If the equations M+Nd ̧y=0, M+Pd ̧z=0,N+Pdx=0,(1) may either of them be derived from the same function u, then d1M=d2N, d ̧M=d ̧P, d ̧N=d ̧P;

and u1, u, u, being the integrals obtained from the equations (1) respectively,

u=u1+f(P-d ̧u1),
=u2+f,(N−d ̧u2),

=u2+ƒ;(M−du). (L. D. C. 282—6.)

(29.) Separation of the variables. All equations of the form X+ Yd,y=0,

in which X and Y are functions of x and y respectively, are immediately integrable; the integral is

S2 X+S,Y=0.

The equations Y+Xd ̧y=0, and X1Y+XY1d ̧y=0 may be reduced to the above form by dividing them by XY. (30.) Homogeneous equations. Any homogeneous equation between two variables, each term of which is of m dimensions may be rendered integrable by dividing every term by a”, and assuming y=zx, then

d ̧y=x+xd ̧%,

and the given equation may be reduced to

Z1 + Z2(x + xd ̧x)=0,

1

in which the variables may readily be separated.

The equation a1 + b1x + c1y + (α2+b2x+c2y) d ̧y=0 may be rendered homogeneous by assuming

whence

а 1⁄2 + b2 x + c2 =V,

(31.) The linear equation, dy+X1y+X2=0.

2

This equation may be rendered integrable by multiplying

every term by ex1: the required integral is

y= e−√xX1{ √xX 2e√xX1 + const.}

(32.) Riccati's equation; dy+ay2 + bxTM=0.

If m=0, the variables are immediately separable: otherwise

substituting these values for y and dy respectively, we obtain x2d2+ax2 + bxm+4=0.

This equation is homogeneous, if m2; and if m=-4, the variables are immediately separable: otherwise assume

and by substituting these values, the equation will be reduced to its original form, and the same process may be repeated.

DIFFERENTIAL EQUATION INTEGRABLE.

(33.) Let M + Nd ̧y=0 be an equation which does not fulfil the condition

d, M=d,N,

and let V=c be the primitive equation, and

then the given equation may be rendered integrable by introducing the factor u. (u): in the simplest case (u) = 1.

To determine u we have

u{d,M—d2N} = Nd ̧u— Md ̧u,

(u)=1.

this however usually presents greater difficulties than the given equation.

G G

If u is a function of a only, then du=0, and

If u is a function of y only, it may be determined in a similar

manner.

If M and N are homogeneous functions, then

u =

1

Mx+Ny

(34.) The required factor may sometimes be conveniently found by transforming the equation

M+Ndy=0 into K+Ldu=0,

in which the relations between x, y, t, and u, are known, and

K
V

finding a factor V such that may be a function of t, and

y,

L

V

of u: then if the values of t and u, in terms of x and be substituted in V, the quantity obtained will render the given equation integrable.

The equation P+Qdu=0 may be rendered integrable by

V being any function of t and u, and T and U, of t and u, respectively.

The equation X1y+(y+X2)d,y=0 may be rendered integrable by the factor

1

y3 + Xy2 + {X− a+b(2 a − X)*} ay

The equation X1y+(X2y+X)dy=0 may be rendered

X1=bX®.d2X + (n − 1) a X" − 2. d ̧X,
X2 = (m-n+2) a X",

2

X ̧=bXm+1+ (m + 1) a Xn−1.

(L. D. C. 316-23; L. C. D. 567-80.)

SINGULAR SOLUTIONS.

(35.) Let f(x,y,c)=0, c being an arbitrary constant, and let p(x,y,y',c)=0, in which y' represents day, be derived from the former by differentiation; from these equations, another of the form (x,y,y)=0 may be obtained by the elimination of c. If u is such a function of x and y, that

p(x,y,y',u)=0,

may be derived by differentiation from

f(x,y,u)=0,

then √(x,y,y')=0 will result from the elimination of u, and f(x,y,u)=0 may be considered as the primitive equation.

Such values of f(x,y,u)=0 as are not included in the general form f(x,y,c)=0 are singular solutions.

The equation du f(x,y,u)=0 must be satisfied by the values of u which give singular solutions.

All singular solutions will be found amongst the equations obtained by substituting in f(x,y,c)=0 those variable values of c which satisfy the equation

dεf(x,y,c)=0.

The values of c which give singular solutions satisfy the equations drc = ∞ dyc=

they also render d2y=

If f(x,y,c)=0, the general solution of a differential equation, represents a class of curves, each of which is determined by assigning a particular value to the arbitrary constant c, then the singular solution, which is independent of c, represents the bounding or circumscribing curve.

If a given differential equation f(x,y,y)=0 can be solved for y', then a singular solution may be obtained by substituting for y'a value which satisfies the equations

For singular solutions of differential equations of higher orders, see L. C. D. 638—58.

(36.) Equations of more than one dimension in yʻ.

If a constant of n dimensions be eliminated between any equation f(x,y)=0 and its derived equation f'(x,y)=0, an equation may be obtained of the form

[n = 1

y'" + a2-1y'n1+ &c. + a1y' + a=0:

let the roots of this be P1, P2, &c. and let the equations

y-p1 =0, y' - p2 = 0, &c.

be integrated separately; then the primitive equation will be the product of the integrals thus obtained.

If we have an equation f(x,y,y) = 0, homogeneous with regard to x and y, then by assuming y=ux, we obtain

whence we may frequently obtain an integrable equation between y and u.

(37) Clairault's formula: y=y'x+f(y').

By differentiating the given equation we obtain
{x+d, f(y') } d2 y=0,

whence day=0, and .. y=c, by substituting which we have

y=cx+f(c).

This is the general solution: a singular solution may be obtained by eliminating y' between the original equation, and

x + dy f(y')=0.

(L. D. C. 342-7; L. C. D. 582-9.)