Integration of a differential equation

D.V. Ionescu
(original), ODEs (theory), paper, works

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D.V. Ionescu, Institutul de Calcul

Dumitru V. Ionescu

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D.V. Ionescu, Integrarea unei ecuații diferențiale (Romanian) Acad. R. P. Romîne. Fil. Cluj. Stud. Cerc. Mat. 8 1957 275–289.
[Integration of a differential equation]

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Studii si Cercetari Matematice

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Academy of the Republic of S.R.

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https://ictp.acad.ro/scm/journal/article/view/16

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INTEGRATION OF A DIFFERENTIAL EQUATION

OFDV IONESCU

In this paper we will deal with the integration of the differential equation
(1) Δ n [ y ] = | y y ′ … y ( n ) y ′ y ′ ′ … y ( n + 1 ) ⋅ ⋅ ⋅ ⋅ y ( n ) y ( n + 1 ) … y ( 2 n ) | = 0 (1) Δ n [ y ] = y y ′ … y ( n ) y ′ y ′ ′ … y ( n + 1 ) ⋅ ⋅ ⋅ ⋅ y ( n ) y ( n + 1 ) … y ( 2 n ) = 0 {:(1)Delta_(n)[y]=|[y,y^('),dots,y^((n))],[y^('),y^(''),dots,y^((n+1))],[*,*,*,*],[y^((n)),y^((n+1)),dots,y^((2n))]|=0:}\Delta_{n}[y]=\left|\begin{array}{cccc} y & y^{\prime} & \ldots & y^{(n)} \tag{1}\\ y^{\prime} & y^{\prime \prime} & \ldots & y^{(n+1)} \\ \cdot & \cdot & \cdot & \cdot \\ y^{(n)} & \ldots & y^{(n+1)} \end{array}\right|=0(1)Δn[y]=|yy′…y(n)y′y′′…y(n+1)⋅⋅⋅⋅y(n)y(n+1)…y(2n)|=0
This problem was posed by Prof. T. Popoviciu during a paper given at the Institute of Computing in Cluj regarding a paper by H. Lö w ne r on monotone matrix functions [1].
We will also integrate the differential equation
(2) Δ n [ y ] = A it is α x (2) Δ n [ y ] = A it is α x {:(2)Delta_(n)[y]=Ae^(alpha x):}\begin{equation*} \Delta_{n}[y]=A e^{\alpha x} \tag{2} \end{equation*}(2)Δn[y]=Ait isαx
where A A AAAand α α alpha\alphaαare constants.
We recall that
(3) Δ 1 [ y ] = 1 (3) Δ 1 [ y ] = 1 {:(3)Delta_(1)[y]=1:}\begin{equation*} \Delta_{1}[y]=1 \tag{3} \end{equation*}(3)Δ1[y]=1
is the equation of the chain and that the equation
(4) Δ 2 [ y ] = 1 (4) Δ 2 [ y ] = 1 {:(4)Delta_(2)[y]=1:}\begin{equation*} \Delta_{2}[y]=1 \tag{4} \end{equation*}(4)Δ2[y]=1
was studied by G. Darboux [2]. Indications for the integration of the differential equation (2) were given by G. Darboux [3]. We will not follow the method given by G. Darboux, but will give a direct method for the integration of equations (1) and (2), which we will then apply to the particular cases (3) and (4.)
  1. Whether y y yyyan integral of the differential equation Δ n ( y ) Δ n ( y ) Delta_(n)(y)\Delta_{n}(y)Δn(y)continuous and with successive derivatives continuous up to the order 2 n 2 n 2n2 n2nincluding within an interval ( α , β α , β alpha,beta\alpha, \betaα,β), which can be ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞)It can identically cancel the coefficient of y ( 2 n ) y ( 2 n ) y^((2n))y^{(2 n)}y(2n)from the equation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0, that is, on Δ n − 1 [ y ] Δ n − 1 [ y ] Delta_(n-1)[y]\Delta_{n-1}[y]Δn−1[y], or not. We will deal further with the first case and now consider the case when Δ n − 1 [ y ] Δ n − 1 [ y ] Delta_(n-1)[y]\Delta_{n-1}[y]Δn−1[y]is not identically null in the interval ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β). Then for a point x 0 x 0 x_(0)x_{0}x0from this interval, we have Δ n − 1 [ y ( x 0 ) ] ≠ 0 Δ n − 1 y x 0 ≠ 0 Delta_(n-1)[y(x_(0))]!=0\Delta_{n-1}\left[y\left(x_{0}\right)\right] \neq 0Δn−1[y(x0)]≠0and Δ n − 1 [ y ] Δ n − 1 [ y ] Delta_(n-1)[y]\Delta_{n-1}[y]Δn−1[y]being a continuous function of x x xxxin the interval ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β)
    , an interval can be determined. ( A , b ) ( A , b ) (a,b)(a, b)(A,b)included in the range ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β)and which contains the point x 0 x 0 x_(0)x_{0}x0, for which Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0We will therefore place ourselves in the interval ( A , b A , b a,ba, bA,b) and we will determine the integrals y y yyyof the differential equation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0, which are such that Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0Once these integrals are determined, we will show that the intervals ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β)and ( A , b ) ( A , b ) (a,b)(a, b)(A,b)can be extended to the range ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞).
Let us determine for such an integral y y yyy, functions λ 0 ( x ) , λ 1 ( x ) … , λ n − 1 ( x ) λ 0 ( x ) , λ 1 ( x ) … , λ n − 1 ( x ) lambda_(0)(x),lambda_(1)(x)dots,lambda_(n-1)(x)\lambda_{0}(x), \lambda_{1}(x) \ldots, \lambda_{n-1}(x)λ0(x),λ1(x)…,λn−1(x)through linear equations
(5) λ 0 ( x ) y + λ 1 ( x ) y ′ + … + λ n − 1 ( x ) y ( n − 1 ) + y ( n ) = 0 λ 0 ( x ) y ′ + λ 1 ( x ) y ′ ′ + … + λ n − 1 ( x ) y ( n ) + y ( n + 1 ) = 0 λ 0 ( x ) y ( n − 1 ) + λ 1 ( x ) y ( n ) + … + λ n − 1 ( x ) y ( 2 n − 2 ) + y ( 2 n − 1 ) = 0 (5) λ 0 ( x ) y + λ 1 ( x ) y ′ + … + λ n − 1 ( x ) y ( n − 1 ) + y ( n ) = 0 λ 0 ( x ) y ′ + λ 1 ( x ) y ′ ′ + … + λ n − 1 ( x ) y ( n ) + y ( n + 1 ) = 0 λ 0 ( x ) y ( n − 1 ) + λ 1 ( x ) y ( n ) + … + λ n − 1 ( x ) y ( 2 n − 2 ) + y ( 2 n − 1 ) = 0 {:(5){:[lambda_(0)(x)y+lambda_(1)(x)y^(')+dots+lambda_(n-1)(x)y^((n-1))+y^((n))=0],[lambda_(0)(x)y^(')+lambda_(1)(x)y^('')+dots+ lambda_(n-1)(x)y^((n))+y^((n+1))=0],[lambda_(0)(x)y^((n-1))+lambda_(1)(x)y^((n))+dots+lambda_(n-1)(x)y^((2n-2))+y^((2n-1))=0]:}:}\begin{array}{lll} \lambda_{0}(x) y+\lambda_{1}(x) y^{\prime}+\ldots+\lambda_{n-1}(x) y^{(n-1)}+y^{(n)}=0 \\ \lambda_{0}(x) y^{\prime}+\lambda_{1}(x) y^{\prime}+\lambda_{n-1}(x) y^{(n-1)}+y^{(n)}=0 y^{(n)}+y^{(n+1)}=0 \tag{5}\\ \lambda_{0}(x) y^{(n-1)}+\lambda_{1}(x) y^{(n)}+\ldots+\lambda_{n-1}(x) y^{(2 n-2)}+y^{(2 n-1)}=0 \end{array}(5)λ0(x)y+λ1(x)y′+…+λn−1(x)y(n−1)+y(n)=0λ0(x)y′+λ1(x)y′′+…+λn−1(x)y(n)+y(n+1)=0λ0(x)y(n−1)+λ1(x)y(n)+…+λn−1(x)y(2n−2)+y(2n−1)=0
