.
.
+
+ +
+
+
+ +
+
+
Art. 26
+ y2 = 13. 2. y2 = 4e~. 3. x = 1 + '\1'3  e2 4. 2x + y2 = 6. 6. x 2 + y2 = Cx, 6. y2 = e2 7. x 2 + y2 = Cx, 8. (a) x 2 + y2 = 5x; (b) x 2 + y2 = lOy. 9. (a) p = 4 cos B; (b) p = (40/3) sin B. 10. y2 = 4(x + 1) + 5eZ. 11. P, 6.93 in.; Q, 13.2 in. 12. 0.697 ft. 13. (b) y = k cosh [(xlk) + C]; area under arc = k times arc length. 14. y = 2 cosh [(xI2) + 0.881]. 1. y2  x 2 = 5 or x 2
1I•
 (z / y) .
16. y = be(za)/k. 16. nynl = (n  l)(x  a). 17. x  y = Cxy. 19. y = 2n cosh [(xI2n) C] or y = 2n. 20. yS  3x 2y = k 3 or y = k. 21. Confocal parabolas: (a) x 2 = 2Cy + C2; (b) p(l  sin B) = C. 22. p = C sin" (Bin), or p sin" (Bin) = C. 23. p2 = C sin 2B, or p = C(l  sin B). 24. (a) y2 = 2Cx + 02; (b) p(l  cos B) = C. 26. (a) (lc 2  1)x2 + k2y2 = 2Cx C2; (b) p = CI(k  cos B); (c) p = 3/(3  2 cos B).
+
+
234
Answers
Art. 26 1. y = Cl:r:&/k. 2. p = CeT fJ/ k. 3. x 2 + y2  1 = ey. 4. 3x2 + y2 = Cx. 6. eZ sin y = C. 6. sin x = C sech y. 7. y2 = 4C(x + C). 8. y2  x 2 = Cx3• 9. x2 + y2 = Ce", 10. x + cos x cosh y = O. 11. 2 sinh y tan (x/2) = C. 12. y2 = i log cos 2x + C. 13. p4 = C cos 2(J. 14. p = 1/(C  sin2 8). 16. p = 2 sin (J + C cos 8. 16. p2  1 = Cp sec (J. 17. 2x2 y2 = 17. 18. xS 3xy2 = 13. 19. p = 64 COS1 6«(J/ 4). 20. p = 4(1  sin 8). 21. v!2 (sin x + sinh y) = cosh y; sinh y = sin x  cos x. 22. ya = C[(a  2)x 2  y2], (a ~ 2); ye:rt/1l' = C, (a = 2). 23. cosh x  a cos y = C sin y. 24. (a) x2 + y2 = 2b log x OJ (b) y = Cz", 26. (a) log p = 1 cos 2(8 + a) + C; 2 (b) (k/p)  (p/k) = (J + C. 26. p(sin n(J)m/n = C. 28. x 2 + y2 1 = Cx. 29. p2 = C(p cos 8  1). 2 30. (x + y2 + 1)2  4%2 = C.
+
+
+
+
+
Art. 29(b)
+
2. (a) 6ez; (b) 4e2:l:; (c) 0; (d) (log a  a)naz• 4. y = C1e kz C26kz; B cosh kx. 5. y = CleGz C2ez • y = A sinh kx 6. y = C1e(v'i4S)z + C26(v'i4+S)z. 7. y = (C1 + C2x)e 2z + (Cs + C4x)e2z. 8. y = C1 + C"e(a/2)z Cse2az. 9. 'II = C1e z + C26z + Cse(1/2)z. 10. y = Ae(3!2)Z sinh [(V'5/2)x a] Be(3!2)Z sinh [(V'5/2)x Pl. 11. y = C1 C2X Cp;e2 + C4ev'2z + Cr,ev'2z. 12. y = (C1 + C2X Cp;e2)e z + C4esz. 13. y = C1e(a/b)z + C26(b/a)z. 14. y = C1ez + C2e2z Cse(2/6)z + C4e(1/2)z. 15. y = C1ez + C2e(l/2)z Cse( 6/6)z + C4e(1/S)z. 16. Y]~=l = (1  e)(2  e)/(3  e); 4.37.
