The integral of which is u = (ht + kx)" F (§) + (ht + kx)" ƒ (§), m and n being the roots of m (m − 1) (h + ka)2 = C. C may be a quasi-constant. c2 Ex. 12. To integrate diu – «.du=0. This falls under the general form considered in Art. 50, Ex. 3. We here assume t1=7, and change the independent from t to T. [The rule for determining in equations of this form what power of t is to be assumed for 7 is this. The power of t in the question is t, take the square root of this (=t2) and integrate this (=). This is the value of T. (If the form of t in the question had been t→, its square root would be t1, and .. 7 = log t.)] d+cdx, Ad- cdx, §=x— CT, &' = x + CT; Operate with (D,A,) ̃1, :. TD,A, (D,A,) ̄1 u = (DA)10 = S110; Ex. 13. To integrate du- ctdu= 0. In this case we must assume =T. •. T (du — c3d ̧u) = 2d,u ; :. τD‚a‚μ = (D,+A‚) U......................... Operate on this equation with (D,A,), :'. T (D,A,)2 u=0; :: u = 8,20 +Σ,20 = = F (§) + τF, (§) +ƒ (§') + +ƒ1 (§')• 1 1 .(1). Substituting F(§)+TF, (§) in equation (1) for u we find F()=cF'(E). Similarly f (§') = — cƒ' (§'). .. u = cт (F'§—ƒ'§) + F§ +ƒ§' consequently the integral of the proposed equation is с * = ei F (§, n. 5 ...) + e ̄ i ƒ (§, n. 5......), and c may be a quasi-constant, or a function of d, dyr... Here D-d,+.d, and a*.-F.; = t = x = = This agrees with the form of the preceding Example, with c for c. ct ct = F(†) + e ̄*ƒ(†). Ex. 16. To integrate du+cu = an integer. Multiply by ť; f A t2 .u; A being :. (ťod22 + ct3) u = A (A − 1) u. For t'd write de (de − 1), (Art. 42, 5); .. {do (do − 1) — A (A − 1)}. u + cť3u = 0 ; Diminish the symbolic factorials by A (Art. 45); .. (de−2.A) (de −1). uta + ct2. uta = 0, or, writing v for uta, (d. — 24) (d。 — 1) v + cť v = 0. We have now to increase the first of the factorials by nk (that is, by 24, for k = 2, and n =- A) to reduce them to the requisite form (Art. 44). We must therefore assume (Art. 49) (t2do) ̄1. vt-21 = (t ̃2)−a. wt-24 = w ; .. de (de − 1) w + cť3w = 0 ; :. f'diw+ctw = 0; .. di̟ w + cw=0. The integral of this equation is known: and u will be known from the equations u=vť1, and v = t2a (t2d ̧)1 w = 1oa (t ̄1d1)1w. For d in this example we may write Dt, and c may be a quasi-constant: if d be used, then c may be a function of dx, dy,.... which falls under the last example. We may write D, for dt, and a quasi-constant for A. Now assume u = vtm, and put m (m − 1) = 2, which gives .. v = st2 0 + S2 0 =tF1(x)+F(x) + tƒ1 (§) +ƒ (§). The supernumerary constants are to be determined by substituting this result in equation (1), (see Ex. 4); |