Page:EB1911 - Volume 08.djvu/244

 particular solution whatever; but if there be one root 1 repeated r times, the terms A1 e1 x + ... + Ar er x must be replaced by (A1 + A2x + ... + Arxr−1)e1 x where A1, ... An are arbitrary constants; the remaining terms in the complementary function will similarly need alteration of form if there be other repeated roots.

To complete the solution of the differential equation we need some method of determining a particular integral u; we explain a procedure which is effective for this purpose in the cases in which R is a sum of terms of the form eax(x), where (x) is an integral polynomial in x; this includes cases in which R contains terms of the form cos bx·(x) or sin bx·(x). Denote d/dx by D; it is clear that if u be any function of x, D(eaxu) = eaxDu + aeaxu, or say, D(eaxu) = eax(D + a)u; hence D2(eaxu), i.e. $d ^{2}⁄dx^{2}$ (eaxu), being equal to D(eaxv), where v = (D + a)u, is equal to eax(D + a)v, that is to eax(D + a)2u. In this way we find Dn(eaxu) = eax(D + a)nu, where n is any positive integer. Hence if (D) be any polynomial in D with constant coefficients, (D) (eaxu) = eax(D + a)u. Next, denoting ∫ udx by D−1u, and any solution of the differential equation $dz⁄dx$ + az = u by z = (d + a)−1u, we have D[eax(D + a)−1u] = D(eaxz) = eax(D + a)z = eaxu, so that we may write D−1(eaxu) = eax(D + a)−1u, where the meaning is that one value of the left side is equal to one value of the right side; from this, the expression D-2(eaxu), which means D−1[D−1(eaxu)], is equal to D−1(eaxz) and hence to eax(D + a)−1z, which we write eax(D + a)-2u; proceeding thus we obtain D-n(eaxu)＝eax(D + a)-nu, where n is any positive integer, and the meaning, as before, is that one value of the first expression is equal to one value of the second. More generally, if (D) be any polynomial in D with constant coefficients, and we agree to denote by [1/(D)]u any solution z of the differential equation (D)z = u, we have, if v = [1/(D + a)]u, the identity (D)(eaxv) = eax(D + a)v = eaxu, which we write in the form $\frac{1}{\psi (\mathrm{D})}(e^{ax}u) = e^{ax}\frac{1}{\psi (\mathrm{D} + a)}u.$ This gives us the first step in the method we are explaining, namely that a solution of the differential equation (D)y = eaxu + ebxv + ... where u, v, ... are any functions of x, is any function denoted by the expression $e^{ax}\frac{1}{\psi (\mathrm{D} + a)}u + e^{bx}\frac{1}{\psi (\mathrm{D} + b)}v + \ldots .$

It is now to be shown how to obtain one value of $$\frac{1}{\psi (\mathrm{D} + a)}u$$, when u is a polynomial in x, namely one solution of the differential equation (D + a)z = u. Let the highest power of x entering in u be xm; if t were a variable quantity, the rational fraction in t, $$\frac{1}{\psi (t + a)}u$$ by first writing it as a sum of partial fractions, or otherwise, could be identically written in the form Krt-r + Kr−1t-r+1 + ... + K1t−1 + H + H1t + ... + Hmtm + tm+1(t)/(t + a), where (t) is a polynomial in t; this shows that there exists an identity of the form 1＝(t + a)(Krt−r + ... + K1t−1 + H + H1t + ... + Hmtm) + (t)tm+1, and hence an identity u＝(D + a) [KrD−r + ... + K1D−1 + H + H1D + ... + HmDm] u + (D) Dm+1u; in this, since u contains no power of x higher than xm, the second term on the right may be omitted. We thus reach the conclusion that a solution of the differential equation (D + a)z = u is given by z＝(KrD−r + ... + K1D−1 + H + H1D + ... + HmDm)u, of which the operator on the right is obtained simply by expanding 1/(D + a) in ascending powers of D, as if D were a numerical quantity, the expansion being carried as far as the highest power of D which, operating upon u, does not give zero. In this form every term in z is capable of immediate calculation.

