(x3t — 2ť1) du + (2x* — t3x) d ̧u = 9 (t3 — x3) u. Comparing this with the formula of Art. 4, we have Du = 9 ( − x ) u, Dạt= t−2t, D = 2 CHAPTER III. DETERMINATION OF QUASI-CONSTANTS. SECOND AND HIGHER ORDERS. EQUATIONS OF THE 11. SINCE Du=du+D ̧x.d2u+D1y. du+... it follows that the single operation denoted by D is equivalent to the compound system of operations denoted by dt + Drx. dx + Dry.dy + ... Hence if Dx, Dy, ... are either constants, or quasi-constants, any function of the one operation D, will be equivalent to the same function of the compound system dr + Dix.d2+ Diy.d1 +... :. F(D¿) = F(d2+ Dix.dz + Diy.dy+ ...)......... (1). Ex. 1. To integrate d3u+2adıd ̧u + a3d ̧3u+ b (du + ad ̧u) +cu = 0. Taking the terms of the second order, we have di2+2adı d2+a2d ̧2 = (dı+ ad„)3. We therefore assume Du = du+adu, which compared with the general formula gives D1x=a; and .. Du+bD ̧u+cu = 0 ; mt .. x = at +§, and u = et F (§) +ent ƒ (§) = €TMt F' (x − at) +ent ƒ (x − at), m and n being the two roots of m2 + bm + c = 0. Ex. 2. To integrate d'u-a'du+b (du+adu) = 0. 2 Since da3d = (dı + ad2) (dr—ada), we shall have (here for the first time) to use two new sets of independent variables, corresponding to the two equations, D1u= d1u+adu, and ▲u= d1u — ad ̧u. These give Dxa, and A,x=-a; and since the symbols Dr, A, are respectively equivalent to the compound symbols d+ada, dt-ad, the proposed equation becomes D1Au+bD¿u= 0. Hence the following system of three simultaneous equations is equivalent to the proposed equation, D1x=α, Ax=-a, and D.Au+bD ̧u = 0...... (1). The first two give by integration, x=at+§, x=— at + §' ; and the third being integrated with St gives ▲ ̧u+bu= F(§) = F(x — at) But the symbols D, and A, are commutative because their equivalents d+ad, de-ad, are so; consequently the equation (1) may be written in the form which being integrated with Σ, gives -bt Du= Adding this equation to equation (2), and writing u for the sum of the left-hand members, we obtain finally for the integral of the proposed equation. 12. The following is the reason why we may write u for D1u + Au+bu. Since the proposed equation is represented by t .(1), differentiate this equation separately with D, and A; and also multiply it by b; the results are (since D and At are commutative) D¿¿. D¿u +b Dr. D¿u = 0, Dƒat · A¿u +bD ̧ . ▲¡μ = 0, DA. bu +bD1. bu = 0; ·.:: Dƒa¿. (D¿u+▲u + bu) +bD1. (D1u+A ̧u+bu) = 0. This equation is of the same form as (1), and we learn from it that Du + ▲u+bu will satisfy the proposed equation quite as well as u will do so. If therefore D1u + Au+bu contain the requisite number of arbitrary functions, it is a complete integral of the proposed equation, and being so we may write u for it. We shall find this principle of great use hereafter; but it must be noticed that it only applies to cases of linear equatious with constant or quasi-constant coefficients and when the symbols of operation are commutative. E. 13. It is well known that the integral of any linear differential equation of the form F (dt, dx, dy, ...)u = W, where W is a function of t, x, y, ..., consists of two independent parts; one of them independent of W is called the absolute part of the integral; the other is dependent on W. These two parts being independent of each other, may be found separately (should it be advantageous to do so); and their sum will constitute the complete integral. The absolute part corresponds to the general supposition, W=0. As an example of this principle, let it be required to integrate the equation du — a3d ̧3u+b(du+ ad ̧u) = W. We first find, as in the preceding example, the absolute integral, i.e. that part of the complete integral which corresponds to W=0. This has already been found to be u = €1t ƒ (x+at) + F(x− at). To find the part that depends on W we have the equation Either of these (whichever may be judged most convenient) will furnish the required result. For integrating the former with St, or the latter with Σ, we have Au+bu=S&W, or Duettbt W. = Integrate, now, the former with Σt, or the latter with S¿; -bt .. u=eet SW, or u= Seebt W. Adding to either of these the absolute part of the integral, we have for the complete result u = €¬btΣ¿€1t St W + € ̄1t ƒ (x+at) + F (x − at), or u = S1eet W+ € ̄bt f (x + at) + F (x — at). |