This is possible because the determinant of the system is Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0Functions λ 0 ( x ) , λ 1 ( x ) , … , λ n − 1 ( x ) λ 0 ( x ) , λ 1 ( x ) , … , λ n − 1 ( x ) lambda_(0)(x),lambda_(1)(x),dots,lambda_(n-1)(x)\lambda_{0}(x), \lambda_{1}(x), \ldots, \lambda_{n-1}(x)λ0(x),λ1(x),…,λn−1(x)are also differentiable. Because of the differential equation (1), we can add to the previous equations the equation
( ′ ) λ 0 ( x ) y ( n ) + λ 1 ( x ) y ( n + 1 ) + … + λ n − 1 ( x ) y ( 2 n − 1 ) + y ( 2 n ) = 0 ( ′ ) λ 0 ( x ) y ( n ) + λ 1 ( x ) y ( n + 1 ) + … + λ n − 1 ( x ) y ( 2 n − 1 ) + y ( 2 n ) = 0 {:('")"lambda_(0)(x)y^((n))+lambda_(1)(x)y^((n+1))+dots+lambda_(n-1)(x)y^((2n-1))+y^((2n))=0:}\begin{equation*} \lambda_{0}(x) y^{(n)}+\lambda_{1}(x) y^{(n+1)}+\ldots+\lambda_{n-1}(x) y^{(2 n-1)}+y^{(2 n)}=0 \tag{$\prime$} \end{equation*}(′)λ0(x)y(n)+λ1(x)y(n+1)+…+λn−1(x)y(2n−1)+y(2n)=0
Deriving each equation (5) in turn and taking into account the next equation, the last one being ( 5 ′ 5 ′ 5^(')5^{\prime}5′), the equations are derived
λ 0 ′ ( x ) y + λ 1 ′ ( x ) y ′ + … + λ n − 1 ′ ( x ) y ( n − 1 ) = 0 λ 0 ′ ( x ) y ′ + λ 1 ′ ( x ) y ′ ′ + … + λ n − 1 ′ ( x ) y ( n ) = 0 λ 0 ′ ( x ) y ( n − 1 ) + λ 1 ′ ( x ) y ( n ) + … + λ n − 1 ′ ( x ) y ( 2 n − 2 ) = 0 λ 0 ′ ( x ) y + λ 1 ′ ( x ) y ′ + … + λ n − 1 ′ ( x ) y ( n − 1 ) = 0 λ 0 ′ ( x ) y ′ + λ 1 ′ ( x ) y ′ ′ + … + λ n − 1 ′ ( x ) y ( n ) = 0 λ 0 ′ ( x ) y ( n − 1 ) + λ 1 ′ ( x ) y ( n ) + … + λ n − 1 ′ ( x ) y ( 2 n − 2 ) = 0 {:[lambda_(0)^(')(x)y+lambda_(1)^(')(x)y^(')+dots+lambda_(n-1)^(')(x)y^((n-1))=0],[lambda_(0)^(')(x)y^(')+lambda_(1)^(')(x)y^('' )+dots+lambda_(n-1)^(')(x)y^((n))=0],[lambda_(0)^(')(x)y^((n-1))+lambda_(1)^(')(x)y^((n))+dots+lambda_(n-1)^(')(x)y^((2n-2))=0]:}\begin{aligned} & \lambda_{0}^{\prime}(x) y+\lambda_{1}^{\prime}(x) y^{\prime}+\ldots+\lambda_{n-1}^{\prime}(x) y^{(n-1)}=0 \\ & \lambda_{0}^{\prime}(x) y^{\prime}(x) \prime}+\ldots+\lambda_{n-1}^{\prime}(x) y^{(n)}=0 \\ & \lambda_{0}^{\prime}(x) y^{(n-1)}+\lambda_{1}^{\prime}(x) y^{(n)}+\ldots+\lambda_{n-1}^{\prime}(x) y^{(2 n-2)}=0 \end{aligned}λ0′(x)y+λ1′(x)y′+…+λn−1′(x)y(n−1)=0λ0′(x)y′+λ1′(x)y′′+…+λn−1′(x)y(n)=0λ0′(x)y(n−1)+λ1′(x)y(n)+…+λn−1′(x)y(2n−2)=0
linear and homogeneous in λ 0 ′ ( x ) , λ 1 ′ ( x ) , … , λ n − 1 ′ ( x ) λ 0 ′ ( x ) , λ 1 ′ ( x ) , … , λ n − 1 ′ ( x ) lambda_(0)^(')(x),lambda_(1)^(')(x),dots,lambda_(n-1)^(')(x)\lambda_{0}^{\prime}(x), \lambda_{1}^{\prime}(x), \ldots, \lambda_{n-1}^{\prime}(x)λ0′(x),λ1′(x),…,λn−1′(x)with the determinant Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0It follows that
λ 0 ′ ( x ) = 0 , λ 1 ′ ( x ) = 0 , … , λ n − 1 ′ ( x ) = 0 λ 0 ′ ( x ) = 0 , λ 1 ′ ( x ) = 0 , … , λ n − 1 ′ ( x ) = 0 lambda_(0)^(')(x)=0,quadlambda_(1)^(')(x)=0,dots,quadlambda_(n-1)^(')(x)=0\lambda_{0}^{\prime}(x)=0, \quad \lambda_{1}^{\prime}(x)=0, \ldots, \quad \lambda_{n-1}^{\prime}(x)=0λ0′(x)=0,λ1′(x)=0,…,λn−1′(x)=0
MEAN
λ 0 ( x ) = A n , λ 1 ( x ) = A n − 1 , … , λ n − 1 ( x ) = A 1 λ 0 ( x ) = A n , λ 1 ( x ) = A n − 1 , … , λ n − 1 ( x ) = A 1 lambda_(0)(x)=A_(n),quadlambda_(1)(x)=A_(n-1),dots,quadlambda_(n-1)(x)=A_(1)\lambda_{0}(x)=A_{n}, \quad \lambda_{1}(x)=A_{n-1}, \ldots, \quad \lambda_{n-1}(x)=A_{1}λ0(x)=An,λ1(x)=An−1,…,λn−1(x)=A1
A 1 , A 2 , … , A n A 1 , A 2 , … , A n A_(1),A_(2),dots,A_(n)A_{1}, A_{2}, \ldots, A_{n}A1,A2,…,Anbeing constants attached to the function y ( x ) y ( x ) y(x)y(x)y(x); doing in the system of equations (5) x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0, we observe that A 1 , A 2 , … , A n A 1 , A 2 , … , A n A_(1),A_(2),dots,A_(n)A_{1}, A_{2}, \ldots, A_{n}A1,A2,…,An, are given by the equations
A n y 0 + A n − 1 y 0 ′ + … + A 1 y 0 ( n − 1 ) + y 0 ( n ) = 0 A n y 0 ′ + A n − 1 y 0 ′ ′ + … + A 1 y 0 ( n ) + y 0 ( n + 1 ) = 0 (6) A n y 0 ( n − 1 ) + A n − 1 y 0 ( n ) + … + A 1 y 0 ( 2 n − 2 ) + y 0 ( 2 n − 1 ) = 0 A n y 0 + A n − 1 y 0 ′ + … + A 1 y 0 ( n − 1 ) + y 0 ( n ) = 0 A n y 0 ′ + A n − 1 y 0 ′ ′ + … + A 1 y 0 ( n ) + y 0 ( n + 1 ) = 0 (6) A n y 0 ( n − 1 ) + A n − 1 y 0 ( n ) + … + A 1 y 0 ( 2 n − 2 ) + y 0 ( 2 n − 1 ) = 0 {:[A_(n)y_(0)+A_(n-1)y_(0)^(')+dots+A_(1)y_(0)^((n-1))+y_(0)^((n))=0],[A_(n)y_(0)^(')+A_(n-1)y_(0)^('')+dots+A_(1)y_(0)^((n))+y_(0)^((n+1))=0],[(6)A_(n)y_(0)^((n-1))+A_(n-1)y_(0)^((n))+dots+A_(1)y_(0)^((2n-2))+y_(0)^((2n-1))=0]:}\begin{align*} & A_{n} y_{0}+A_{n-1} y_{0}^{\prime}+\ldots+A_{1} y_{0}^{(n-1)}+y_{0}^{(n)}=0 \\ & A_{n} y_{0}^{\prime}+A_{n-1} y_{0}^{\prime \prime}+\ldots+A_{1} y_{0}^{(n)}+y_{0}^{(n+1)}=0 \\ & A_{n} y_{0}^{(n-1)}+A_{n-1} y_{0}^{(n)}+\ldots+A_{1} y_{0}^{(2 n-2)}+y_{0}^{(2 n-1)}=0 \tag{6} \end{align*}Any0+An−1y0′+…+A1y0(n−1)+y0(n)=0Any0′+An−1y0′′+…+A1y0(n)+y0(n+1)=0(6)Any0(n−1)+An−1y0(n)+…+A1y0(2n−2)+y0(2n−1)=0
where
(7) y ( k ) ( x 0 ) = y 0 ( k ) ( k = 0 , 1 , 2 , … , 2 n − 1 ) (7) y ( k ) x 0 = y 0 ( k ) ( k = 0 , 1 , 2 , … , 2 n − 1 ) {:(7)y^((k))(x_(0))=y_(0)^((k))quad(k=0","1","2","dots","2n-1):}\begin{equation*} y^{(k)}\left(x_{0}\right)=y_{0}^{(k)} \quad(k=0,1,2, \ldots, 2 n-1) \tag{7} \end{equation*}(7)y(k)(x0)=y0(k)(k=0,1,2,…,2n−1)
are Cauchy's conditions.
It follows that any integral of the differential equation (1) for which Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0, is the integral of the differential equation
(8) y ( n ) + A 1 y ( n − 1 ) + … + A n y = 0 (8) y ( n ) + A 1 y ( n − 1 ) + … + A n y = 0 {:(8)y^((n))+A_(1)y^((n-1))+dots+A_(n)y=0:}\begin{equation*} y^{(n)}+A_{1} y^{(n-1)}+\ldots+A_{n} y=0 \tag{8} \end{equation*}(8)y(n)+A1y(n−1)+…+Any=0
with constant coefficients A 1 , A 2 , … , A n A 1 , A 2 , … , A n A_(1),A_(2),dots,A_(n)A_{1}, A_{2}, \ldots, A_{n}A1,A2,…,Andetermined by the system of linear equations (6) by Cauchy's conditions.
Conversely, any integral of the equation with constant coefficients (8), for which Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0, is the integral of the differential equation (1), which is immediately proven.
