vary continuously with the other. Equation (201) which gives the relation between y and y, must be consistent with such continuous variation; and this can only be the case when the ydifferential of the coefficient of (x — Xo) on the right-hand side of (201) is not infinite, that is, when“, hon dof (x, y) does not become in " dy finite for any value of the variables between the limits. Hence if x — Xo = h, since y = F(Q), F(x) = f(xo) +hf{xo+0h, F(xq) +oah}; and we infer, that if f(x,y) and“ (w) and .J lit, y) are finite and con dy tinuous for all values of the variables between x and Xo, there is always a function of x, viz. F(x), capable of satisfying the given differential equation, and of becoming a definite value, viz. Yo = F(x), when x = xy. Thus the result contains a general undetermined constant. It may also be observed that as these are the circumstances requisite for a general integral, it is thus proved that every differential equation of the first order has an integral. 404.] Now consistently with these conditions we can shew that there is only one general form of function which will satisfy the given equation ; for suppose y = f(x) to be a general form of function which satisfies the equation dy = f (x,y) dx ; and suppose another form which satisfies it to be y = F (x) ++ (x). Then we have simultaneously yo = F(xo), Yo = F(x) + + (xo), f'(x) = f {x, F()}; F'(x) + *'(x) = f {x, F(x) + +(x)}; .: +(xv) = 0; F"(x) ++(x) = f {x, F()}+6(x) ., f {2, F(x) +0° ()} ; :: $'(x) = 6(3) 6,1 {x, f(x) +00(x)}; and as this is to be true for all values between the limits, it is true when x = x,, in which case + (x ) = 0, and d’rn) does not generally vanish ; and therefore we have *'(.) = 0 x f(n, y); .du f (xo, Yo) = -0 ; which result is inconsistent with the given conditions. The general integral therefore involves only one general form of function. Consequently if we can discover a second form of function which satisfies a differential equation, besides a general integral, we may be sure that this second form is not a general integral. 405.] Let us suppose then y = f(x) to be a function of x which satisfies the given differential equation dy = f(x,y) dx ; then if each of the functions f{r, F(x)}, , f (x, y) is finite and continuous for all values of x, or, at least, for all values of x between certain assigned limits, we may take any one of these for that which we have represented by Xo; and thus y = f(x) will be a function of x which will satisfy all the conditions stated in Art. 403, and therefore will be either the general or a particular integral of the given differential equation. But if, on the other hand, f{x, F(x)} or f(x, y) is infinite or discontinuous or indeterminate for all values of x, then the conditions necessary for a general or particular integral cannot be fulfilled, although the value is such as to satisfy the given differential equation : the case of discontinuity we may at once dismiss as beyond our province; and f{x, F(x)}, which is equal to f'(x), cannot be infinite for all values of x unless f'(x) is, and therefore unless F(X) is, and thus this is another case which we may exclude : and therefore the cases which remain are 0 0 d (1) f{0,F(x)} = 8; (2) f(x,y) = ; (3) de f(x,y) = 20; and when y = f(x) is such as to satisfy the given differential equation and at the same time to satisfy one or other of these conditions, the integral is not general. Yet such a value satisfies the differential equation, and is therefore either a particular integral or a singular solution; and to determine these it is necessary to investigate the relations between x and y which will render f (x,y), or 2, f(x, y), indeter minate, and which will render f(x,y) = 0; if they satisfy at the same time the given differential equation, they are either singular solutions or particular integrals ; if the general integral is known, there is no difficulty in determining whether any particular constant value of the arbitrary constant will reduce the general integral to the form y = F(x), but if only the differential equation is given, we must apply the criterion which will be investigated in Article 407. The last of the above-mentioned conditions may be conveniently applied in the following way. Let us use Lagrange's notation of derived functions: then, since f(x, y) = 1 = ; is er yf(x, y) = (202) If therefore ay is found from the given differential equation, and equated to oo, and a functional relation between x and y is thereby determined and of the form y = f(x), this is a singular solution if it satisfies the given differential equation. Of this method of discovering singular solutions some examples are added. Ex. 1. * (he) – y + m = 0; (y2 -4mx)} 2 x ay = 00, if y= 0, or if x = 0; dy 2 (y x) and as either of these satisfies the given differential equation, they are singular solutions. Ex. 3. de ter um 1.0. dx (x* + y2 — a)} the singular solution is x: + y2 - a= 0. Ex. 4. Y-X = 0 is a singular solution of +(y—~)}–1 = 0. Ex. 5. Prove that x2 + 4y ='0 is a singular solution of + com-y = 0. Ex. 6. Find the singular solution of any = y log y. 406.] If the differential equation is in the implicit form $(, y, '%) = 0; (203) then, considering x, y, and y' as three independent variables, we have and not simultaneously equal to 0. And if we eliminate ý between (203) and (204), the resulting equation in terms of X and y will be a singular solution, if it satisfies (203); but if simultaneously C) = 0, and (203) is satisfied, the result may be a particular integral or a singular solution. Ex. 1. c ( ) +(a-o=y) + y = 0 = f (x, y, aume); ) = 1–y'; ) = 2 vý'+a-r—y = 0, 2 xy' + a-X-y = 0, substituting which in the given equation we have after reduction 4xy = (x +y-a)?; or 2 + y = a, which satisfies the differential equation ; and as a ) does not vanish, it is a singular solution. Ex. 2. y - xy'- () = f (x, y, y') = 0; .: (M)= —r—P(x) = 0; the elimination of y' between which and the given differential . equation will give the singular solution, if the result satisfies the equation. This is the equation known as Clairaut's equation. Ex. 3. x2 + 2 xyy' +(a— xo)y? = 0 = f(x, y, y): () = 2.xy'; () = 2 xy+2 (a? – x2) jʻ= 0, substituting which in the original equation we have y2 + x2 = az; and as 2) is not equal to 0, and as the above satisfies the given differential equation, it is a singular solution. Ex. 4. 42 + yy' + x = f (x, y, y) = 0; Gal = y; (i)= 2y +y = 0, if y'= ; substituting which in the given equation, we have y2 = 4x; but as this does not satisfy the given differential equation, it is not a solution at all. Ex. 5. Prove that y2 –4x3 = 0 is a singular solution of +2y2 + x3 = 0. integral, hurticular Ex. 6. Find the singular solution of y 3 – 4x yy' +8y2 = 0. 407.] We proceed now to another question : y = f(x) satisfies a given differential equation dy = f(x,y) dx ; is it a singular solution or a particular integral ? If the general integral of the equation is known, we can determine whether y = f(x) is a particular integral or a singular solution. If it is a particular integral, the substitution of F (x ) for y in the general integral will yield a particular value for the arbitrary constant: but if it is a singular solution, the arbitrary constant will be equal to a function of the variables : to this subject we shall return in the following Articles, and therefore I merely subjoin an example. Ex. 1. The general integral of y ( 4 )* + 2x -y = 0 is ya=2 cx + c2; other solutions are (1) y’=2x+1; (2) y2 + x2 =0; are they singular solutions, or particular integrals ? (1) Comparing y2 = 2cx + c2 and y2 = 2 x+1, it is manifest that c =1; the solution therefore is a particular integral. |