(8) This transformation fails when m + n = 0 while y is not equal to 0. In this case the following method may The equation may evidently be put under the form

be used.

then considering a as a function of ≈ and y,

The transformations in the three preceding examples are given by a writer who sigus himself "G. C." in the Cambridge Mathematical Journal, Vol. 1. p. 162.

SECT. 4. Equations integrable by various methods.

Lagrange's Method.

Let a partial Differential Equation between three variables be of the form

where P, Q, R are functions of x, y and x; then if we can integrate two of the following equations,

so as to obtain two integrals,

$ (x, y, x) = ß, † (x, y, z)

the integral of the given equation will be

B = f (a).

= α,

Lagrange, Mémoires de Berlin, 1774, p. 197; 1779, p. 152.

For the success of this method it is necessary either that one of the three auxiliary equations should contain only the two variables the differentials of which it involves, or that by their combination such an equation should be obtained. By integrating it we obtain an equation by means of which one of the variables may be eliminated from either of the other auxiliary equations.

The last of these alone is immediately integrable and gives x2 + y2 = a2.

(x − a x) dz + (bx ay) dy = 0 ....................... (3).

Multiply (2) by a, and

by br

(3)

.........

by b; subtract and divide

dz + adx + bdy = 0, or ≈ + ax + by

= α.

Again multiply (2) by x, and (3) by y; subtract and divide by ba ay: there results

xdx + ydy + z dz

Therefore

=

0, whence a2 + y2 + x2 = ß.

x22 + y2 + ≈2 = ƒ (≈ + ax + by)

is the integral of the proposed equation. equation to surfaces of revolution.

This is the general

Equation (1) may be put under the form

wdy - yd + x d +ydy = 0 ;

Multiplying (2) by x, (3) by y and adding, we have

U = a, V = b, W = c,

the integral of the proposed equation is

U = ƒ (V, W),

or 4 (U, V, W) = 0.

(u + y + z) du − (x + y + z) dx = 0.

Adding these three equations we have

(u + y + z) (du + dx + dy + d≈) − 3 (u + x + y + z) d x

Putting u + x + y + ≈ = v, this gives

= 0.

Subtracting the second equation from the first, we have

ƒ{v (u - z), v (x − ≈)3, v (y − ≈)3} = 0.