2. The characteristic equation of the differential equation (8) is obtained by eliminating A 1 , A 2 , … , A n A 1 , A 2 , … , A n A_(1),A_(2),dots,A_(n)A_{1}, A_{2}, \ldots, A_{n}A1,A2,…,An, between equations (6) and equation
(9) r n + A 1 r n − 1 + … + A n = 0 (9) r n + A 1 r n − 1 + … + A n = 0 {:(9)r^(n)+A_(1)r^(n-1)+dots+A_(n)=0:}\begin{equation*} r^{n}+A_{1} r^{n-1}+\ldots+A_{n}=0 \tag{9} \end{equation*}(9)Rn+A1Rn−1+…+An=0
This is possible because it was assumed Δ n − 1 [ y ( x 0 ) ] ≠ 0 Δ n − 1 y x 0 ≠ 0 Delta_(n-1)[y(x_(0))]!=0\Delta_{n-1}\left[y\left(x_{0}\right)\right] \neq 0Δn−1[y(x0)]≠0The characteristic equation is therefore
( ′ ) φ ( r ) = | 1 r r 2 … r n y 0 y 0 ′ y 0 ′ ′ … y 0 ( n ) y 0 ′ y 0 ′ ′ y 0 ′ ′ ′ … y 0 ( n + 1 ) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ y 0 ( n − 1 ) y 0 ( n ) y 0 ( n + 1 ) … y 0 ( 2 n − 1 ) | = 0 ( ′ ) φ ( r ) = 1 r r 2 … r n y 0 y 0 ′ y 0 ′ ′ … y 0 ( n ) y 0 ′ y 0 ′ ′ y 0 ′ ′ ′ … y 0 ( n + 1 ) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ y 0 ( n − 1 ) y 0 ( n ) y 0 ( n + 1 ) … y 0 ( 2 n − 1 ) = 0 {:('")"varphi(r)=|[1,r,r^(2),dots,r^(n)],[y_(0),y_(0)^('),y_(0)^(''),dots,y_(0)^((n))],[y_(0)^('),y_(0)^(''),y_(0)^('''),dots,y_(0)^((n+1))],[*,*,*,*,*],[*,*,*],[y_(0)^((n-1)),y_(0)^((n)),y_(0)^((n+1)),dots,y_(0)^((2n-1))]|=0:}\varphi(r)=\left|\begin{array}{ccccc} 1 & r & r^{2} & \ldots & r^{n} \tag{$\prime$}\\ y_{0} & y_{0}^{\prime} & y_{0}^{\prime \prime} & \ldots & y_{0}^{(n)} \\ y_{0}^{\prime} & y_{0}^{\prime \prime} & y_{0}^{\prime \prime \prime} & \ldots & y_{0}^{(n+1)} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ y_{0}^{(n-1)} & y_{0}^{(n)} & y_{0}^{(n+1)} & \ldots & y_{0}^{(2 n-1)} \end{array}\right|=0(′)φ(R)=|1RR2…Rny0y0′y0′′…y0(n)y0′y0′′y0′′′…y0(n+1)⋅⋅⋅⋅⋅⋅⋅⋅y0(n−1)y0(n)y0(n+1)…y0(2n−1)|=0
Let us assume that the initial conditions (7) are chosen such that the characteristic equation ( 9 ′ 9 ′ 9^(')9^{\prime}9′) to have all the roots r 1 , r 2 , … , r n r 1 , r 2 , … , r n r_(1),r_(2),dots,r_(n)r_{1}, r_{2}, \ldots, r_{n}R1,R2,…,Rndistinct. In this case
(10) y = C 1 e r 1 x + C 2 e r 2 x + … + C n e r n x (10) y = C 1 e r 1 x + C 2 e r 2 x + … + C n e r n x {:(10)y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)+dots+C_(n)e^(r_(n)x):}\begin{equation*} y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+\ldots+C_{n} e^{r_{n} x} \tag{10} \end{equation*}(10)y=C1it isR1x+C2it isR2x+…+Cnit isRnx
is an integral of the differential equation (8) and will be an integral of the differential equation (1), if we show that Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0, what happens when
(11) C 1 C 2 … C n ≠ 0 (11) C 1 C 2 … C n ≠ 0 {:(11)C_(1)C_(2)dotsC_(n)!=0:}\begin{equation*} C_{1} C_{2} \ldots C_{n} \neq 0 \tag{11} \end{equation*}(11)C1C2…Cn≠0
Indeed
Δ n − 1 [ y ] = | C 1 e r 1 x C 2 e r 2 x … C n e r n x C 1 r 1 e r 1 x C 2 r 2 e r 2 x … C n r n e r n x ⋅ ⋅ ⋅ ⋅ C 1 r 1 n − 1 e r 1 x C 2 r 2 n − 1 e r 2 x … C n r n n − 1 e r n x | ⋅ | 1 r 1 … r 1 n − 1 1 r 2 … r 2 n − 1 ⋅ ⋅ ⋅ ⋅ 1 r n … r n n − 1 | Δ n − 1 [ y ] = C 1 e r 1 x C 2 e r 2 x … C n e r n x C 1 r 1 e r 1 x C 2 r 2 e r 2 x … C n r n e r n x ⋅ ⋅ ⋅ ⋅ C 1 r 1 n − 1 e r 1 x C 2 r 2 n − 1 e r 2 x … C n r n n − 1 e r n x ⋅ 1 r 1 … r 1 n − 1 1 r 2 … r 2 n − 1 ⋅ ⋅ ⋅ ⋅ 1 r n … r n n − 1 Delta_(n-1)[y]=|[C_(1)e^(r_(1)x),C_(2)e^(r_(2)x),dots,C_(n)e^(r_(n)x)],[C_(1)r_(1)e^(r_(1)x),C_(2)r_(2)e^(r_(2)x),dots,C_(n)r_(n)e^(r_(n)x)],[*,*,*,*],[C_(1)r_(1)^(n-1)e^(r_(1)x),C_(2)r_(2)^(n-1)e^(r_(2)x),dots,C_(n)r_(n)^(n-1)e^(r_(n)x)]|*|[1,r_(1),dots,r_(1)^(n-1)],[1,r_(2),dots,r_(2)^(n-1)],[*,*,*,*],[1,r_(n),dots,r_(n)^(n-1)]|\Delta_{n-1}[y]=\left|\begin{array}{cccc} C_{1} e^{r_{1} x} & C_{2} e^{r_{2} x} & \ldots & C_{n} e^{r_{n} x} \\ C_{1} r_{1} e^{r_{1} x} & C_{2} r_{2} e^{r_{2} x} & \ldots & C_{n} r_{n} e^{r_{n} x} \\ \cdot & \cdot & \cdot & \cdot \\ C_{1} r_{1}^{n-1} e^{r_{1} x} & C_{2} r_{2}^{n-1} e^{r_{2} x} & \ldots & C_{n} r_{n}^{n-1} e^{r_{n} x} \end{array}\right| \cdot\left|\begin{array}{cccc} 1 & r_{1} & \ldots & r_{1}^{n-1} \\ 1 & r_{2} & \ldots & r_{2}^{n-1} \\ \cdot & \cdot & \cdot & \cdot \\ 1 & r_{n} & \ldots & r_{n}^{n-1} \end{array}\right|Δn−1[y]=|C1it isR1xC2it isR2x…Cnit isRnxC1R1it isR1xC2R2it isR2x…CnRnit isRnx⋅⋅⋅⋅C1R1n−1it isR1xC2R2n−1it isR2x…CnRnn−1it isRnx|⋅|1R1…R1n−11R2…R2n−1⋅⋅⋅⋅1Rn…Rnn−1|
MEAN
(12) Δ n − 1 [ y ] = C 1 C 2 … C n V 2 ( r 1 , r 2 , … , r n ) e ( r 1 + r 2 + … + r n ) x (12) Δ n − 1 [ y ] = C 1 C 2 … C n V 2 r 1 , r 2 , … , r n e r 1 + r 2 + … + r n x {:(12)Delta_(n-1)[y]=C_(1)C_(2)dotsC_(n)V^(2)(r_(1),r_(2),dots,r_(n))e^((r_(1)+r_(2)+dots+r_(n))x):}\begin{equation*} \Delta_{n-1}[y]=C_{1} C_{2} \ldots C_{n} V^{2}\left(r_{1}, r_{2}, \ldots, r_{n}\right) e^{\left(r_{1}+r_{2}+\ldots+r_{n}\right) x} \tag{12} \end{equation*}(12)Δn−1[y]=C1C2…CnV2(R1,R2,…,Rn)it is(R1+R2+…+Rn)x
where V ( r 1 , r 2 , … , r n ) V r 1 , r 2 , … , r n V(r_(1),r_(2),dots,r_(n))V\left(r_{1}, r_{2}, \ldots, r_{n}\right)V(R1,R2,…,Rn)is the Vandermonde determinant of distinct numbers r 1 , r 2 , … , r n r 1 , r 2 , … , r n r_(1),r_(2),dots,r_(n)r_{1}, r_{2}, \ldots, r_{n}R1,R2,…,Rn. When condition (11) is satisfied, formula (10) is an integral of the differential equation (1). This integral is defined in the interval ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞)and the condition Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0is valid in the interval ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞).
The integral that corresponds to Cauchy's conditions (7) is obtained by determining A 1 , A 2 , … , A n A 1 , A 2 , … , A n A_(1),A_(2),dots,A_(n)A_{1}, A_{2}, \ldots, A_{n}A1,A2,…,Anfrom equations (6) and then putting
(13) C 1 e r 1 x 0 = C 1 ′ , C 1 e r 2 x 0 = C 2 ′ , … , C n e r n x 0 = C n ′ (13) C 1 e r 1 x 0 = C 1 ′ , C 1 e r 2 x 0 = C 2 ′ , … , C n e r n x 0 = C n ′ {:(13)C_(1)e^(r_(1)x_(0))=C_(1)^(')","quadC_(1)e^(r_(2)x_(0))=C_(2)^(')","dots","C_(n)e^(r_(n)x_(0))=C_(n)^('):}\begin{equation*} C_{1} e^{r_{1} x_{0}}=C_{1}^{\prime}, \quad C_{1} e^{r_{2} x_{0}}=C_{2}^{\prime}, \ldots, C_{n} e^{r_{n} x_{0}}=C_{n}^{\prime} \tag{13} \end{equation*}(13)C1it isR1x0=C1′,C1it isR2x0=C2′,…,Cnit isRnx0=Cn′
it remains to be determined C 1 ′ , C 2 ′ , … , C n ′ C 1 ′ , C 2 ′ , … , C n ′ C_(1)^('),C_(2)^('),dots,C_(n)^(')C_{1}^{\prime}, C_{2}^{\prime}, \ldots, C_{n}^{\prime}C1′,C2′,…,Cn′from the system of equations
C 1 ′ + C 2 ′ + … + C n ′ = y 0 (14) C 1 ′ v 1 + C 2 ′ v 2 + … + C n ′ v n = y 0 ′ C 1 ′ v 1 n − 1 + C 2 ′ v 2 n − 1 + … + C n ′ v n n − 1 = y 0 ( n − 1 ) C 1 ′ + C 2 ′ + … + C n ′ = y 0 (14) C 1 ′ v 1 + C 2 ′ v 2 + … + C n ′ v n = y 0 ′ C 1 ′ v 1 n − 1 + C 2 ′ v 2 n − 1 + … + C n ′ v n n − 1 = y 0 ( n − 1 ) {:[C_(1)^(')+C_(2)^(')+dots+C_(n)^(')=y_(0)],[(14)C_(1)^(')v_(1)+C_(2)^(')v_(2)+dots+C_(n)^(')v_(n)=y_(0)^(')],[C_(1)^(')v_(1)^(n-1)+C_(2)^(')v_(2)^(n-1)+dots+C_(n)^(')v_(n)^(n-1)=y_(0)^((n-1))]:}\begin{align*} & C_{1}^{\prime}+C_{2}^{\prime}+\ldots+C_{n}^{\prime}=y_{0} \\ & C_{1}^{\prime} v_{1}+C_{2}^{\prime} v_{2}+\ldots+C_{n}^{\prime} v_{n}=y_{0}^{\prime} \tag{14}\\ & C_{1}^{\prime} v_{1}^{n-1}+C_{2}^{\prime} v_{2}^{n-1}+\ldots+C_{n}^{\prime} v_{n}^{n-1}=y_{0}^{(n-1)} \end{align*}C1′+C2′+…+Cn′=y0(14)C1′V1+C2′V2+…+Cn′Vn=y0′C1′V1n−1+C2′V2n−1+…+Cn′Vnn−1=y0(n−1)
We note that it is not possible for all numbers C 1 ′ , … , C n ′ C 1 ′ , … , C n ′ C_(1)^('),dots,C_(n)^(')C_{1}^{\prime}, \ldots, C_{n}^{\prime}C1′,…,Cn′to be null. It would follow that y 0 = y 0 ′ = … = y 0 ( n − 1 ) = 0 y 0 = y 0 ′ = … = y 0 ( n − 1 ) = 0 y_(0)=y_(0)^(')=dots=y_(0)^((n-1))=0y_{0}=y_{0}^{\prime}=\ldots=y_{0}^{(n-1)}=0y0=y0′=…=y0(n−1)=0and so as Δ n − 1 [ y ( x 0 ) ] = 0 Δ n − 1 y x 0 = 0 Delta_(n-1)[y(x_(0))]=0\Delta_{n-1}\left[y\left(x_{0}\right)\right]=0Δn−1[y(x0)]=0, which is contrary to the hypothesis. Let us prove that it is not possible for even one of C 1 ′ , … , C n ′ C 1 ′ , … , C n ′ C_(1)^('),dots,C_(n)^(')C_{1}^{\prime}, \ldots, C_{n}^{\prime}C1′,…,Cn′to be null.
