Page images
PDF
EPUB

arbitrary constants being thus 2n + 4. But there are relations subsisting between the constants which reduce the number of independent constants to 2n. In the first place, the constant ß will only alter gi, , gh,, &c., and it may therefore be neglected, so that the number of arbitrary constants is reduced to 2n + 3.

Again, since

[ocr errors]

we have by squaring and adding equations (13)

cos' PE(go) + sin? PE (ho) + 2 sin q cos Q (gh).

1=

In order that this equation may subsist for all values of we must have the conditions

Σ(g) = 1, Σ (%) = 1, Σ(gh) = 0.

These three conditions reduce the number of arbitrary constants to 2n.

It is to be observed that the integrals for determining t and are not independent: for if we assume a function

[blocks in formation]

CHAPTER VIII.

SINGULAR SOLUTIONS OF DIFFERENTIAL EQUATIONS.

[ocr errors]

By a singular solution of a differential equation, is meant a certain relation between the variables which satisfies the differential equation, but does not satisfy the general integral. Solutions of this kind have long attracted the attention of mathematicians, and the memoirs in which they are discussed are very numerous. Their existence was first pointed out by Taylor, in his Methodus Incrementorum, p. 27, and afterwards they were noticed by Clairaut, in the Mémoires de l'Académie des Sciences for 1734. But Euler, in the Mémoires de l'Académie de Berlin for 1756, was the first who considered the subject in its bearing on the general Theory of Integration ; and in his Integral Calculus, Vol. 1. Sect. 2, Chap. IV., he gave a test for discovering whether a given solution be or be not included in the general integral. Lagrange, in the Mémoires de l'Académie de Berlin, and afterwards in his Théorie des Fonctions, and his Calcul des Fonctions, discussed the theory of these solutions, and shewed the connection between them and the general integral, and their relative geometric interpretations. Other points of the theory have been elucidated by Laplace (Mémoires de l'Académie des Sciences, 1772), Legendre (Ib. 1790), and Poisson, Journal de l'Ecole Polytechnique, Cahier xili.

Having given a differential equation, to find its singular solutions if it have any. Let

U = 0 be a differential equation of the first order between x and y

dy cleared of radicals and fractions, then if we represent

dx the relations between x and y found by eliminating p between

by P,

U = 0,

= 0.

where p =

= 0.*

2

[ocr errors]
[ocr errors]

dU
and 0,

dp are singular solutions of U = 0, provided they satisfy that

dU equation, and do not at the same time make

We

dy might also deduce the singular solutions from eliminating dv

between dy

dU
U = 0,
and

0,

dp. du

provided that they do not at the same time dy dU make

dx
Ex. (1) Let the equation be

dy dy
· y

+ m = 0.
dx dx

dU Here

= 2.rp - y = 0;

dp and eliminating p between this and the preceding equation, we find

y 4 m 2 0, as the singular solution.

dy ) + (y – x x +

a

= 0,

d.c be the given equatior. Then

(x + y) 4ay = 0) is the singular solution. (3) Lety – 2xy

=1 dx

dv be the given equation : the singular solution is

y = 1 + x2.

Laplace, Mémoires de l'Académic, 1772.

[ocr errors]

+ (1 + ) ()

[ocr errors]

וי

2

This is the equation with respect to which Taylor first made the remark that it admitted of a solution not involved in the general integral“ singularis quædam solutio,” as he terms it. See his Methodus Incrementorum, p. 27.

dy

dy (4) Letx? + 2xy + (ai? – w*)

= 0. dc The singular solution of this is

x? + y2 Q= 0.

dy Let

+ y

+ X = 0. \dx

dx)

dy

[ocr errors]

dx

dy

The equation resulting from the elimination of between this equation and

[merged small][ocr errors]
[blocks in formation]

is

y4x = 0; but as this does not satisfy the given equation it is not a singular solution.

dy (6) Let 1+

Y
da

dx
dU
In this case

= ( gives us

2

dP

a

[ocr errors]
[ocr errors][ocr errors]

2

(1 + )! Eliminating p by means of this equation we find as the singular solution

xi + y = .
dy

dy
(7) Let
Lety

+ ax + by = 0.
\dx
dU
Here = o gives us p

and the result of the elimination of p is

ax + by - 22

[ocr errors]

dp

[ocr errors]

0;

but as this does not satisfy the given equation it is no solution at all.

(8) Let the equation be

dy
(
yay-2y) + x = 0.

a
dx

dx

The singular solution is

ya – 4x3 = 0. When a solution of an equation is given, and it is required to find whether it be a singular solution or a particular

dy integral, we must deduce from it the value of

р

and

dx

dU see whether when substituted in it make it vanish, and

dp

dU do not at the same time make vanish. If this be the

dy case it is a singular solution, otherwise it is a particular integral. (9) Are y

Are y' = 2x + 1, and y + x2 = 0, singular solutions or particular integrals of the equation

dy + 2 x

do

2

dy dx

y = 0?

[blocks in formation]

Now from the first of the given solutions we find

1

р

y

dU which does not make vanish: it is therefore a particular

dp integra).

From the second of the given solutions we find

р

« PreviousContinue »