We shall conclude by some indication of the principal steps of the application of the calculus of operations to the solution of differential equations and equations of differences.
However :much the calculus of operations may throw light upon the character and principles of algebra, it would at one time have been thought unlikely that it should much facilitate actual processes. It does this, nevertheless, and nowhere more than in the subject we are now going to describe. Solutions which by the usual method OY PARAMETERS] would never have been considered fit examples for an elementary work, on account of their complexity, may be represented with ease, and obtained in full with very little trouble.
When rules of algebra are true of the meanings of any symbols, all consequences of the use of these rules, all relations which are legiti mate deductions from them, also represent truths. Not that these truths are always intelligible without subsequent interpretation : nor do we DICI1D to say that, in the present state of the science, the inter pretations are always attainable. And further, it may happen that theorems can be pointed out, derived from proceesee in which sonic only, and not all, of the fundamental rules of algebra are true. This does not prevent our right to deduce conclusions from such theorems, so long as we use no fundamental rules except those which are true of the expressions in question. For instance, we have seen that the operations of our calculus are not convertible with the operation of multiplying by a function of the variable. Thus if E stand for the direc tion to change x into x +1, A for that of forming the difference thence arising, and D for the direction to take the differential coefficient with respect to x, we have no right to say Ecpx(tpx)=cp.rE(P.r), or Atpx(tpx)= cp.rA(fx), or rep.r (+4= cpxn(fx); in which st:x is the function operated upon. But, when we thus use another function, cp.; besides the one operated on, 4'r, this convertibility of operations is the only rule of algebra which fails ; it is therefore the only one the use of which we must avoid.
The operations E, A, and D, are closely counected with E-Fa, A + a, D+ a, of which they are particular cases ; a being a constant, positive, negative, or nothing. We have (E - a). cp.r = 0.r) (A - a) ipx = (a + ((a + cpx) (n - a) cpx = e" D cps).
The first sides of these equations being representatives of ca(r +1) -aox, Aq5x-acpx, and ˘'x-actr. If these operations be repeated, we have (E - a)ncpx = a.+PA.1 (a-'$x) (A - a)'"cpx = (a + ((a + ˘x) (15 - a)'55x = 6"D' cpx).
These results will also be found to be true when in is negative, by which means we are enabled to interpret (A- and and their repetitions.
These same forms may be extended, as follows :-Let E. and as severally denote the operations of changing x into x + 1 and y into y + 1; and let D., A., be similarly interpreted with respect to the differentiations and differences. We have then - a)" (e - 6)* = a.+" " A: (a - (A - 6)* = (a + 1)'+' (6 + 1)Y +.*A" A" (a - a)n - 6)* = e=+is in which the function first operated upon is left out to save room. Here as and n may be either positive or negative integers. And even a or 6 may be symbols of operation, but not with respect to x or y. Thus (n. - a D, )." = in which the second side is to be thus interpreted. Changing y into y - a.r, differentiate in times with respect to x, and then change y into y +a.e.
We shall now give the heads of some methods of solution, observing that this article is intended only for those who can already master the :same solutions by other methods.
Take the common linear d"y d.c. + 6 d.r" ' = in which a and 6 are constants, and x a function of x. The operation performed upon y is an" + + ...: if this be called a, then y is the result of performing the inverse operation upon x. By the method explained in FRACTIONS, DECOMPOSITION OF, transform (an" + into s(n - +B(n - +, &c., where &c. are the roots of the algebraical equation as" + + = 0. Then y is A() - x + B(n - o)-! x + or At"fe"" xd.c + xd.c + substitnting for its usual mode of expression. The arbitrary parts of the solutions will be obtained by the constants of integration in the usual manner. But the arbitrary part may always be obtained, in all inverse operations, by considering the function operated upon as x+ 0, and operating separately upon x and 0. Thus (n- x may be com pletely expressed by ta.(j d.r)? x + s=( dr)! 0, the second term of which is e" (r + r, Q, and n being any constants.