Indeed, suppose we had C 1 ′ = 0 C 1 ′ = 0 C_(1)^(')=0C_{1}^{\prime}=0C1′=0Then the numbers r 2 , … , r n r 2 , … , r n r_(2),dots,r_(n)r_{2}, \ldots, r_{n}R2,…,Rnbeing distinct, between the equations
C 2 ′ + C 3 ′ + … + C n ′ = y 0 C 2 ′ r 2 + C 3 ′ r 3 + … + C n ′ r n = y 0 ′ C 2 ′ r 2 n − 1 + C 3 ′ r 3 n − 1 + … + C n ′ r n n − 1 = y 0 ( n − 1 ) C 2 ′ + C 3 ′ + … + C n ′ = y 0 C 2 ′ r 2 + C 3 ′ r 3 + … + C n ′ r n = y 0 ′ C 2 ′ r 2 n − 1 + C 3 ′ r 3 n − 1 + … + C n ′ r n n − 1 = y 0 ( n − 1 ) {:[C_(2)^(')+C_(3)^(')+dots+C_(n)^(')=y_(0)],[C_(2)^(')r_(2)+C_(3)^(')r_(3)+dots+C_(n)^(')r_(n)=y_(0)^(')],[C_(2)^(')r_(2)^(n-1)+C_(3)^(')r_(3)^(n-1)+dots+C_(n)^(')r_(n)^(n-1)=y_(0)^((n-1))]:}\begin{aligned} & C_{2}^{\prime}+C_{3}^{\prime}+\ldots+C_{n}^{\prime}=y_{0} \\ & C_{2}^{\prime} r_{2}+C_{3}^{\prime} r_{3}+\ldots+C_{n}^{\prime} r_{n}=y_{0}^{\prime} \\ & C_{2}^{\prime} r_{2}^{n-1}+C_{3}^{\prime} r_{3}^{n-1}+\ldots+C_{n}^{\prime} r_{n}^{n-1}=y_{0}^{(n-1)} \end{aligned}C2′+C3′+…+Cn′=y0C2′R2+C3′R3+…+Cn′Rn=y0′C2′R2n−1+C3′R3n−1+…+Cn′Rnn−1=y0(n−1)
can be removed C 2 ′ , … , C n ′ C 2 ′ , … , C n ′ C_(2)^('),dots,C_(n)^(')C_{2}^{\prime}, \ldots, C_{n}^{\prime}C2′,…,Cn′. We obtain a relation of the form
(15) y 0 ( n − 1 ) + p 1 y 0 ( n − 2 ) + p 2 y 0 ( n − 3 ) + … + p n − 1 y 0 = 0 (15) y 0 ( n − 1 ) + p 1 y 0 ( n − 2 ) + p 2 y 0 ( n − 3 ) + … + p n − 1 y 0 = 0 {:(15)y_(0)^((n-1))+p_(1)y_(0)^((n-2))+p_(2)y_(0)^((n-3))+dots+p_(n-1)y_(0)=0:}\begin{equation*} y_{0}^{(n-1)}+p_{1} y_{0}^{(n-2)}+p_{2} y_{0}^{(n-3)}+\ldots+p_{n-1} y_{0}=0 \tag{15} \end{equation*}(15)y0(n−1)+p1y0(n−2)+p2y0(n−3)+…+pn−1y0=0
where
p 1 = − ( r 2 + r 3 + … + r n ) , p 2 = r 2 r 3 + … + r n − 1 r n , … p n − 1 = ( − 1 ) n − 1 r 2 r 3 … r n p 1 = − r 2 + r 3 + … + r n , p 2 = r 2 r 3 + … + r n − 1 r n , … p n − 1 = ( − 1 ) n − 1 r 2 r 3 … r n p_(1)=-(r_(2)+r_(3)+dots+r_(n)),quadp_(2)=r_(2)r_(3)+dots+r_(n-1)r_(n),dotsp_(n-1)=(-1)^(n-1)r_(2)r_(3)dotsr_(n)p_{1}=-\left(r_{2}+r_{3}+\ldots+r_{n}\right), \quad p_{2}=r_{2} r_{3}+\ldots+r_{n-1} r_{n}, \ldots p_{n-1}=(-1)^{n-1} r_{2} r_{3} \ldots r_{n}p1=−(R2+R3+…+Rn),p2=R2R3+…+Rn−1Rn,…pn−1=(−1)n−1R2R3…Rn
In equations (6) let us replace
A 1 = p 1 − r 1 , A 2 = p 2 − p 1 r 1 , A 3 = p 3 − p 2 r 1 , … , A n − 1 = p n − 1 − p n − 2 r 1 A n = − p n − 1 r 1 A 1 = p 1 − r 1 , A 2 = p 2 − p 1 r 1 , A 3 = p 3 − p 2 r 1 , … , A n − 1 = p n − 1 − p n − 2 r 1 A n = − p n − 1 r 1 A_(1)=p_(1)-r_(1),A_(2)=p_(2)-p_(1)r_(1),A_(3)=p_(3)-p_(2)r_(1),dots,A_(n-1)=p_(n-1)-p_(n-2)r_(1)A_(n)=-p_(n-1)r_(1)A_{1}=p_{1}-r_{1}, A_{2}=p_{2}-p_{1} r_{1}, A_{3}=p_{3}-p_{2} r_{1}, \ldots, A_{n-1}=p_{n-1}-p_{n-2} r_{1} A_{n}=-p_{n-1} r_{1}A1=p1−R1,A2=p2−p1R1,A3=p3−p2R1,…,An−1=pn−1−pn−2R1An=−pn−1R1The first
equation (6) is written
y 0 ( n ) + ( p 1 − r 1 ) y 0 ( n − 1 ) + ( p 2 − p 1 r 1 ) y 0 ( n − 2 ) + … + ( p n − 1 − p n − 2 r 1 ) y 0 ′ − p n − 1 r 1 y 0 = 0 y 0 ( n ) + p 1 − r 1 y 0 ( n − 1 ) + p 2 − p 1 r 1 y 0 ( n − 2 ) + … + p n − 1 − p n − 2 r 1 y 0 ′ − p n − 1 r 1 y 0 = 0 y_(0)^((n))+(p_(1)-r_(1))y_(0)^((n-1))+(p_(2)-p_(1)r_(1))y_(0)^((n-2))+dots+(p_(n-1)-p_(n-2)r_(1))y_(0)^(')-p_(n-1)r_(1)y_(0)=0y_{0}^{(n)}+\left(p_{1}-r_{1}\right) y_{0}^{(n-1)}+\left(p_{2}-p_{1} r_{1}\right) y_{0}^{(n-2)}+\ldots+\left(p_{n-1}-p_{n-2} r_{1}\right) y_{0}^{\prime}-p_{n-1} r_{1} y_{0}=0y0(n)+(p1−R1)y0(n−1)+(p2−p1R1)y0(n−2)+…+(pn−1−pn−2R1)y0′−pn−1R1y0=0
and taking into account equation (15) the coefficient of r 1 r 1 r_(1)r_{1}R1is zero, and the equation reduces to
( ′ ) y 0 ( n ) + p 1 y 0 ( n − 1 ) + p 2 y 0 ( n − 2 ) + … + p n − 1 y 0 ′ = 0 . ( ′ ) y 0 ( n ) + p 1 y 0 ( n − 1 ) + p 2 y 0 ( n − 2 ) + … + p n − 1 y 0 ′ = 0 . {:('")"y_(0)^((n))+p_(1)y_(0)^((n-1))+p_(2)y_(0)^((n-2))+dots+p_(n-1)y_(0)^(')=0.:}\begin{equation*} y_{0}^{(n)}+p_{1} y_{0}^{(n-1)}+p_{2} y_{0}^{(n-2)}+\ldots+p_{n-1} y_{0}^{\prime}=0 . \tag{$\prime$} \end{equation*}(′)y0(n)+p1y0(n−1)+p2y0(n−2)+…+pn−1y0′=0.
Further, taking this into account, the second equation (6) reduces to
( ′ ′ ) y 0 ( n + 1 ) + p 1 y 0 ( n ) + p 2 y 0 ( n − 1 ) + … + p n − 1 y 0 ′ ′ = 0 ( ′ ′ ) y 0 ( n + 1 ) + p 1 y 0 ( n ) + p 2 y 0 ( n − 1 ) + … + p n − 1 y 0 ′ ′ = 0 {:(('')")"y_(0)^((n+1))+p_(1)y_(0)^((n))+p_(2)y_(0)^((n-1))+dots+p_(n-1)y_(0)^('')=0:}\begin{equation*} y_{0}^{(n+1)}+p_{1} y_{0}^{(n)}+p_{2} y_{0}^{(n-1)}+\ldots+p_{n-1} y_{0}^{\prime \prime}=0 \tag{$\prime\prime$} \end{equation*}(′′)y0(n+1)+p1y0(n)+p2y0(n−1)+…+pn−1y0′′=0
and so on, until the penultimate equation is obtained
( ′ ′ ′ ) y 0 ( 2 n − 2 ) + p 1 y 0 ( 2 n − 3 ) + p 2 y 1 ( 2 n − 4 ) + … + p n − 1 y 1 ( n − 1 ) = 0 ( ′ ′ ′ ) y 0 ( 2 n − 2 ) + p 1 y 0 ( 2 n − 3 ) + p 2 y 1 ( 2 n − 4 ) + … + p n − 1 y 1 ( n − 1 ) = 0 {:((''')")"y_(0)^((2n-2))+p_(1)y_(0)^((2n-3))+p_(2)y_(1)^((2n-4))+dots+p_(n-1)y_(1)^((n-1))=0:}\begin{equation*} y_{0}^{(2 n-2)}+p_{1} y_{0}^{(2 n-3)}+p_{2} y_{1}^{(2 n-4)}+\ldots+p_{n-1} y_{1}^{(n-1)}=0 \tag{$\prime\prime\prime$} \end{equation*}(′′′)y0(2n−2)+p1y0(2n−3)+p2y1(2n−4)+…+pn−1y1(n−1)=0
But in this case the determinant
Δ n − 1 [ y ( x 0 ) ] = | y 0 y 0 ′ … y 0 ( n − 1 ) y 0 ′ y 0 ′ ′ … y 0 ( n ) ⋅ ⋅ ⋅ ⋅ y 0 ( n − 1 ) y 0 ( n ) … y 0 ( 2 n − 2 ) | Δ n − 1 y x 0 = y 0 y 0 ′ … y 0 ( n − 1 ) y 0 ′ y 0 ′ ′ … y 0 ( n ) ⋅ ⋅ ⋅ ⋅ y 0 ( n − 1 ) y 0 ( n ) … y 0 ( 2 n − 2 ) Delta_(n-1)[y(x_(0))]=|[y_(0),y_(0)^('),dots,y_(0)^((n-1))],[y_(0)^('),y_(0)^(''),dots,y_(0)^((n))],[*,*,*,*],[y_(0)^((n-1)),y_(0)^((n)),dots,y_(0)^((2n-2))]|\Delta_{n-1}\left[y\left(x_{0}\right)\right]=\left|\begin{array}{llll} y_{0} & y_{0}^{\prime} & \ldots & y_{0}^{(n-1)} \\ y_{0}^{\prime} & y_{0}^{\prime \prime} & \ldots & y_{0}^{(n)} \\ \cdot & \cdot & \cdot & \cdot \\ y_{0}^{(n-1)} & y_{0}^{(n)} & \ldots & y_{0}^{(2 n-2)} \end{array}\right|Δn−1[y(x0)]=|y0y0′…y0(n−1)y0′y0′′…y0(n)⋅⋅⋅⋅y0(n−1)y0(n)…y0(2n−2)|
is null because between the elements of the lines there is the same linear combination expressed by formulas (15), ( 15 ′ ) , ( 15 ′ ′ ) , ( 15 ′ ′ ′ ) 15 ′ , 15 ′ ′ , 15 ′ ′ ′ (15^(')),(15^('')),(15^('''))\left(15^{\prime}\right),\left(15^{\prime \prime}\right),\left(15^{\prime \prime \prime}\right)(15′),(15′′),(15′′′)This, however, contradicts the hypothesis that
Δ n − 1 [ y ( x ) ] ≠ 0 . Δ n − 1 [ y ( x ) ] ≠ 0 . Delta_(n-1)[y(x)]!=0.\Delta_{n-1}[y(x)] \neq 0 .Δn−1[y(x)]≠0.