+
+
+
+
+ +
+
+ +
+
+
Art. 29(c)
1. y = A cos kx + B sin kx. 2. y = eZ(A cos V3x + B sin V3x). 3. y = ez/2[A cos (x/3) + B sin (x/3)]. 4. 8 = e2t(A cos 3t + B sin 3t). 5. x = eat (A cos Vk 2  a2t + B sin Vk 2  a2t). 6. y = 01 + C26~ + ez/2(Cs cos v!2x + C4 sin v!2x). 7. y = A sinh (x + a) + Be z / 2 sin [(V3/2)x+PJ+Cez/ 2sin[(V3/2)x+"Y] 8. y = C1e6z + eSz[C2 cos (x/2) + Cs sin (x/2)]. 9. y = (C1 + C2X)e z + (Cs + C4X)e Z + Cr, cos V 2x + C6 sin V2x. 10. 8 = A sinh (2'V2t + a) + B sin (2.y2t + P). 11. 8 = Ae 2t sin (2t + a) + Be:" sin (2t + P). 12. y = C1e 2z + C2e2z + (Cs + C4X) cos .y2x + (C6 + CaX) sin y'2x.
235
Answers
13. y = eCl + C2x)eZ + eZ(Ca cos x + C4 sin z), 14. y = (Cl + CttX + Caa: 2) cos 2x + (C4 + Cr;t + C6X 2 ) sin 2x. 16. y = Cle 2z + C2 cos X + Ca sin x + eZ(C4 cos 2x + Co sin 2x). 16. y = Cle b + (C2 + CsX) cos x + (C4 + CoX) sin x. 17. (a) No; (b) yes. Art.30(b)
2. (a) 9x2 + 4v2 = 225; (b) 3 ft, 6 ft/sec; (e) 3.76 sec. 3. 2.63 units. 4. (a) 0040 in.; (b) 13.0 in.zsec downward. 6. (a) 1.77 sec; (b) 2.36 ft above water. 6. 6.67 in. 7. 0.636. 8. 0.783 sec; x = i cos 8.02t. 9. 3.44 ft. 10. (a) 42.2 min; (b) 14.1 min. 11. 2rVLJg; (J = €X cos V g/ Lt. 12. (J = VLJgwo sin v'"iT£t. 13. (a) 3.26 ft; (b) 0.219 rad/sec. 14. i. 1. (a) 0.300a; (b) 0.375a.
Art. 30(') 3. 4. 7. 10. 12. 14. 16. 17.
20.
t = [l/(rl  T2)] log [(r2vO  b2xo)/(rlvO  b2xo)]. t = [log (r2!rl)]/(rl  r2). 6. (D2 + 0.0353D + 21.7)x = O.
e4.68t ; (D2 + 9.37D + 39.5)x = O. 8. 1.56 ft/sec 2. 9. 0.198 sec. K = 1; 0.172 sec. 11. (a) 0.549 sec; (b) 2.16 ft; 0.0895 ft/sec. (a) 0.206 sec; (b) 3.12 ft; 1.74 ft/sec. 13. 1.63 ft; 0.369 ft/sec. (a) 0.223 sec; (b) 4.09 ft; 8.86 ft/sec. 16. (a) 0.154 ft; (b) 0.315 sec. (a) 0.259 in. above equilibrium; (b) 4.94 in. below equilibrium. (a) 1.16 sec; (b) 60.5 per cent; (e) 8.35 in. 18. 0.565 ft above. 19. 40.2'. 93.5 per cent. 21. T = v' ap(47r 2r2 + log2n)/2rV3g(p  1); 1.15 sec. Art. 32
1. (a) Ely = 20x(x2  300); (b) Ely = 20(20  x)(100  40x + x 2); (e) (4 X 104)/EI ft. 2. 102,500/EI ft. 3. (a) Ely = ire qx 4  4Wx3 + (4qL3 + 12WL2)x]; (b) ~W lb. 4. (5PL3 + 6qL4)/48EI ft. 6. (a) wL4/192EI ft; (b) 54.16wL4/104EI ft. 6. 4.22 ft. 7. 1 ft from each end. 8. kwL6/6Eh3 ft.
