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-x) on the right-hand side of does not become in (201) is not infinite, that is, when d.f(x, y) finite for any value of the variables between the limits. Hence if x-xh, since y = F(x), = F(x) = F(x。)+hƒ{x。+0h, F(x)+0xh}; and we infer, that if f(x, y) and d.f (x, y) are finite and con tinuous for all values of the variables between x and x, there is always a function of a, viz. F(x), capable of satisfying the given differential equation, and of becoming a definite value, viz. Yo = F(x), when x = x。• 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 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 4′(x) does not generally vanish; and therefore we have The which result is inconsistent with the given conditions. 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 æ which satisfies the given differential equation dy = f (x, y) dx ; d then if each of the functions ƒ {x, F(x)}, dyf(x, y) is finite and continuous for all values of x, or, at least, for all values of æ between certain assigned limits, we may take any one of these for that which we have represented by ; and thus y = = F(x) will be a function of a 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. d But if, on the other hand, ƒ {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 ƒ{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 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 d between x and y which will render f(x, y), or f(x, y), indeter d dy minate, and which will render f(x, y) = ∞; if they satisfy at dy 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) = dx dy dy If therefore and equated to ∞, 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. is found from the given differential equation, and as this satisfies the given differential equation, it is a singular solution. and as either of these satisfies the given differential equation, they are singular solutions. Ex. 5. Prove that x2+ 4y =0 is a singular solution of dy Ex. 6. Find the singular solution of = y log y. dx 406.] If the differential equation is in the implicit form then, considering x, y, and y' as three independent variables, we df and is not simultaneously equal to 0. And if we eliminate dy y' 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 = 0, and (203) is satisfied, the result may df dy be a particular integral or a singular solution. dy dy ( 1/24 ) 2 + ( a − x − y) 1/4 + y = 0 = f (x, y, dy); dx +(a−x = 1-y'; if y= dx = 2xy + a—xy = 0, x + y-a 2x substituting which in the given equation we have after reduction 4xy = (x+y—a)2; or x+y= a}, which satisfies the differential equation; and as vanish, it is a singular solution. Ex. 2. y-y- $ (y'′) = ƒ (x, y, y') = 0; (df) = − x — p'(y') = 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+2xyy′+ (a2 — x2) y22 = 0 = ƒ (x, y, y′): and as ( y2+x2 = a2; is not equal to 0, and as the above satisfies the given differential equation, it is a singular solution. Ex. 4. y2+yy' + x = ƒ (x, y, y') = 0; df = 2y+y=0, if y' = 2 dy 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 y2-4x30 is a singular solution of Ex. 6. Find the singular solution of y'3-4xyy +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 ? (x) 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 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 (dy)2+2x dx dy dx y = 0 is y2=2cx+c2; other solutions are (1) y2=2x+1; (2) y2 + x2 =0; are they singular solutions, or particular integrals? (1) Comparing y2 = 2cx + c2 and y2 = 2x+1, it is manifest that c = 1; the solution therefore is a particular integral. |