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= a?

By means of these eliminating p and q, we find

* + y + xứ:
as the singular solutìon.
(22) Let the equation be

(px - qy)' q + 4 m 03 (x – px) = 0.
dU

=9 (px - qy) - 2m x3 = 0,
dp
dU

= (px - qy) (px - 3qy) = 0.

dq These two equations agree with the original one, if we

assume

px - 399 = 0; and the singular solution found by eliminating p and q is

2 - 4mry = 0. Legendre, Mémoires de l'Académie, 1790, p. 238. (23) Let

(č px qy)" = Apoq.
dU

= - mx (: - px - qy)"-1 - a Apa-'q' = 0,
dp
dU

= - my (x px - qy)"-1 - 6 A pq-1=0.
dq
Dividing the first of these by the second we find

aa

bp y Dividing the original equation by the first we have

px - 94 р

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By means of these values, eliminating p and q, we find the singular solution to be

arber x"y" ze = (-)"+b A.

m"

If the partial differential equation be of the second order and we put

d’z
d?
d x2 dx dy dy

1',

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the conditions which must be satisfied in order that an equation should be the singular solution of the first order of the equation U = 0 are

DU
dU

DU
0,

: 0.
dr.
ds

dt

= 0,

If the function is to be a singular solution belonging to the final integral, it must in addition satisfy the equations

dU

dU 0,

da

= 0,

dp

(24)

Let the given equation be

mo - 29r (

(1- 11.) + (-)

y = 0.

Here

dU

= 0 is the only condition, and we have

dr

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is the singular solution required. The integral of this is

(1 + x) ° (y). Poisson, Jour. de l'Ecole Polyt. Cah. XIII. p. 113.

(25) Let the equation be

2 - 2 +

(

*) (* – – – y) = 0.

It will be found that

X = x + y

satisfies the original equation as well as
dU

U
dU

dU
: 0,
0,
0,

0;
dr
dt

dq it is therefore a singular solution corresponding to the final integral

dp

CHAPTER IX.

QUADRATURE OF AREAS AND SURFACES, RECTIFICATION

OF CURVES AND CUBATURE OF SOLIDS.

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When an area is referred to rectangular co-ordinates x and y, the double integral Ssd xdy taken between the proper limits gives the value of the area. One of the integrations may always be performed, so that we have either

Sydx + C or sædy + C, and these integrals are to be taken between the limits of y or x, which form the boundaries of the area. If we take the first of these expressions, the limiting values of y must cither be constants or functions of a given by the equation to the bounding curve: therefore on substituting these values we obtain a function of x alone, which is to be integrated, and taken between the limits of that variable which are required by the problem. If after the first integration we suppose C = 0, the integral A = Sydx expresses the area included between the axis of x, the curve, and two ordinates corresponding to the limits of x.

In taking the integral Sydä between the final limits of X', it is necessary that the interval should not contain a value of x which causes y to vanish or become infinite, as in that case we might be led to an erroneous conclusion. Thus if we suppose a curve to be symmetrically situate in the first and third quadrants, and to intersect the axis at the origin; and if we were to integrate from x = a to x =

a we should obtain zero as our result, instead of finding the area to be double of that from a = 0 to x = a, or that from x = 0 to

Therefore when any interval from a to b contains a value c of a which makes y vanish or become infinite, we must break it up into two intervals, one from b to c and the other from c to a, and add the integrals corresponding

W = a.

to these. In like manner if the interval contain several values of x which make y vanish or become infinite, we must split it up into as many smaller intervals, each having one of these values of x as a limit, and add them all together.

If the co-ordinates be not rectangular, and a be the angle between them, we must multiply the integral by sin a to obtain the value of the area.

Ex. (1) If we take the general equation to a parabola of any order

y*** = a" a",

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(2) The general equation to hyperbolas referred to their asymptotes is

01"y" a"+",

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This formula fails when m = n, in which case

A = C + a2 log æ.

(3)

When the common hyperbola is referred to its axes, the sectorial area ACP (fig. 53) is easily found. For ACP = NCP - ANP = { vy Sydä.

b Now

y = - (v – a),

a

b

and Sydv = . {v} (x® – a°)}} – ao log {x + (no – a°)}} + C.

Determining the constant by the condition that the area vanishes when x = a, we have

x2 Sy dx = kxy - ab

?

a

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