9. (a) Ely = (q/48)x2(L  x)(3L  2x). 10. (a) (6EIL/W)y = (L  e)x(x 2  2Lc e2) and e(L  x)(e2  2Lx
+
respectively; (b) (eW/3EIL)[(L2  e2)/3]s/2 ft.
+ x 2)
Art. 34
y = (CI + C2x)eZ + e2z. 2.' y = C1e z + C2ez  x 2  2x  2. y = Cle6z + C2ez  ie3z  ie2Z• y = Cle 2z + (02 + 2x)e3z  !ue 2X. y = C: + C2x + Cse z + I\X 4  lxs + ix2• 6. y = Cse" + e(S/8)x[C2 sin (y23/8)x + C« cos (v"23/8)x] + Xl  x.
1. 3. 4. 6.
236
Answers
7. y = (01 + 2x)ez + 0~2z + ix2 + ix + i.
8. y = 01e4z + 02e 3z + lItre9z + !ez  h. 9. y = [01 + C2X  (x2/6)]ez + 03e2z  1ez + 2. 10. y = Ole(l+vI2)Z + 02e(lv2)z + 1(cos x  sin x). 11. y = [01 + (x/4)] sin 2x + O2 cos 2x + f sin x. 12. y = 01 + 02ez + Oaez  2 sin x + cos x  3x2• 13. y = [01 + (x/7)]e z/2 + 02e 3z  !e z/2  i. 14. y = 01 + 0~z/2  4x2 + 20x + ez12  2 cos 2x + 1 sin 2x. 16. x = (01 + t) sin t + (02  t2) cos t. 16. y = Ole2z + (0 2  Ix  Ix 2  x3)e2z• 17. y = [01  (x/4) + (x2/8)]ez + (0 2 + OaX + ! sin x + ! cos x)ez , 18. y = 01 + 02X + OaX2  120x3  30x4  6x5  x 6 + 04ez + (20/9)e3z• 19. y = eZ(Ol sin x + 02 cos X + tx sin x) + ieZ(cos x  sin x). 20. y = 01 + 02X + e(4/3)z[03 sin (0/3)x + 0 4 cos (y2/3)x] + (x3  6x2 + 12x)ez• 21. y = 01e4t + 02e2t + (Iv cos 2t  h sin 2t]e t • 22. y = [01 + l(sin x + cos x)]ez + eZ/ 2[02 sin (0/2)x + 0 3 cos (0/2)x]. 23. y = 01ez + [02 + OaX  (x2/2)  (x3/2)]e2z  x. 24. y = Aez + B sin (2x + a)  sin (1r/6)]  1. 2z Z 26. y = e (Cl sin 20x + O2 cos 20x) + e sin 20x. 26. y = (01 + 02X + x3)ez + (0 3 + x)e2z• 27. y = 01ez/ 2 + [02  (x/4)  (x2/ 8) + ! sin (x/2) + i cos (x/2)]ez / 2• 28. x = et  fe2 t + le3t• 29. 8y = x 2 + 2x + 4  2ez• 30. y = (x + 2)2 + (x 2  4)ez • 31. 0.0749.
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Art. 36
= = 3. y = 4. y = 6. y = 6. y = 1. y 2. y
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(01 + log sin x) sin x + (02  x) cos x. (01 + 1 log tan x) sin 2x + O2 cos 2x. [01 + (x/3)] sin (x/3) + [02 + log cos (x/3)] cos (x/3). [01 + log (sec x + tan x)] sin x + O2 cos X  2. OleZ + C~z ! ! sin 2 x. [(01 + x) sin x + (C2 + log cos x) cos x]ez/2 + iez/2•
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y = Olez tf2/2. y = {Ol sin x [02 log (sec x tan x)] cos x}ez y = [01 (x2/4)] sin x [C2 (x/4)] cos x. z y = Ole(02 2 sin x 3 cos x)e 2Z Cae3z• y = 01e z (02  sin eZ)e 2z.