So under the only condition that the characteristic equation ( 9 ′ 9 ′ 9^(')9^{\prime}9′) to have distinct roots, formula (10) in which C 1 C 2 … C n ≠ 0 C 1 C 2 … C n ≠ 0 C_(1)C_(2)dotsC_(n)!=0C_{1} C_{2} \ldots C_{n} \neq 0C1C2…Cn≠0is the integral of the differential equation (1).
Observation. Determination of constants C 1 , C 2 , … , C n ; r 1 , r 2 , … , r n C 1 , C 2 , … , C n ; r 1 , r 2 , … , r n C_(1),C_(2),dots,C_(n);r_(1),r_(2),dots,r_(n)C_{1}, C_{2}, \ldots, C_{n} ; r_{1}, r_{2}, \ldots, r_{n}C1,C2,…,Cn;R1,R2,…,Rnfrom formula (10), so that y ( x ) y ( x ) y(x)y(x)y(x)to verify Cauchy's conditions (7), is done by putting
C 1 e r 1 x 0 = C 1 ′ , C 2 e r 2 x 0 = C 2 ′ , … , C n e r n x 0 = C n ′ C 1 e r 1 x 0 = C 1 ′ , C 2 e r 2 x 0 = C 2 ′ , … , C n e r n x 0 = C n ′ C_(1)e^(r_(1)x_(0))=C_(1)^('),quadC_(2)e^(r_(2)x_(0))=C_(2)^('),dots,C_(n)e^(r_(n)x_(0))=C_(n)^(')C_{1} e^{r_{1} x_{0}}=C_{1}^{\prime}, \quad C_{2} e^{r_{2} x_{0}}=C_{2}^{\prime}, \ldots, C_{n} e^{r_{n} x_{0}}=C_{n}^{\prime}C1it isR1x0=C1′,C2it isR2x0=C2′,…,Cnit isRnx0=Cn′
and solving the system of equations
C 1 ′ + C 2 ′ + … + C n ′ = y 0 (16) C 1 ′ r 1 + C 2 ′ r 2 + … + C n ′ r n = y 0 ′ C 1 ′ r 1 2 n − 1 + C 2 ′ r 2 2 n − 2 + … + C n ′ r n 2 n − 1 = y 0 ( 2 n − 1 ) C 1 ′ + C 2 ′ + … + C n ′ = y 0 (16) C 1 ′ r 1 + C 2 ′ r 2 + … + C n ′ r n = y 0 ′ C 1 ′ r 1 2 n − 1 + C 2 ′ r 2 2 n − 2 + … + C n ′ r n 2 n − 1 = y 0 ( 2 n − 1 ) {:[C_(1)^(')+C_(2)^(')+dots+C_(n)^(')=y_(0)],[(16)C_(1)^(')r_(1)+C_(2)^(')r_(2)+dots+C_(n)^(')r_(n)=y_(0)^(')],[C_(1)^(')r_(1)^(2n-1)+C_(2)^(')r_(2)^(2n-2)+dots+C_(n)^(')r_(n)^(2n-1)=y_(0)^((2n-1))]:}\begin{align*} & C_{1}^{\prime}+C_{2}^{\prime}+\ldots+C_{n}^{\prime}=y_{0} \\ & C_{1}^{\prime} r_{1}+C_{2}^{\prime} r_{2}+\ldots+C_{n}^{\prime} r_{n}=y_{0}^{\prime} \tag{16}\\ & C_{1}^{\prime} r_{1}^{2 n-1}+C_{2}^{\prime} r_{2}^{2 n-2}+\ldots+C_{n}^{\prime} r_{n}^{2 n-1}=y_{0}^{(2 n-1)} \end{align*}C1′+C2′+…+Cn′=y0(16)C1′R1+C2′R2+…+Cn′Rn=y0′C1′R12n−1+C2′R22n−2+…+Cn′Rn2n−1=y0(2n−1)
in relation cu C 1 ′ , C 2 ′ , … , C n ′ cu C 1 ′ , C 2 ′ , … , C n ′ cuC_(1)^('),C_(2)^('),dots,C_(n)^(')\mathrm{cu} C_{1}^{\prime}, C_{2}^{\prime}, \ldots, C_{n}^{\prime}withC1′,C2′,…,Cn′and r 1 , r 2 , … , r n r 1 , r 2 , … , r n r_(1),r_(2),dots,r_(n)r_{1}, r_{2}, \ldots, r_{n}R1,R2,…,Rn.
His elimination C 1 ′ , C 2 ′ , … , C n ′ C 1 ′ , C 2 ′ , … , C n ′ C_(1)^('),C_(2)^('),dots,C_(n)^(')C_{1}^{\prime}, C_{2}^{\prime}, \ldots, C_{n}^{\prime}C1′,C2′,…,Cn′between these equations leads to equations (6), where
A 1 = − Σ r 1 , A 2 = Σ r 1 r 2 … , A n = ( − 1 ) n r 1 r 2 … r n A 1 = − Σ r 1 , A 2 = Σ r 1 r 2 … , A n = ( − 1 ) n r 1 r 2 … r n A_(1)=-Sigmar_(1),A_(2)=Sigmar_(1)r_(2)dots,A_(n)=(-1)^(n)r_(1)r_(2)dotsr_(n)A_{1}=-\Sigma r_{1}, A_{2}=\Sigma r_{1} r_{2} \ldots, A_{n}=(-1)^{n} r_{1} r_{2} \ldots r_{n}A1=−ΣR1,A2=ΣR1R2…,An=(−1)nR1R2…Rn
The equation that determines r 1 , r 2 , … , r n r 1 , r 2 , … , r n r_(1),r_(2),dots,r_(n)r_{1}, r_{2}, \ldots, r_{n}R1,R2,…,Rnis equation (9').
Example. Given the function
Y ( x ) = ∫ α β p ( s ) e s x d s Y ( x ) = ∫ α β   p ( s ) e s x d s Y(x)=int_(alpha)^(beta)p(s)e^(sx)dsY(x)=\int_{\alpha}^{\beta} p(s) e^{s x} d sY(x)=∫αβp(S)it isSxdS
where p ( s ) p ( s ) p(s)p(s)p(S)is a positive function in the interval ( α , β α , β alpha,beta\alpha, \betaα,β), which can be cancelled in α α alpha\alphaαand β β beta\betaβ, one can determine an integral of the differential equation (1) that satisfies Cauchy's conditions.
y ( x 0 ) = Y ( x 0 ) , y ′ ( x 0 ) = Y ′ ( x 0 ) , … , y ( 2 n − 1 ) ( x 0 ) = Y ( 2 n − 1 ) ( x 0 ) y x 0 = Y x 0 , y ′ x 0 = Y ′ x 0 , … , y ( 2 n − 1 ) x 0 = Y ( 2 n − 1 ) x 0 y(x_(0))=Y(x_(0)),y^(')(x_(0))=Y^(')(x_(0)),dots,y^((2n-1))(x_(0))=Y^((2n-1))(x_(0))y\left(x_{0}\right)=Y\left(x_{0}\right), y^{\prime}\left(x_{0}\right)=Y^{\prime}\left(x_{0}\right), \ldots, y^{(2 n-1)}\left(x_{0}\right)=Y^{(2 n-1)}\left(x_{0}\right)y(x0)=Y(x0),y′(x0)=Y′(x0),…,y(2n−1)(x0)=Y(2n−1)(x0)
Indeed, the system of equations (16) corresponding to this case is
C 1 ′ + C 2 ′ + … + C n ′ = ∫ α β p ( s ) e s x 0 d s C 1 ′ r 1 + C 2 ′ r 2 + … + C n ′ r n = ∫ α β s p ( s ) e s x 0 d s C 1 ′ r 1 2 n − 1 + C 2 ′ r 2 2 n − 1 + … + C n ′ r n 2 n − 1 = ∫ α β s 2 n − 1 p ( s ) e s x 0 d s C 1 ′ + C 2 ′ + … + C n ′ = ∫ α β   p ( s ) e s x 0 d s C 1 ′ r 1 + C 2 ′ r 2 + … + C n ′ r n = ∫ α β   s p ( s ) e s x 0 d s C 1 ′ r 1 2 n − 1 + C 2 ′ r 2 2 n − 1 + … + C n ′ r n 2 n − 1 = ∫ α β   s 2 n − 1 p ( s ) e s x 0 d s {:[C_(1)^(')+C_(2)^(')+dots+C_(n)^(')=int_(alpha)^(beta)p(s)e^(sx_(0))ds],[C_(1)^(')r_(1)+C_(2)^(')r_(2)+dots+C_(n)^(')r_(n)=int_(alpha)^(beta)sp(s)e^(sx_(0))ds],[C_(1)^(')r_(1)^(2n-1)+C_(2)^(')r_(2)^(2n-1)+dots+C_(n)^(')r_(n)^(2n-1)=int_(alpha)^(beta)s^(2n-1)p(s)e^(sx_(0))ds]:}\begin{aligned} & C_{1}^{\prime}+C_{2}^{\prime}+\ldots+C_{n}^{\prime}=\int_{\alpha}^{\beta} p(s) e^{s x_{0}} d s \\ & C_{1}^{\prime} r_{1}+C_{2}^{\prime} r_{2}+\ldots+C_{n}^{\prime} r_{n}=\int_{\alpha}^{\beta} s p(s) e^{s x_{0}} d s \\ & C_{1}^{\prime} r_{1}^{2 n-1}+C_{2}^{\prime} r_{2}^{2 n-1}+\ldots+C_{n}^{\prime} r_{n}^{2 n-1}=\int_{\alpha}^{\beta} s^{2 n-1} p(s) e^{s x_{0}} d s \end{aligned}C1′+C2′+…+Cn′=∫αβp(S)it isSx0dSC1′R1+C2′R2+…+Cn′Rn=∫αβSp(S)it isSx0dSC1′R12n−1+C2′R22n−1+…+Cn′Rn2n−1=∫αβS2n−1p(S)it isSx0dS
This system is classical; it is found in the theory of quadrature formulas. It is shown [5] that the numbers r 1 , r 2 , … , r n r 1 , r 2 , … , r n r_(1),r_(2),dots,r_(n)r_{1}, r_{2}, \ldots, r_{n}R1,R2,…,Rnare all real, distinct, and contained within α α alpha\alphaαand β β beta\betaβ, and the numbers C 1 ′ , C 2 ′ , … , C n ′ C 1 ′ , C 2 ′ , … , C n ′ C_(1)^('),C_(2)^('),dots,C_(n)^(')C_{1}^{\prime}, C_{2}^{\prime}, \ldots, C_{n}^{\prime}C1′,C2′,…,Cn′, that is C 1 , C 2 , … , C n C 1 , C 2 , … , C n C_(1),C_(2),dots,C_(n)C_{1}, C_{2}, \ldots, C_{n}C1,C2,…,Cnare all positive.
