Page images
PDF
[merged small][ocr errors][merged small]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

(22 - y2 - a2)

" (0202 + y2 – a )' and each of these = oo, if x2 + y2 - a2 = 0; in which case c = y, and therefore the solution is singular.

The same singular solution may also be found by the methods previously investigated. The differential equation of which the given equation is the integral is

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

and as this relation satisfies the differential equation given above, it is a singular solution.

Also if we find the c-differential of the general integral, and then eliminate c according to the method of Art. 409, we have

c=y; ii ya +202 — a? = 0; and thus all the methods for finding singular solutions lead to the same result.

SECTION 8.Differential equations of the first order and of any

degree. 411.] Order of differential equation depends on the index of the symbol of differentiation with which the highest differential or differential coefficient is affected, and degree on the power to which such highest differential or differential coefficient is raised. Thus a differential expression of the first order and nth degree is that which involves (Gly)", but no higher derived-function, and

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

where F1, F2, ... F, are symbols for functions of x and y.

Let us suppose the equation (214) to be resolved into n factors of the form,

[ocr errors]

where f1, fa, ... fn are the roots of (214) and are generally functions of x and y: let each of these equations be integrated separately, and let their integrals be

01(x,y,c7) = 0, $2(x,y,cy) = 0, ... (x,y,cm) = 0, (216) where C1, C2, ... Cm are arbitrary constants. Then the equation

01 (X, Y, C) 02(x, y, cz) x ... x 0(X, Y, Cn) = 0 (217) will contain all the integrals of (214), because it and (214) vanish simultaneously for each of the n functions. And the truth of this final equation will not be affected if the arbitrary constants are equal, that is, if c = cq = ... = CH = c, because c is arbitrary, and therefore will pass through the values C1, C2, ... Cn, if it receives all the values of which it is capable.

The following are examples of this mode of integration.
Ex. 1. mg) – 4a? «= 0;

\dx'

[ocr errors][ocr errors]

Y-C = axz; y-C2 = aw. and these two solutions may be combined into the single equation

(y-ax? –) (y + axé — C) = 0; either of which factors satisfies the given differential equation; and if c = cg = 0, we have

g-c)^-ao c* = 0. Now this is equally true with the former equation, as the primitive from which the differential equation is derived; for it may

be derived either from this latter by the elimination of c, or from either of the former by the elimination of cy or Cg.

The singular solution of this equation is x=0; and considered geometrically the general integral represents two parabolas which have a common axis, viz. that of y, and a common vertex on the axis of y at a distance c from the origin; and the singular solution represents a point on the axis of y, through which all the parabolas pass, and which is consequently an envelope. Ex. 2. 43 )-(a+2bx +3 ex>) (

+(6bex® +3 aexo + 2abx) -6 abex = 0; dy = a; :: y = ax +C; han = 2bx ; y = bx2 +6z;

on = 3ex®; y = ex®+cz ; and the integral is

(y-ax-C)(y-622 - cz)(y-ex3 — Cz) = 0; and which may be simplified if c, = cg = = . In this case also the singular solution is a point on the axis of y.

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

.. x dy + y dx = + (y2 + x2) dx, which are homogeneous, and may be integrated by the methods explained above.

412.] Certain forms of differential equations of the first order and of any degree are capable of integration without resolution into factors according to the method of the preceding Article, and these we proceed to explain. The first form is xf (4') +(4') + F(y') = 0;

(218)

dy where y' according to Lagrange's notation denotes and f, ¢, and F are symbols of given functions.

Let this equation be divided by $(y); so that making obvious substitutions it becomes y = x (4')+f(y);

(219) and let us in the first place take the simple case in which $(y) = ý': so that y = xy +f(');

(220) this form is known by the name of Clairaut's equation. Let this be differentiated; then we have, resubstituting,

{o+8"(er 0. (221) Now this may be satisfied in two ways; (1) maig = 0; interesate = c;

(222) and substituting this in (220) we have

y = cx + f(c), which is the general integral, containing the arbitrary constant c.

We might of course integrate (222) immediately; whereby we have y = cx +CI

(223) where cy is a new arbitrary constant: but as (223) is to satisfy (220), C1 = f(c). This result is also manifest from the fact that (220) is a differential equation of the first order, and therefore its integral must contain only one arbitrary constant.

(2) x+f" (hy) = 0; i ele = $(x);

and substituting this value n the equation (220), an expression results which of course satisfies the differential equation, and is independent of c the arbitrary constant, and is therefore either a particular integral or a singular solution; and it is manifestly the latter, because c which is equal to Oy, is replaced by a function of x, viz. $(x).

The following are differential equations solved by this process, which generally is called Integration by means of Differentiation. Ex. 1. y = no en tu

Home dumne+ {2-4(e)an which is satisfied either by

(1) = 0; i due = c; .. y = ca+ and this is the general integral; or by : (2) = 0)*; : y = (ax)* + (ax)# ;

- y2 = 4ax ; and this is the singular solution, since it involves no arbitrary constant, and is not a particular integral, because the constant is replaced by (m)", which is a function of æ.

Ex. 2. y = voin - Care

[merged small][ocr errors][merged small][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors]
« PreviousContinue »