Example.—For the equation $\frac{d^{4}v}{dx^{4}}+ 2\frac{d^2y}{dx^{3}} + y = x^3 \cos x$ or $\displaystyle(\mathrm{D}^2 + 1)^2 = x^3 \cos x,$|undefined the roots of the associated algebraic equation (2 + 1)2 = 0 are = ±i, each repeated; the complementary function is thus (A + Bx)eix + (C + Dx)e−ix, where A, B, C, D are arbitrary constants; this is the same as (H + Kx) cos x + (M + Nx) sin x, where H, K, M, N are arbitrary constants. To obtain a particular integral we must find a value of (1 + D2)−2x3 cos x; this is the real part of (1 + D2)−2 eixx3 and hence of eix [1 + (D + i)2]−2 x3 or eix [2iD(1 + iD)]−2 x3, or −eix D−2 (1 + iD − D2 − iD3 + D4 + iD5 ...)x3, or −eix (x5 + ix4 − x3 − ix2 + x +  i); the real part of this is − ( x5 − x2 + x) cos x + (x4 − x2 + ) sin x. This expression added to the complementary function found above gives the complete integral; and no generality is lost by omitting from the particular integral the terms −x cos x + sin x, which are of the types of terms already occurring in the complementary function.

The symbolical method which has been explained has wider applications than that to which we have, for simplicity of explanation, restricted it. For example, if (x) be any function of x, and a1, a2, ... an be different constants, and [(t + a1) (t + a2) ... (t + an)]−1 when expressed in partial fractions be written Σcm(t + am)−1, a particular integral of the differential equation (D + a1)(D + a2) ... (D + an)y = (x) is given by y＝Σcm(D + am)−1 (x)＝Σcm (D + am)−1 e−amxeamx (x)＝Σcme−amxD−1 (eamx(x) )＝Σcme−amx &int; eamx(x)dx. The particular integral is thus expressed as a sum of n integrals. A linear differential equation of which the left side has the form $x^{n}\frac{d^{n}y}{dx^{n}} + \mathrm{P}_{1}x^{n- 1}\frac{d^{n- 1}y}{dx^{n- 1}} + \ldots + \mathrm{P}_{n- 1}x\frac{dy}{dx} + \mathrm{P}_{n}y,$|undefined where P1, ... Pn are constants, can be reduced to the case considered above. Writing x = et we have the identity $x^{m}\frac{d^{m}u}{dx^{m}} = \theta (\theta - 1)(\theta - 2) \ldots (\theta - m + 1)u,$ where ＝d/dt|undefined

When the linear differential equation, which we take to be of the second order, has variable coefficients, though there is no general rule for obtaining a solution in finite terms, there are some results which it is of advantage to have in mind. We have seen that if one solution of the equation obtained by putting the right side zero, say y1, be known, the equation can be solved. If y2 be another solution of $\frac{d^2y}{dx^2} + \mathrm{P}\frac{dy}{dx} + \mathrm{Q} = 0,$ there being no relation of the form my1 + ny2 = k, where m, n, k are constants, it is easy to see that $\frac{d}{dx}(y_{1}'y_{2} - y_{1}y_{2}') = \mathrm{P}(y_{1}'y_{2} - y_{1}y_{2}'),$ so that we have y1′y2 − y1y2′＝A exp. (&int; Pdx), where A is a suitably chosen constant, and exp. z denotes ez. In terms of the two solutions y1, y2 of the differential equation having zero on the right side, the general solution of the equation with R = (x) on the right side can at once be verified to be Ay1 + By2 + y1u − y2v, where u, v respectively denote the integrals u＝&int; y2(x) (y1′y2 − y2′y1)−1dx, v＝&int; y1(x) (y1′y2 − y2′y1)−1dx.

The equation $\frac{d^2y}{dx^2} + \mathrm{P}\frac{dy}{dx} + \mathrm{Q} = 0,$ by writing y = v exp. (− &int; Pdx), is at once seen to be reduced to d 2v/dx2 + v = 0, where = Q − dP/dx − P2. If = − 1/v dv/dx, the equation d 2v/dx2 + v = 0 becomes d/dx =  + 2, a non-linear equation of the first order.

More generally the equation $\frac{d\eta}{dx} = \mathrm{A} + \mathrm{B}\eta + \mathrm{C}\eta^2$ where A, B, C are functions of x, is, by the substitution $\eta = - \frac{1}{\mathrm{C}y}\frac{dy}{dx}$ reduced to the linear equation $\frac{d^2y}{dx^2} - \left(\mathrm{B} + \frac{1}{\mathrm{C}}\frac{d\mathrm{C}}{dx}\right)\frac{dy}{dx} + \mathrm{AC} = 0.$|undefined The equation $\frac{d\eta}{dx} = \mathrm{A} + \mathrm{B}\eta + \mathrm{C}\eta^2$ known as Riccati’s equation, is transformed into an equation of the same form by a substitution of the form = (aY + b)/(cY + d), where a, b, c, d are any functions of x, and this fact may be utilized to obtain a solution when A, B, C have special forms; in particular if any particular solution of the equation be known, say 0, the