3. The integrals of the differential equation (1) depend on the initial conditions (7). They can be such that the characteristic equation ( 9 ′ 9 ′ 9^(')9^{\prime}9′) of the differential equation (8) to have all distinct roots, and this case was studied in the previous point. However, it is possible that the initial conditions (9) are such that the characteristic equation ( 9 ′ 9 ′ 9^(')9^{\prime}9′) to have multiple roots.
Let us assume that the initial conditions (7) are such that the characteristic equation ( 9 ′ 9 ′ 9^(')9^{\prime}9′) has the root r 1 r 1 r_(1)r_{1}R1multiple of the order p 1 , r 2 p 1 , r 2 p_(1),r_(2)p_{1}, r_{2}p1,R2, multiple of the order p 2 , … , r k p 2 , … , r k p_(2),dots,r_(k)p_{2}, \ldots, r_{k}p2,…,Rkmultiple of the order p k p k p_(k)p_{k}pkwhere
p 1 + p 2 + … + p k = n . p 1 + p 2 + … + p k = n . p_(1)+p_(2)+dots+p_(k)=n.p_{1}+p_{2}+\ldots+p_{k}=n .p1+p2+…+pk=n.
In this case the differential equation (8) has the integral
y = ( C 1 ( p 1 − 1 ) ! x p 1 − 1 + C 11 ( p 1 − 2 ) ! x p 1 − 2 + … + C 1 , p 1 − 1 ) e r 1 x (17) + ( C 2 ( p 2 − 1 ) ! x p 2 − 2 + C 21 ( p 2 − 2 ) ! x p 2 − 2 + … + C 2 , p 2 − 1 ) e r 2 x + ( C k ( p k − 1 ) ! x p k − 1 + C k 1 ( p k − 2 ) ! x p k − 2 + … + C k , p k − 1 ) e r k x y = C 1 p 1 − 1 ! x p 1 − 1 + C 11 p 1 − 2 ! x p 1 − 2 + … + C 1 , p 1 − 1 e r 1 x (17) + C 2 p 2 − 1 ! x p 2 − 2 + C 21 p 2 − 2 ! x p 2 − 2 + … + C 2 , p 2 − 1 e r 2 x + C k p k − 1 ! x p k − 1 + C k 1 p k − 2 ! x p k − 2 + … + C k , p k − 1 e r k x {:[y=((C_(1))/((p_(1)-1)!)x^(p_(1)-1)+(C_(11))/((p_(1)-2)!)x^(p_(1)-2)+dots+C_(1,p_(1)-1))e^(r_(1)x)],[(17)+((C_(2))/((p_(2)-1)!)x^(p_(2)-2)+(C_(21))/((p_(2)-2)!)x^(p_(2)-2)+dots+C_(2,p_(2)-1))e^(r_(2)x)],[+((C_(k))/((p_(k)-1)!)x^(p_(k)-1)+(C_(k1))/((p_(k)-2)!)x^(p_(k)-2)+dots+C_(k,p_(k)-1))e^(r_(k)x)]:}\begin{align*} y & =\left(\frac{C_{1}}{\left(p_{1}-1\right)!} x^{p_{1}-1}+\frac{C_{11}}{\left(p_{1}-2\right)!} x^{p_{1}-2}+\ldots+C_{1, p_{1}-1}\right) e^{r_{1} x} \\ & +\left(\frac{C_{2}}{\left(p_{2}-1\right)!} x^{p_{2}-2}+\frac{C_{21}}{\left(p_{2}-2\right)!} x^{p_{2}-2}+\ldots+C_{2, p_{2}-1}\right) e^{r_{2} x} \tag{17}\\ & +\left(\frac{C_{k}}{\left(p_{k}-1\right)!} x^{p_{k}-1}+\frac{C_{k 1}}{\left(p_{k}-2\right)!} x^{p_{k}-2}+\ldots+C_{k, p_{k}-1}\right) e^{r_{k} x} \end{align*}y=(C1(p1−1)!xp1−1+C11(p1−2)!xp1−2+…+C1,p1−1)it isR1x(17)+(C2(p2−1)!xp2−2+C21(p2−2)!xp2−2+…+C2,p2−1)it isR2x+(Ck(pk−1)!xpk−1+Ck1(pk−2)!xpk−2+…+Ck,pk−1)it isRkx
and we have
Δ n − 1 [ y ] = ( − 1 ) k C 1 p 1 C 2 p 2 … C k p k V 2 ( r 1 , … , r 1 ⏟ p 1 ori r 2 , … , r 2 ⏟ p 2 ori , , r k , … , r k ⏟ p k ori ) e ( p 1 r 1 + … + p k r k x Δ n − 1 [ y ] = ( − 1 ) k C 1 p 1 C 2 p 2 … C k p k V 2 ( r 1 , … , r 1 ⏟ p 1  ori  r 2 , … , r 2 ⏟ p 2  ori  , , r k , … , r k ⏟ p k  ori  ) e p 1 r 1 + … + p k r k x Delta_(n-1)[y]=(-1)^(k)C_(1)^(p_(1))C_(2)^(p_(2))dotsC_(k)^(p_(k))V^(2)(ubrace(r_(1),dots,r_(1))_(p_(1)" ori ")ubrace(r_(2),dots,r_(2))_(p_(2)" ori "),quad,ubrace(r_(k),dots,r_(k))_(p_(k)" ori "))e^((p_(1)r_(1)+dots+p_(k)r_(k)x:})\Delta_{n-1}[y]=(-1)^{k} C_{1}^{p_{1}} C_{2}^{p_{2}} \ldots C_{k}^{p_{k}} V^{2}(\underbrace{r_{1}, \ldots, r_{1}}_{p_{1} \text { ori }} \underbrace{r_{2}, \ldots, r_{2}}_{p_{2} \text { ori }}, \quad, \underbrace{r_{k}, \ldots, r_{k}}_{p_{k} \text { ori }}) e^{\left(p_{1} r_{1}+\ldots+p_{k} r_{k} x\right.}Δn−1[y]=(−1)kC1p1C2p2…CkpkV2(R1,…,R1⏟p1 OR R2,…,R2⏟p2 OR ,,Rk,…,Rk⏟pk OR )it is(p1R1+…+pkRkx
where the determinant V ( r 1 , … , r 1 ; r 2 , … , r 2 ⏟ p 1 ori , … , r k , … , r k ⏟ p k ori ) V ( r 1 , … , r 1 ; r 2 , … , r 2 ⏟ p 1  ori  , … , r k , … , r k ⏟ p k  ori  ) V(r_(1),dots,r_(1);ubrace(r_(2),dots,r_(2))_(p_(1)" ori "),dots,ubrace(r^(k),dots,r_(k))_(p_(k)" ori "))\mathrm{V}(r_{1}, \ldots, r_{1} ; \underbrace{r_{2}, \ldots, r_{2}}_{p_{1} \text { ori }}, \ldots, \underbrace{r^{k}, \ldots, r_{k}}_{p_{k} \text { ori }})V(R1,…,R1;R2,…,R2⏟p1 OR ,…,Rk,…,Rk⏟pk OR )has the first column consisting of 1 , r 1 , … , r 1 n − 1 1 , r 1 , … , r 1 n − 1 1,r_(1),dots,r_(1)^(n-1)1, r_{1}, \ldots, r_{1}^{n-1}1,R1,…,R1n−1, the second of the derivatives of these elements with respect to r 1 r 1 r_(1)r_{1}R1divided by 1!, the third with the second-order derivatives divided by 2 ! , … , a k 2 ! , … , a k 2!,dots,ak2!, \ldots, a k2!,…,Ak-a of the derivatives of the order k k kkk-1 of the elements of the first column divided by ( k − 1 k − 1 k-1k-1k−1)!. The following k 2 k 2 k_(2)k_{2}k2columns are formed in the same way, replacing r 1 r 1 r_(1)r_{1}R1with r 2 … r 2 … r_(2)dotsr_{2} \ldotsR2…, and so on.
It follows that if C 1 C 2 … C k ≠ 0 , y C 1 C 2 … C k ≠ 0 , y C_(1)C_(2)dotsC_(k)!=0,yC_{1} C_{2} \ldots C_{k} \neq 0, yC1C2…Ck≠0,yis the integral of the differential equation (1). And this integral is defined in the interval ( − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞), and the condition Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0is also valid in the interval ( − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞).
Formula (18) can be proven directly, but the calculations are complicated.
However, it is known that the integrals of a differential equation with constant coefficients corresponding to a multiple root can be obtained from the integrals corresponding to distinct roots, by a passage to the limit. We will use this passage to the limit to prove formula (8).
Suppose we are dealing with the case when the root r 1 r 1 r_(1)r_{1}R1is double, the other roots being distinct.
