Each particular form assigned to ƒ gives a distinct par

Y

Y

ticular integral. If we assume f X =a+b, we find

x=Y+b, y=X+a, z=XY,

from which, eliminating X and Y, we have z = (x-b) (y — a), and this is one form of the complete primitive assigned in Chap. XIV. Art. 7. We may observe that the elimination may be so effected as to lead to general primitives.

11. In equations of the second order we should have, in addition to the above transformations, to change

in order to form the reciprocal equation. Then the second integral of either being found in the form z = ↓ (x, y), that of the other will be found as before by eliminating X and Y from (49). For since

therefore dx = RdX+ SdY, dy = SdX+TdY,

The extension of the theorem to higher orders involves no difficulty.

12. It is an immediate consequence of the above, that any equation of the form

$ (p, q) r + † (p, q) s + x (p, q) t = 0.......................(52)

can be reduced to an equation of the form

x (x, y) r — f (x, y) s + $ (x, y) t = 0.........(53), usually more convenient for solution. Legendre's solution of the equation

(1 + q2) r − 2pqs + (1 + p2) t = 0,

by the aid of the above transformation, will be found in Lacroix (Tom. II. p. 623).

The same transformation makes the solution of any equation of the form Rr + Ss + Tt = V (rt — s2) dependent on that of an equation of the form

Rr+ Ss+ Tt= V,

but with different coefficients. The subject of these transformations has been most fully treated by Prof. De Morgan (Cambridge Philosophical Transactions, Vol. VIII. p. 606).

13. Legendre also shews how, by a transformation formally resembling the above, to integrate the equation

r = f (s, t).

Assuming s and t as independent variables, and v = sx+ty − q as dependent variable, the equation is reduced to the form

dr dr

where S and T are the values of and furnished by the

ds dt

given equation. Lacroix, Tom. II. p. 631.

EXERCISES.

1. To what condition must u and v be subject, in order that uf(v) may be a first integral of an equation of the form Rr+ Ss + Tt = V?

Integrate by Monge's method the following equations:

2. x2r + 2xys + y2t = 0.

3. qr-2pqs+ p2t = 0.

4. Integrate ps — qr = 0.

5. Integrate by Monge's method the equation

q (1 + q) r − (p + q + 2pq) s +p (1+p) t = 0.

6. The solution of Ex. 3 may, by the law of reciprocity, be made to depend on that of Ex. 2.

7. Monge's method would not enable us to solve the

equation r - t =

2p

x

8. Deduce by Poisson's method a particular integral of (1 + q3) r — 2pqs + (1 + p2) t = 0.

9. Shew that the equations

rt-s2=f(p, q), and rt-s2= {ƒ (x, y)} ̄',

are connected by the law of reciprocity.

10. The solution of the equation r―t=

may be derived from that of the equation r t + Art. 7, Ex. 4.

CHAPTER XVI.

SYMBOLICAL METHODS.

1. THE term symbolical is, by a restriction of its wider meaning, applied more peculiarly to those methods in Analysis in which operations, separated by a mental abstraction from the subjects upon which they are performed, are expressed by symbols in whose laws the laws of the operations themselves are represented.

du

d dx

Thus is written symbolically in the form u, the sym

d

dx

bol denoting an operation of which u is the subject. In dx thus expressing an operation by a symbol, in studying the laws of that symbol, and in founding processes and methods upon those laws, we introduce no strange or novel principle of Language; for it is the very office of Language to express by symbols the procedure of Thought.

and so on.

dǝ

It will be observed that the symbol precedes the

subject on which it operates.

Operations may be performed in succession. Thus

denotes that we first perform on the subject u the operation

+b, and then on the result effect the operation

+

dx

a. Thus a and b being constant, we have

d

(a + a) (a + b) u = (a + a) (d+bu)

dx

dx

dx

When an operation is repeated, the number of times which it is understood to be performed is expressed by an index attached to the symbol of operation. Thus

If in the second member of (3), as in the first, we separate the symbols from their subject, we have

d

d dx

d

(~ + a) (a + b) u = { 1 x + (a + b) a + ab } u...(.5).

dx

dx2

dx

Now the symbolic expressions for the equivalent operations performed upon u in the two members of this equation are in formal analogy with the algebraic equation

(m + a) (m + b) u = {m2 + (a+b) m + ab} u,

and this is a particular illustration of a general theorem to the statement and demonstration of which we shall now proceed.