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Art. 37 1. 1.70 ft. 2. (a) x = 1 sin V3Yt  (V3Y/2)t cos V3Yt; (b) 1.57 ft above equilibrium position. 3. (a) 0.138 ft above equilibrium position; (b) ±0.866 ft. 4. (a) 0.171 ft above equilibrium position; (b) 0.792 sec.
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6. 1.06 ft below equilibrium position. 6. 2.19 ft below equilibrium position. 7. 0.095 ft below equilibrium position. 9. 0.272 amp. 10. (a) 0.638 amp; (b) i amp. 11. (a) 0.909 amp; (b) 1 amp. 12. (a) 0.0868 amp; (b) 0.16 and 0.09 amp. 13. 1.22 amp; 0.00245 coulomb. 14. 0.0323 amp. 16. 0.546 amp. 16. 0.0354 coulomb; 1.03 amp. 1'1. x = Xo cosh wt + [(vo/w)  (g/2( 2)] sinh wt + (g/2( 2) sin wt; Xo = 0, Vo = g/2w. 18. (b) 1.10 in. 19. (b) 1.41 in.
Art. 38 3. y = C1x + C2Xl +va + Carva + 2x3. 4. y = X(C1 sin log x + C2 cos log x + log x).
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6. y = C1X C2X 2 4x2log x itr sin log x h cos log x. S 6. y = (C1 log" X  ! log x)X (C2/X)  2. 7. y = (C1 C2 sin log x C3 cos log x)(l/x) (x2/30) (3x/l0)  2. 8. y = C1 C2X CaX(5+3V5) 12 C4X(53V6) /2 log x To log2 X. 9. y = (C1 2 log x) (l/x) (C2 sin log x Cs cos log x)x. 10. y = C1 C2 log x [C3 sin log x) C4 cos log x)]X S i log" X h log2 x. 11. y = [C1 C2log x C3log2 X (log" x)/48]x. 12. y = (C1 C2log x i log2 x)0  /r;x3. 13. y = C1X C2X2+V4.25 + CaX2V4.25  fix log x + log" X 29 log x 475. 14. y = (Cl/X) (C2/X2) (x/6)  (sin x)/x2• 16. y = [C1 C2log (2x  3)]V2x  3 2x. . 16. y = C1 C2 log (1 3x) C3(l 3x)3  i log2 (1 3x)  3x/2. 1'1. y = C1(2x  1) C2(2x  1)1+(V3/2) C3(2x  1)1(...;3/2) !(2x  1) log (2x  1)  1. 18. y = ¥  3x ¥X2 lx2log x. 19. U = U1  [CUI  U2) log (r/r1)]/log (r2/r1). 20. U = U1  [(U1  U2)(r  r1)/(r2  r1)](r2/r).
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Art. 40
1. 3. 6. 7. 10. 14. 17. 20. 26.
y = log (x2 + C1) + C2 • 2. Y = x + C1e:r?/2 + C2. x 2 + (y + C2) 2 = C1. 4. Y = x 2 + C1log (x2  C1) + C2• y = x 3 + C1x2log x + C2X2 + CaX + C4 • 6. Y = (C1 + x)/(C2  x). y = (C1X + C2)2. 8. y3 = (C1x + C2)2. 9. y = C1 sinh" (x + C2). xy + C1X + C2y = Cs. 12. y = log cos (2x + C1) + C2. 2y = sec [x + (r/3)]. 16. x = 3  (2/VY). 16. y = 2 sin1 X + 1. x = 2(eY/2  1). 18. y = (x  2)e~ + 2. 19. x = jlog 2  log sin y. sin y = V2 sin x. 21. 1.73. 22. 27. 23. 1.42. 24. t. 26. 3.21. 1.83; 0.330.