Putting r 2 = r 1 + h r 2 = r 1 + h r_(2)=r_(1)+hr_{2}=r_{1}+hR2=R1+h, for h ≠ 0 h ≠ 0 h!=0h \neq 0h≠0, the differential equation (8) has the integral
(19) y = C 1 e ( r 1 + h ) x − e r 1 x h + C 2 e r 1 x + C 3 e r 3 x + … + C n e r n x (19) y = C 1 e r 1 + h x − e r 1 x h + C 2 e r 1 x + C 3 e r 3 x + … + C n e r n x {:(19)y=C_(1)(e^((r_(1)+h)x)-e^(r_(1)x))/(h)+C_(2)e^(r_(1)x)+C_(3)e^(r_(3)x)+dots+C_(n)e^(r_(n)x):}\begin{equation*} y=C_{1} \frac{e^{\left(r_{1}+h\right) x}-e^{r_{1} x}}{h}+C_{2} e^{r_{1} x}+C_{3} e^{r_{3} x}+\ldots+C_{n} e^{r_{n} x} \tag{19} \end{equation*}(19)y=C1it is(R1+h)x−it isR1xh+C2it isR1x+C3it isR3x+…+Cnit isRnx
which we can write in the form
(20) y = C 1 ′ e r 1 x + C 2 ′ e ( r 1 + h ] x + C 3 e r 3 x + … + C n e r n x (20) y = C 1 ′ e r 1 x + C 2 ′ e r 1 + h x + C 3 e r 3 x + … + C n e r n x {:(20)y=C_(1)^(')e^(r_(1)x)+C_(2)^(')e^((r_(1)+h]x)+C_(3)e^(r_(3)x)+dots+C_(n)e^(r_(n)x):}\begin{equation*} y=C_{1}^{\prime} e^{r_{1} x}+C_{2}^{\prime} e^{\left(r_{1}+h\right] x}+C_{3} e^{r_{3} x}+\ldots+C_{n} e^{r_{n} x} \tag{20} \end{equation*}(20)y=C1′it isR1x+C2′it is(R1+h]x+C3it isR3x+…+Cnit isRnx
where
(21) C 1 ′ = C 2 − C 1 h , C 2 ′ = C 1 h (21) C 1 ′ = C 2 − C 1 h , C 2 ′ = C 1 h {:(21)C_(1)^(')=C_(2)-(C_(1))/(h)","quadC_(2)^(')=(C_(1))/(h):}\begin{equation*} C_{1}^{\prime}=C_{2}-\frac{C_{1}}{h}, \quad C_{2}^{\prime}=\frac{C_{1}}{h} \tag{21} \end{equation*}(21)C1′=C2−C1h,C2′=C1h
When h → 0 h → 0 h rarr0h \rightarrow 0h→0, function y y yyyhas a limit
(22) y 1 = ( C 1 x + C 2 ) e r 1 x + C 3 e r 3 x + … + C n e r n x (22) y 1 = C 1 x + C 2 e r 1 x + C 3 e r 3 x + … + C n e r n x {:(22)y_(1)=(C_(1)x+C_(2))e^(r_(1)x)+C_(3)e^(r_(3)x)+dots+C_(n)er_(n)x:}\begin{equation*} y_{1}=\left(C_{1} x+C_{2}\right) e^{r_{1} x}+C_{3} e^{r_{3} x}+\ldots+C_{n} e r_{n} x \tag{22} \end{equation*}(22)y1=(C1x+C2)it isR1x+C3it isR3x+…+Cnit isRnx
and y 1 y 1 y_(1)y_{1}y1is the integral of the differential equation (8). It is easy to prove that the derivative of any order p p pppof the function y y yyy, tends to the derivative of the same order of the function y 1 y 1 y_(1)y_{1}y1when h → 0 h → 0 h rarr0h \rightarrow 0h→0, from which it follows that
(23) lim h → 0 Δ n → 1 [ y ] = Δ n − 1 [ y 1 ] . (23) lim h → 0   Δ n → 1 [ y ] = Δ n − 1 y 1 . {:(23)lim_(h rarr0)Delta_(n rarr1)[y]=Delta_(n-1)[y_(1)].:}\begin{equation*} \lim _{h \rightarrow 0} \Delta_{n \rightarrow 1}[y]=\Delta_{n-1}\left[y_{1}\right] . \tag{23} \end{equation*}(23)limh→0Δn→1[y]=Δn−1[y1].
But applying formula (12) we have
Δ n − 1 [ y ] = C 1 ′ C 2 ′ C 3 … C n V 2 ( r 1 , r 1 + h , r 3 , … , r n ) e ( 2 r 1 + h + r 3 + … + r n ) x Δ n − 1 [ y ] = C 1 ′ C 2 ′ C 3 … C n V 2 r 1 , r 1 + h , r 3 , … , r n e 2 r 1 + h + r 3 + … + r n x Delta_(n-1)[y]=C_(1)^(')C_(2)^(')C_(3)dotsC_(n)V^(2)(r_(1),r_(1)+h,r_(3),dots,r_(n))e^((2r_(1)+h+r_(3)+dots+r_(n))x)\Delta_{n-1}[y]=C_{1}^{\prime} C_{2}^{\prime} C_{3} \ldots C_{n} V^{2}\left(r_{1}, r_{1}+h, r_{3}, \ldots, r_{n}\right) e^{\left(2 r_{1}+h+r_{3}+\ldots+r_{n}\right) x}Δn−1[y]=C1′C2′C3…CnV2(R1,R1+h,R3,…,Rn)it is(2R1+h+R3+…+Rn)x
Replacing C 1 ′ , C 2 ′ cu C 1 ′ , C 2 ′ cu C_(1)^('),C_(2)^(')cuC_{1}^{\prime}, C_{2}^{\prime} \mathrm{cu}C1′,C2′withformulas (21), we will have
Δ n − 1 [ y ] = C 1 ( C 2 h − C 1 ) C 3 … C n [ V ( r 1 , r 1 + h , r 3 , … , r n ) h ] 2 e ( 2 r 1 + h + r 3 + … + r n ) x Δ n − 1 [ y ] = C 1 C 2 h − C 1 C 3 … C n V r 1 , r 1 + h , r 3 , … , r n h 2 e 2 r 1 + h + r 3 + … + r n x Delta_(n-1)[y]=C_(1)(C_(2)h-C_(1))C_(3)dotsC_(n)[(V(r_(1),r_(1)+h,r_(3),dots,r_(n)))/(h)]^(2)e^((2r_(1)+h+r_(3)+dots+r_(n))x)\Delta_{n-1}[y]=C_{1}\left(C_{2} h-C_{1}\right) C_{3} \ldots C_{n}\left[\frac{V\left(r_{1}, r_{1}+h, r_{3}, \ldots, r_{n}\right)}{h}\right]^{2} e^{\left(2 r_{1}+h+r_{3}+\ldots+r_{n}\right) x}Δn−1[y]=C1(C2h−C1)C3…Cn[V(R1,R1+h,R3,…,Rn)h]2it is(2R1+h+R3+…+Rn)x
When we do h → 0 h → 0 h rarr0h \rightarrow 0h→0, we obtain
lim h → 0 Δ n − 1 [ y ] = − C 1 2 C 3 … C n V 2 ( r 1 , r 1 , r 3 , … , r n ) e ( 2 r 1 + r 3 + … + r n ) x lim h → 0   Δ n − 1 [ y ] = − C 1 2 C 3 … C n V 2 r 1 , r 1 , r 3 , … , r n e 2 r 1 + r 3 + … + r n x lim_(h rarr0)Delta_(n-1)[y]=-C_(1)^(2)C_(3)dotsC_(n)V^(2)(r_(1),r_(1),r_(3),dots,r_(n))e^((2r_(1)+r_(3)+dots+r_(n))x)\lim _{h \rightarrow 0} \Delta_{n-1}[y]=-C_{1}^{2} C_{3} \ldots C_{n} V^{2}\left(r_{1}, r_{1}, r_{3}, \ldots, r_{n}\right) e^{\left(2 r_{1}+r_{3}+\ldots+r_{n}\right) x}limh→0Δn−1[y]=−C12C3…CnV2(R1,R1,R3,…,Rn)it is(2R1+R3+…+Rn)x
because
lim h → 0 1 h | 1 1 1 … 1 r 1 r 1 + h r 3 … r n r 1 2 ( r 1 + h ) 2 r 3 2 … r n 2 r 1 3 ( r 1 + h ) 3 r 3 3 … r n 3 ⋯ ⋯ ⋯ ⋯ ⋯ r 1 n − 1 ( r 1 + h ) n − 1 r 3 n − 1 … r n n − 1 | = V ( r 1 , r 2 , r 3 … , r n ) . lim h → 0   1 h 1 1 1 … 1 r 1 r 1 + h r 3 … r n r 1 2 r 1 + h 2 r 3 2 … r n 2 r 1 3 r 1 + h 3 r 3 3 … r n 3 ⋯ ⋯ ⋯ ⋯ ⋯ r 1 n − 1 r 1 + h n − 1 r 3 n − 1 … r n n − 1 = V r 1 , r 2 , r 3 … , r n . lim_(h rarr0)(1)/(h)|[1,1,1,dots,1],[r_(1),r_(1)+h,r_(3),dots,r_(n)],[r_(1)^(2),(r_(1)+h)^(2),r_(3)^(2),dots,r_(n)^(2)],[r_(1)^(3),(r_(1)+h)^(3),r_(3)^(3),dots,r_(n)^(3)],[cdots,cdots,cdots,cdots,cdots],[r_(1)^(n-1),(r_(1)+h)^(n-1),r_(3)^(n-1),dots,r_(n)^(n-1)]|=V(r_(1),r_(2),r_(3)dots,r_(n)).\lim _{h \rightarrow 0} \frac{1}{h}\left|\begin{array}{ccccc} 1 & 1 & 1 & \ldots & 1 \\ r_{1} & r_{1}+h & r_{3} & \ldots & r_{n} \\ r_{1}^{2} & \left(r_{1}+h\right)^{2} & r_{3}^{2} & \ldots & r_{n}^{2} \\ r_{1}^{3} & \left(r_{1}+h\right)^{3} & r_{3}^{3} & \ldots & r_{n}^{3} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ r_{1}^{n-1} & \left(r_{1}+h\right)^{n-1} & r_{3}^{n-1} & \ldots & r_{n}^{n-1} \end{array}\right|=V\left(r_{1}, r_{2}, r_{3} \ldots, r_{n}\right) .limh→01h|111…1R1R1+hR3…RnR12(R1+h)2R32…Rn2R13(R1+h)3R33…Rn3⋯⋯⋯⋯⋯R1n−1(R1+h)n−1R3n−1…Rnn−1|=V(R1,R2,R3…,Rn).
Taking into account formula (23), we thus deduce that
(24) Δ n − 1 [ y 1 ] = − C 1 2 C 3 … C n V 2 ( r 1 , r 1 , r 3 , … , r n ) e ( 2 r 1 + r 3 + … + r n ) x (24) Δ n − 1 y 1 = − C 1 2 C 3 … C n V 2 r 1 , r 1 , r 3 , … , r n e 2 r 1 + r 3 + … + r n x {:(24)Delta_(n-1)[y_(1)]=-C_(1)^(2)C_(3)dotsC_(n)V^(2)(r_(1),r_(1),r_(3),dots,r_(n))e^((2r_(1)+r_(3)+dots+r_(n))x):}\begin{equation*} \Delta_{n-1}\left[y_{1}\right]=-C_{1}^{2} C_{3} \ldots C_{n} V^{2}\left(r_{1}, r_{1}, r_{3}, \ldots, r_{n}\right) e^{\left(2 r_{1}+r_{3}+\ldots+r_{n}\right) x} \tag{24} \end{equation*}(24)Δn−1[y1]=−C12C3…CnV2(R1,R1,R3,…,Rn)it is(2R1+R3+…+Rn)x
Let us now suppose that we are dealing with the case when the root r 1 r 1 r_(1)r_{1}R1is triple, the others being distinct.
