This is the equation to the elastic curve. Jac. Bernoulli, Opera, p. 576. which is integrable when divided by (1 + p2)3. The complete integral is y = (a2 + b2 − x2)3 – blog b + {(a2 + b2 − x2) } } c(x - a) v and the complete integral is a y + c' = log (a2 − c3) — —-log c and c' being arbitrary constants. If the independent variable (x) be wanting, we put day dy dp = dx2 dx dy = Ρ dy and then we have an equation be tween p and y from which by integration we find p in terms of y, or y in terms of P, and then is known from the equation (,。 dy + 1) dp 1 + p31 a linear equation in y, which being integrated gives whence by eliminating p we obtain a relation between a d2y\ 'dy 2 = ny dx dx + y2 - y da2 ) ・ which is of Clairaut's form. The general integral is there fore p = Cy+n (1 + a2 C2), whence Ca + C' = log { Cy + n (1 + a°C2)}}. The singular solution is The examples in this section are taken chiefly from Euler, Calc. Integ. Vol. 1. Sect. 111. and Vol. 11. Sect. I. Cap. 2 and 3. CHAPTER V. INTEGRATION OF DIFFERENTIAL EQUATIONS BY SERIES. THE method employed for integrating Differential Equations by series, is to assume an expression for the dependent variable in terms of the independent variable with indeterminate coefficients and indices, and then to determine them by the condition of the given equation. -8 Assume y=xa (A + x2 + 2 + А2 x2+1 + А3 x3n+8 + &c.) Whence we find day dx2 2 = a (a− 1) Axa2 + (a + n + 2) (a + n + 1) A ̧ xa+” +&c. a Аxa + n + a A1 xa+2n+2 + &c. 1 and aay= Substituting these values in the equation, and equating to zero the coefficients of the powers of x, we have a (a − 1) A = 0, (a + n + 2) (a + n + 1) A ̧ + a A = 0, a = 1. The first of these is satisfied either by a = 0 or Taking a = 0 and substituting it in the other equations, we find 1.2.3 (n + 1) (2n + 3) (3n + 5) (n + 2)3 &c. &c. + (n + 1) (n + 2) 1 . 2 (n + 1) (2 n + 3) (n + 2)2 &c.} But as this contains only one arbitrary constant 4, it is not the complete solution. Let us take a = 1 and call A', A, A, &c. the corresponding coefficients; we then find in the same way as before y = A' {x ax2+3 + a2x2n+5 - &c.} (n+3) (n + 2) 1. 2 (n + 3) (2n + 5) (n + 2)2 which is another incomplete integral with one arbitrary conThe sum of these two series is the complete integral stant. of the equation. When n = 2 both the series fail, as the denominators are then infinite: but the true integral is easily found. a. a (a− 1) = 0. This is a quadratic equation, which gives two values for The first of the preceding series will fail when n= 2r-1 ber: the complete integral may however be found by the following process. Assume where v is the particular integral furnished by the series which does not fail. On substituting this value of y in the original equation we obtain the system of equations the second of which serves to determine u. Euler, Calc. Integ. Vol. 11. Chap. VII. |