[A]?+1= (A— nr) [4]*=A[A—r];; so that if in (1) we replace y by its equivalent [A-2] [A+2n]" {A(A−1)—▼(▼—1) x2} Y=0, Hence the solution of (1) is reduced to that of (A−1±vx)(A+уx) Y=0. But a particular integral of (2) is Y=x ̄ß (x+1)ß−α-1. Consequently (1) has a particular integral of the form (x+1)`R(x)=y, where R(a) is rational. 3. This form is included in e(x)dx, where p(x) is rational. So, too, is that of the solution of the first case of the second description of regular forms, viz. {(A−2n) (D+3)+A(D+a) x2} y=0, where A=D+ any constant. For if we replace y by then (3) reduces to (3) [A]; y, where Hence a particular integral of (3) will be given by Consequently (3) will have a particular integral of the form y= (x2+1)`R(x), where R(x) is rational, and therefore of the form e()dx, where (x) is rational. 4. The second case of the second description, viz. p. (7) which last is solved, with redundant constants (conf. op. cit. 421, et Suppl. p. 189), by y=xm(Co+C1x2+..+C2n-2x2n-2), if we make A=D—m. And the redundancy is got rid of if in (5) we substitute for y the dexter of (7). For A and A-2n respectively annul the terms Com and C2n-2x+2n, leaving all the n quantities Co, C2, Can-2 to satisfy the remaining n-1 homogeneous conditions. Thus (5) has a particular integral, which is of the form R(x), and therefore of the form ef¤(x)dx; R(x) being rational and entire, and (a) rational. 5. The first primary form of Boole (op. cit. p. 428) may, without loss of real generality, be written But this form is not truly primary when n is an integer or the half of an integer, the latter case corresponding to a quadratic resolvent. And this accords with what precedes. The complete integral of (8) or (9) is C+1(x2+√x2+1)2 +С_1 (x−√x2+1)"=y. -1 Take n a positive integer and C+1=C-1. Then y is rational and entire. Take n the half of a positive integer. Then √x2 + 1 dy n dx = · C+ 1 (x + √ x2 + 1)” — C_1(x−√√/x2 + 1)"; and if C+,=1 and C_1=−1, then √x2+11 dy will become n y dx (x+√ x2 + 1)22 + (x−√ x2 + 1) 22+2(−1)” 2n 1 dy In either case* y dx is rational, and dy dx in the former rational and entire. In both cases, therefore, y is of the form es¤(x)dx ̧ where (x) is rational. The proof may be extended to the cases of n, a negative integer, or the half thereof. In no other cases has (8) or (9) a particular integral of the form ef(x)dx. And since (8) and (9), if soluble through Boole's reduction, must fall under one of the three forms (1), (3), or (5), the first primary form is not in general so soluble. The same may be proved for his second primary form. 6. Let where z is any function of x. Suppose that 2a= dz where p= Then (10) is reducible to an equation with condx stant coefficients by a change of the independent variable. The primary (8) is thus soluble. But it may be otherwise solved. Let SU mean SUda, and let P(S) mean the double operation SSSS... continued in infinitum. Also let y=4(x) be a particular integral of (10). Then if § be determined from the equation of the casura, viz. +2ağ=0, and ʼn from * The same is true of every case. the three forms dę da For (1), (3), and (5), together with {(A-2m)(A-1)—a2x2}y=0, {(A-2n) (D+B)+Ax2}y=0, 1 and the six other forms deduced from these by the change of x into stitute the twelve forms which the reductions of Boole solve. Both here and in the text m and n are integers. The arguments for (a), (b), and (c) respectively are analogous to those for (1), (3), and (5), and show that there is at least one particular integral of the form ef(x)dx. †This method of synthetical solution may give a finite result, a series summable or otherwise, or a suggestion of the form of (x), in which a constant or constants are to be determined by substitution. For the terordinal d3y dx2 dy where a, 6, and c may be variable, we assume y=PS San S25.(x), where (x) is a particular integral, say zero, and έ is determined by the casura where R(x) is rational, and therefore of the form ep(x)dx, w] (x) is rational. 4. The second case of the second description, viz. {A(D+B)+(A−2n) (D+a)x2} [4] reduces to or y=0, [4]2 which last is solved, with redundant constants (conf. op. cit 421, et Suppl. p. 189), by if we make ▲=D-m. And the redundancy is got rid of i (5) we substitute for y the dexter of (7). For A and Arespectively annul the terms Com and C2n-22m+2n, leaving all n quantities Co, C2, .., C2n-2 to satisfy the remaining n-1 mogeneous conditions. Thus (5) has a particular integral, wh is of the form R(x), and therefore of the form ep(x)dr; R being rational and entire, and (x) rational. 5. The first primary form of Boole (op. cit. p. 428) may, wi out loss of real generality, be written But this form is not truly primary when n is an integer or half of an integer, the latter case corresponding to a quadra resolvent. And this accords with what precedes. The compl integral of (8) or (9) is n C+1(x+√x2+1)" +С_1 (x−√x2+1)"=y. Take n a positive integer and C+1=C-1. Then y is ration and entire. Take n the half of a positive integer. Then x2 + 1 dy = C + 1 ( x + √ x2+1)* −C_1(x−√/x2+1)" ; n dx and if C+,=1 and C_1=-1, then √x2+11 dy will beco 2n 2n n y dx (x+√ x2 + 1)22 + (x−√√/x2+1)2+2(−1)” and entire. In both cases, therefore, y is of the form ex)ds, where (x) is rational. The proof may be extended to the cases of n, a negative integer, or the half thereof. In no other cases has (8) or (9) a particular integral of the form e¤(x)dx. And since (8) and (9), if soluble through Boole's reduction, must fall under one of the three forms (1), (3), or (5), the first primary form is not in general so soluble. The same may be proved for his second primary form. 6. Let where z is any function of x. Suppose that 2a= Р dz dx 2z +C√z, where p= Then (10) is reducible to an equation with constant coefficients by a change of the independent variable. The primary (8) is thus soluble. But it may be otherwise solved. Let SU mean SUda, and let P(S§§,n) operation SSSS... continued in infinitum. Also let y=↓(x) be a particular integral of (10). Then if § be determined from +2ağ=0, and ʼn from the equation of the casura, viz. * The same is true of every case. the three forms de mean the double For (1), (3), and (5), together with {(A-2m)(A-1)-a2x2}y=0, {(A−2n) (D+ß)+Ax2}y=0, and the six other forms deduced from these by the change of x into stitute the twelve forms which the reductions of Boole solve. Both here and in the text m and n are integers. The arguments for (a), (b), and (c) respectively are analogous to those for (1), (3), and (5), and show that there is at least one particular integral of the form e(x)dx ̧ This method of synthetical solution may give a finite result, a series summable or otherwise, or a suggestion of the form of (x), in which a constant or constants are to be determined by substitution. For the terordinal where a, b, and c may be variable, we assume y=PSSS(), where (a) is a particular integral, say zero, and έ is determined by the casura |