(38.) Let the equation be f(day,x)=0. If this equation can be solved for day, another may be obtained of the form whence and d2y= X; d2y=f,X+c1, y=f2 X+c1 x + C2· Let the equation be f(day,y)=0. If this can be solved for day, we obtain d2y=Y; whence, since d2y=y'd,y', (d ̧y)2=2fy Y+c, and x=√(2,Y+c)−ŝ +¤ ̧· the primitive equation between x and y may be obtained by eliminating between these two equations, each of which involves one arbitrary constant. Let the equation be f(day,dy,x)=0. This is a differential equation of the first order with respect to x and dy, the integral of which is of the form If (1) can be solved for a only, then we have and by integrating by parts, x=f(y'), y=y'x-Syf(y'): (1) from these two equations the primitive equation may be obtained by elimination. Let the equation be f(day,d,y,y) = 0. If this equation can be solved for day, we have d2y=y'd,y=(y',y), which is an equation of the first order, from which we may obtain (y',y,c)=0. If this can solved for y', we have y'Y, which is immediately integrable: if it can be solved for y only, then we obtain y=(y'), hence y'd,(y'), and a = ƒ dy (y); y' y from these the primitive equation may be obtained by eliminating y'. (L. D. C. 353-60; Tr. L. 273-7.) (39.) Let the equation be day+n2y+X=0; X being a function of a and constants only. By multiplying by cos na, and integrating by parts, we obtain cos nx.dy+ny sin nx + cos nx. X=0: this being divided by (cos nx)o and integrated, the result is (40.) Let day+X1d ̧y+X2y=0: assume y=e", then by substitution d ̧u+u2 + X1u+X2=0; an equation of the first order, which is not however generally integrable, unless X, and X, are constants. 2 Let day+X1d ̧y+X2y+X ̧=0. assume y=tu, and substitute for day, day, and y, their values in terms of t and u; then u being determined by assuming the condition and this last equation, by assuming v=1 becomes (41.) If the given equation is homogeneous with respect to x and y, assume then the equation becomes f(y', v, u)=0: from this last equation solved for v, and the equations (1), we have 1 du!' — y — u fi(Y', u) = 0; hence by integration y' =ƒ, (u, c1), and finally (42.) Differential equations which do not contain either variable, and are of the form by assuming d"-1y=u, whence dry=du. Equations of the form f(dry, d"-2y)=0 may in a similar manner by assuming d"-2y=u be reduced to f(džu, u)=0. By the same assumption the equation f(dry, d"-1y, d ̧n-2y)=0 may be reduced to the form f(džu, d ̧u, u)=0: this and the two former equations in u may be integrated by preceding methods. (43.) Solution of the equation n-1 dry + A1d "-1y + Ad ̧”−2y + &c. + An-1dxy + A„y=0. (1) by substituting these values of u and its differential coefficients, and dividing by e", we obtain where du + B1d "-1u+ &c. + B-1d2u + A„=0; 1 1 Ꮖ -2 n (2) B, B2-1=(d ̧u)"1 + A1 (d ̧u)"− 2 + &c. + An−1· This equation is seldom integrable; if however A1, A, &c. the coefficients of (1) are constant, (2) will be satisfied by a constant value of du; in which case since du = 0, du=0, &c. -2 ・n-1 (d ̧u)" + 4, (d ̧u)”−1 +42(d ̧u)” −2 + ... A2- 1 d ̧u + A„=0. let the roots of this equation be a1 α, &c., then from du=a, we obtain y=C1 €1*, If two roots, as aj, a2, a+b√√√-1, then the corresponding part of the value of y k1e sin (bx+ k2). If a1=a2=&c. = αp› the corresponding part of the value of y e°, * {K1+K2x+ ... + K2∞”-1}. If there are two equal impossible roots, the corresponding part of the value of y ‡c°1*{ (F1+F2x) cos bx + (G1 + G2∞) sin bæ}. 1 2 (44.) Solution of the equation 2 -1 =ek* {dn-1y + B1d-2y+ ... + B2-19}; by differentiating which, and equating coefficients, B1, &c. may be determined in terms of 41, &c. and k will have n values determined by the equation -1 19 k” + A1 k2 −1 + ... + Aμ- 2k + An=0, 1 the equation (1) will therefore have n immediate integrals of the form dn-1y+B1dn-2y+... +B2-1y=c** { X1+ K}, 1 1 the value of X1 depending on that of k: from these n equations the n-1 differential coefficients may be eliminated, and the result will be the primitive equation. |