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Art. 43
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1. y = ClX  2x2  x 3 C2Xe:e. 2. y = C1(2x l)e:e C2e:e. 3. y = C1(X2 2x 2) [(2x3/3) C2]e:e. 4. y = [Ci/(x 1)] C2X. 6. y = (C1log x C2)e:e e:e/2. 6. xy = C1 C2e:e  sin x  cos x. 7. 0.843. 8. 0.51 ft; 0.49 ft. 9. 2.39 sec. 10. 116 hr. 11. 9.90 hr. 12. 6.95 mi/seo. 13. 11.6 min. 14. 45.1 min. 16. (a) 15.9 ft; (b) 61.1 ft; (c) 136 lb, 144 lb. 16. 58.4 ft. 18. (a) catenaries; (b) circles; (c) parabolas; (tl) cycloids. 20. y = log (cos x  sin x); 0,1,0. 21. y = log (cos x sin x). 22. The family of catenaries: y  b = k cosh [(x  a)/k]. 23. 1.38; 3.63. 24. y = ±[(x2  1)/4 llogx]. 26. y = Cl(Sin x)/VX C2(cos x)/0. 26. (7) 0.775; (10) 810 hr; (11) 13.5 hr; (12) 4.91 mi/seo. 28. 0.910. 29. 2.36 sec. 31. (a) 1.02 sec; (b) 0.723 sec. 32. t = VL/g(1  sin 8) cosh"! (L/a). 34. (a) x = [a kyl+k/2(1 k)]  [a kylk/2(1  k)] ak/(1  k2) ; (b) x = 1[(y2/2a)  (a/2) a log (a/y)].
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Art. 44
+ X)2 + (z  3x)2 = 1; (b) (3y + Z)2 = 3x  z. 2. y = x log x + 2x, z = x log x + 3; [1, 3, 1]. 3. y = 0 cos [x 1 + (71'/4)], z = 0 sin [x 1 + (71'/4)]. 2 2 4. C1e:e + ce:: and C1e:e  C~:e. 6. x = t + Cl, Y = C2(X + t). 6. (1 + y)2 = (1 + X)2 + Cl, Z = C2(x + y + 2). 7. y = C1e:e + C2e:e, Z = C1e:e  C~:e + sin x. 1. (a) (y
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8. (a) x ;:= eeat , y = [ac/(a  b)](ebt  eat ) , Z = e + [e/(a  b)](beat  aebt ) ; (b) x = ce at , y = acteat , Z = e[1  (1 + at)e at ] . 9. 161 min, 4.51 lb. 10. 18.3 min. 11. 9.43 lb. Art. 46
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1. y = C1e2:e C2e2:e l sin 2x  1 cos 2x, Z = C1e2:e  3C2e2:e  i cos 2x. 2. y = C1eb C2e:e  2 cos x  sin x, 3:e Z = 2Cle 2C2e:e  cos x 8 sin x. t(C1 3. x = esin t C2 cos t) (t/2)  1, t[(C2 y = e C1) sin t  (C2 C1) cos t] 1. 4. x = C1 sin t C2 cos t sin 2t, y = C1 cos t  C2 sin t cos 2t. 6. y = C1 sin x C2 cos X e:e lx sin x, Z = [C2  C1 l (x/2)] sin x  [C2 C1  (x/2)] cos x  e", 6. y = C1 sin 0 x C2 cos 0 x 5 sin x  cos x, Z = [(C1 20C2)/3] sin 0 x [(C 2  20C1)/3] cos 0 x sin x  3 cos x.
