Page images
PDF
EPUB

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

d =

Y = ";

d then considering x as a function of and y,

(2)(C)

dy

dia
da
dr =
dat

dy
d z

da dy

dy dz =

1

and therefore

dix

[ocr errors]
[ocr errors]
[merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small]

By substituting these values the equation becomes (since m + n = 0)

d

= (-)"0"; \d

Y the integral of which is

X = - cyž "+ P(;').

'd
(9) Let

dy
The transformed equation is

[ocr errors]

2

2

VX =

[ocr errors]

y

[merged small][merged small][subsumed][merged small][ocr errors][ocr errors][subsumed][ocr errors][merged small]

The transformations in the three preceding examples are given by a writer who signs 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

[ocr errors][merged small][merged small][merged small][merged small]

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

Pdy Qdx = 0,
Pd: - Rdx = 0,

Qdx Rdy = 0, so as to obtain two integrals,

(x, y, x) = B, y(x, y, z) = a, the integral of the given equation will be

B = f (a). 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.

d: dx
+y

+ x = 0.
dx

dy
In this case the auxiliary equations are

ady - zdx = 0,
xdx + yd x = 0,

xdx + ydy = 0.
The last of these alone is immediately integrable and gives

x + y = a'.

(1) Let

[merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small]

sin

[ocr errors]

= f (y + 2), is the integral of the proposed equation.

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

ax) dx

= 0

= a.

The auxiliary cquations are,
(y bx) dy + (x

0....... (1)
(y bx) dx (bx - ay) dx = 0 ......... (2)

(x - ax) dx + (bx - ay) dy ...... (3). Multiply (2) by a, and (3) by b; subtract and divide by bx ay: we find

dx + adx + bdy = 0, or z + ax + by Again multiply (?) by X, and (3) by y; subtract and divide by bx - ay: there results

xd x + ydy + zdx = 0, whence 2? + y + = ß. Therefore

+ y + m = f (z + ax + by) is the integral of the proposed equation. This is the general equation to surfaces of revolution.

[merged small][merged small][ocr errors][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors][ocr errors]
[merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small]

......

The auxiliary equations are,

(x + y) dy - (y x) dx = 0 ............ (1)

(x + y) dz zdx = 0 .......... (2)

(y x) dx - xdy = 0 ....... (3). Equation (1) may be put under the form

ædy - ydx + xda + ydy = 0;

[ocr errors]

-1

= a.

whence tan - log (x2 + y)2

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

dx

« dx + ydy

w? + y

[ocr errors]
[ocr errors][merged small][merged small][ocr errors]
[ocr errors]

(x2 + y')}} {tan

log (x2 + y)}}

Y is the required integral.

d&

dx (7) Let (x 2y) + (2x – 3y)

dv The integral is

(*- y) = (x - y) This method may be extended to functions of more variables. Thus if

du du

du P

S, dy

dx and if from three equations such as

+Q

dx

+R

U = a,

W = C,

Pdy - Qdx = 0,
Pd: - Rdx = 0,

Pdu - Sdx = 0, we can obtain three integrals,

V = b,
the integral of the proposed equation is

U = f(V, W), or p (U, V, W) = 0.
(8) Let
du
du

du (1 + y + x) + (u + x +) + (u + x + y) = x + y +2. dx

dy
The auxiliary equations are,

(1 + y + x) dy - (u + x + x) dx = 0,
(u + y + x) dx - (u + x + y) dx = 0,

(u + y +) du (x + y + x) dx = 0. Adding these three equations we have (u + y + x) (du + dx + dy + dx) – 3 (u + x + y + x) dx = 0. Putting u + x + y + z = 1', this gives

dr du

dz

[ocr errors][merged small]

Subtracting the second equation from the first, we have
(u + y + x) (dy - dx) = (z y) dx;
dx

dy - da

or

[ocr errors][merged small][merged small][merged small][ocr errors]

and v (y - x) = a. From the symmetry of the expressions it is obvious that we must have also

v (0 - 2) = b, v (u - 2) = c. Therefore

f {v(u – 2), v (- ), v (y – )}} = 0.

« PreviousContinue »