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7. y = 01e 2z + ! sin 2x  ! cos 2x, Z =  (Ol/2)e2:e + i cos 2x  t sin 2x. 8. x = 01 sin t + O2 cos t + et + (t/2) sin t; y = [01  O2  t + (t/2)] sin t + [01 + O2 + (t/2)] cos t + 2et• 9. y = (01 + 02x)eZ + (Os  x)eZ, Z = 1  02ez  (20 3 + 3  2x)ez• 10. y = (01 + 02X)eZ + Osez + 3e2z  1, Z =  02ez  20se z  8e2z. 11. x = C1 + 02t2 t, y = 01  02t2 + t. 12. x = [01  02 + (01/2) log t]t  (Os/t) + log t + 1, y = [02  (01/2) log t]t + (Os/t)  2 log t  2. 13. ~ = 0: y = 1  2ez , Z = eZ + xI. n = 1: y = x  eZ, Z = Oe z + !eZ + x  2. n = i: Z = 01e sz + 02e(3/2)z + /rrez + x  t, y = t01e3z  102e(3/2)z  fez + (x/3) + i. 14. x = t sin t  if: sin 3t, y = 1.,p sin t + A sin 3t; 0.955, 1.96. 16. x = A sin (t + a) + B sin (9t + (j) + (at/9) + (h/9) + (8b/81), y = A cos (t + a) + B cos (9t + po) + (bt/9) + (k/9)  (&/81). 16. y = 01 + 02e t, x = 01 + (02t + ,03)et, Z = 01 + (Ott 02 + 03)et•
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Art. 46 1. (a) y =  [gx2/(2v~ cos2 a)] + x tan ai (b) (v~ sin2 a)/2g ft, (v~ sin 2a)/g ft. 2. (a) [(wvo sin a)/kg]  (w2/ k 2g) log [(kvo sin a w)/w] ft. 3. (a) 1.24, 1.33; (b)  i < x < i, 0 < Y < 0. 4. x = cos (t/y'2) cosh (t/0), y = sin (t/y'2) sinh (t/y'2); (3.83, 3.70). 6. (b) 2.37 amp, 0.061 sec; (c) 2.15 amp, 2.85 amp. 6. (c) R = [(I/ro)  (k/h 2)] cos 8 + (k/h 2).
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Art. 49 3. x = Ae tlt + Betltl2 sin [( V3/2) at + a], y = Ae tlt + Betltl2 sin [( y'3/2) at + a + (21r/3)], Z = Ae tlt + Betltl2 sin [(V3/2)at + a + (41r/3)]. 4. x = Ae3 t + B, y = Ae St + 0, Z = Ae St  (B + 0). 6. x = Ae t/2 + Be t, y = Aet/ 2 + Ce:', Z = Ae tl2  (B 6. x = Ae t + Betl2  t/2 !, y = Aet + Oetl2  t/2 Z = Ae t  (B + C)e t / 2  t/2  !. 7. x = At2 BtI , y = At2 + Otl, Z = At2  (B + C)t1• 8. Xl = Acz + Bttl + K sin (a log t + a), X2 = Acz  Bttl + K cos (a log t + a), Xs = Acz + Bt:»  K sin (a log t + a), X4 = Attl  Bttl  K cos (a log t + a). 12. Ji' = (1 + wG)/(1 + w2G).
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Art. 60
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1. y = ao(l x) CZ2X2 • 2. y = ao(l  x) lZ2(X2  x3 x4  ••• ) = A(l  x) B/(l x). 2 3 li 3. y = ao(l  2x ) aleX  ix  (1/2·4)x  (1·3/2·4·6)x7  ••• ] . = ao(l  2x2) alxVl  x2• 2 2x4/2·4) 3x 6/2·4·6) 4. y = ao[l  (4x / 2)  (1·4  (l·3·4 (l·3·S·44x8/2·4·6·S) _ ... ] alX = aoVl  4x2 alX. 6. y = ao(l x 2) a4[x4 (x6/3) (x 8/ 3 ·4) (xl°/3·4·S) 2 = A(l x ) B~. 6. y = ao[l  (x4/21) (xSj4f)  ••. ] lZ2[x2  (x 6j3f) (xlOjSI)  ... ] 2 2 = ao cos x lZ2 sin x • 2/2) 7. y = ao[l (x + (x4/2·4) (x6/2·4·6) +
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