In the integral (22) of the differential equation (8), let us put r 3 = r 1 + h r 3 = r 1 + h r_(3)=r_(1)+hr_{3}=r_{1}+hR3=R1+hand let's write this integral in the form
y = C 1 e ( r 1 + h ) x − e r 1 x − h x e r 1 x h 2 + C 2 e ( r 1 + h ) x − e r 1 x h + C 3 e r 1 x + C 4 e r 4 x + … + C n e r n x y = C 1 e r 1 + h x − e r 1 x − h x e r 1 x h 2 + C 2 e r 1 + h x − e r 1 x h + C 3 e r 1 x + C 4 e r 4 x + … + C n e r n x y=C_(1)(e^((r_(1)+h)x)-e^(r_(1)x)-hxe^(r_(1)x))/(h^(2))+C_(2)(e^((r_(1)+h)x)-e^(r_(1)x))/(h)+C_(3)e^(r_(1)x)+C_(4)e^(r_(4)x)+dots+C_(n)e^(r_(n)x)y=C_{1} \frac{e^{\left(r_{1}+h\right) x}-e^{r_{1} x}-h x e^{r_{1} x}}{h^{2}}+C_{2} \frac{e^{\left(r_{1}+h\right) x}-e^{r_{1} x}}{h}+C_{3} e^{r_{1} x}+C_{4} e^{r_{4} x}+\ldots+C_{n} e^{r_{n} x}y=C1it is(R1+h)x−it isR1x−hxit isR1xh2+C2it is(R1+h)x−it isR1xh+C3it isR1x+C4it isR4x+…+Cnit isRnx
or
y = ( C 1 ′ x + C 2 ′ ) r 1 x + C 3 ′ e ( r 1 + h ) x + C 4 e r 4 x + … + C n e r n x y = C 1 ′ x + C 2 ′ r 1 x + C 3 ′ e r 1 + h x + C 4 e r 4 x + … + C n e r n x y=(C_(1)^(')x+C_(2)^('))^(r_(1)x)+C_(3)^(')e^((r_(1)+h)x)+C_(4)e^(r_(4)x)+dots+C_(n)e^(r_(n)x)y=\left(C_{1}^{\prime} x+C_{2}^{\prime}\right)^{r_{1} x}+C_{3}^{\prime} e^{\left(r_{1}+h\right) x}+C_{4} e^{r_{4} x}+\ldots+C_{n} e^{r_{n} x}y=(C1′x+C2′)R1x+C3′it is(R1+h)x+C4it isR4x+…+Cnit isRnx
where
(25) C 1 ′ = − C 1 h , C 2 ′ = − C 1 h 2 − C 2 h + C 3 , C 3 ′ = C 1 h 2 + C 2 h (25) C 1 ′ = − C 1 h , C 2 ′ = − C 1 h 2 − C 2 h + C 3 , C 3 ′ = C 1 h 2 + C 2 h {:(25)C_(1)^(')=-(C_(1))/(h)","quadC_(2)^(')=-(C_(1))/(h^(2))-(C_(2))/(h)+C_(3)","quadC_(3)^(')=(C_(1))/(h^(2))+(C_(2))/(h):}\begin{equation*} C_{1}^{\prime}=-\frac{C_{1}}{h}, \quad C_{2}^{\prime}=-\frac{C_{1}}{h^{2}}-\frac{C_{2}}{h}+C_{3}, \quad C_{3}^{\prime}=\frac{C_{1}}{h^{2}}+\frac{C_{2}}{h} \tag{25} \end{equation*}(25)C1′=−C1h,C2′=−C1h2−C2h+C3,C3′=C1h2+C2h
When h → 0 , y h → 0 , y h rarr0,yh \rightarrow 0, yh→0,ytends towards the integral
(26) y 2 = ( C 1 x 2 2 ! + C 2 x 1 ! + C 3 ) e r 1 x + C 4 e r 4 x + … + C n e r n x (26) y 2 = C 1 x 2 2 ! + C 2 x 1 ! + C 3 e r 1 x + C 4 e r 4 x + … + C n e r n x {:(26)y_(2)=(C_(1)(x^(2))/(2!)+C_(2)(x)/(1!)+C_(3))e^(r_(1)x)+C_(4)e^(r_(4)x)+dots+C_(n)e^(r_(n)x):}\begin{equation*} y_{2}=\left(C_{1} \frac{x^{2}}{2!}+C_{2} \frac{x}{1!}+C_{3}\right) e^{r_{1} x}+C_{4} e^{r_{4} x}+\ldots+C_{n} e^{r_{n} x} \tag{26} \end{equation*}(26)y2=(C1x22!+C2x1!+C3)it isR1x+C4it isR4x+…+Cnit isRnx
of the differential equation (8) and it is shown that
(27) lim h → 0 Δ n − 1 [ y ] = Δ n − 1 [ y 2 ] . (27) lim h → 0   Δ n − 1 [ y ] = Δ n − 1 y 2 . {:(27)lim_(h rarr0)Delta_(n-1)[y]=Delta_(n-1)[y_(2)].:}\begin{equation*} \lim _{h \rightarrow 0} \Delta_{n-1}[y]=\Delta_{n-1}\left[y_{2}\right] . \tag{27} \end{equation*}(27)limh→0Δn−1[y]=Δn−1[y2].
On the other hand, taking into account equation (24) and formulas (25), we can write
Δ n − 1 [ y ] = − C 1 2 ( C 1 + h C 2 ) C 4 … C n [ [ V ( r 1 , r 1 , r 1 + h , r 4 , … , r n ) ] h 2 e ( 3 r 1 + h + r 4 + … + r n ) x Δ n − 1 [ y ] = − C 1 2 C 1 + h C 2 C 4 … C n V r 1 , r 1 , r 1 + h , r 4 , … , r n h 2 e 3 r 1 + h + r 4 + … + r n x Delta_(n-1)[y]=-C_(1)^(2)(C_(1)+hC_(2))C_(4)dotsC_(n)[([V(r_(1),r_(1),r_(1)+h,r_(4),dots,r_(n))])/(h^(2))e^((3r_(1)+h+r_(4)+dots+r_(n))x):}\Delta_{n-1}[y]=-C_{1}^{2}\left(C_{1}+h C_{2}\right) C_{4} \ldots C_{n}\left[\frac{\left[V\left(r_{1}, r_{1}, r_{1}+h, r_{4}, \ldots, r_{n}\right)\right]}{h^{2}} e^{\left(3 r_{1}+h+r_{4}+\ldots+r_{n}\right) x}\right.Δn−1[y]=−C12(C1+hC2)C4…Cn[[V(R1,R1,R1+h,R4,…,Rn)]h2it is(3R1+h+R4+…+Rn)x
Taking into account that
lim h → 0 1 h 2 | 1 0 1 1 … 1 r 1 1 r 1 + h r 4 … r n r 1 2 C 2 1 r 1 ( r 1 + h ) 2 r 4 2 … r n 2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ r 1 n − 1 C n − 1 1 r 1 n − 2 ( r 1 + h ) n − 1 r 4 n − 1 … r n n − 1 | = V ( r 1 , r 1 , r 1 , r 4 , … , r n ) lim h → 0   1 h 2 1 0 1 1 … 1 r 1 1 r 1 + h r 4 … r n r 1 2 C 2 1 r 1 r 1 + h 2 r 4 2 … r n 2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ r 1 n − 1 C n − 1 1 r 1 n − 2 r 1 + h n − 1 r 4 n − 1 … r n n − 1 = V r 1 , r 1 , r 1 , r 4 , … , r n lim_(h rarr0)(1)/(h^(2))|[1,0,1,1,dots,1],[r_(1),1,r_(1)+h,r_(4),dotsr_(n)],[r_(1)^(2),C_(2)^(1)r_(1),(r_(1)+h)^(2),r_(4)^(2),dotsr_(n)^(2)],[cdots,cdots,cdots,cdots,cdots,cdots],[r_(1)^(n-1),C_(n-1)^(1)r_(1)^(n-2),(r_(1)+h)^(n-1),r_(4)^(n-1),dotsr_(n)^(n-1)]|=V(r_(1),r_(1),r_(1),r_(4),dots,r_(n))\lim _{h \rightarrow 0} \frac{1}{h^{2}}\left|\begin{array}{cccccc} 1 & 0 & 1 & 1 & \ldots & 1 \\ r_{1} & 1 & r_{1}+h & r_{4} & \ldots r_{n} \\ r_{1}^{2} & C_{2}^{1} r_{1} & \left(r_{1}+h\right)^{2} & r_{4}^{2} & \ldots r_{n}^{2} \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ r_{1}^{n-1} & C_{n-1}^{1} r_{1}^{n-2} & \left(r_{1}+h\right)^{n-1} & r_{4}^{n-1} & \ldots r_{n}^{n-1} \end{array}\right|=V\left(r_{1}, r_{1}, r_{1}, r_{4}, \ldots, r_{n}\right)limh→01h2|1011…1R11R1+hR4…RnR12C21R1(R1+h)2R42…Rn2⋯⋯⋯⋯⋯⋯R1n−1Cn−11R1n−2(R1+h)n−1R4n−1…Rnn−1|=V(R1,R1,R1,R4,…,Rn)
and taking the limit in the previous formula, we deduce that
lim h → 0 Δ n → 1 [ y ] = − C 1 3 C 4 … C n V 2 ( r 1 , r 1 , r 1 r 4 , … , r n ) e ( 3 r 1 + r 4 + … + r n ) x lim h → 0   Δ n → 1 [ y ] = − C 1 3 C 4 … C n V 2 r 1 , r 1 , r 1 r 4 , … , r n e 3 r 1 + r 4 + … + r n x lim_(h rarr0)Delta_(n rarr1)[y]=-C_(1)^(3)C_(4)dotsC_(n)V^(2)(r_(1),r_(1),r_(1)r_(4),dots,r_(n))e^((3r_(1)+r_(4)+dots+r_(n))x)\lim _{h \rightarrow 0} \Delta_{n \rightarrow 1}[y]=-\mathrm{C}_{1}^{3} C_{4} \ldots C_{n} V^{2}\left(r_{1}, r_{1}, r_{1} r_{4}, \ldots, r_{n}\right) e^{\left(3 r_{1}+r_{4}+\ldots+r_{n}\right) x}limh→0Δn→1[y]=−C13C4…CnV2(R1,R1,R1R4,…,Rn)it is(3R1+R4+…+Rn)x
and according to formula (27), we will have
Δ n − 1 [ y 2 ] = − C 1 3 C 4 … C n V 2 ( r 1 , r 1 , r 1 , r 4 , … , r n ) e ( 3 r 1 + r 4 + … + r n ) x Δ n − 1 y 2 = − C 1 3 C 4 … C n V 2 r 1 , r 1 , r 1 , r 4 , … , r n e 3 r 1 + r 4 + … + r n x Delta_(n-1)[y_(2)]=-C_(1)^(3)C_(4)dotsC_(n)V^(2)(r_(1),r_(1),r_(1),r_(4),dots,r_(n))e^((3r_(1)+r_(4)+dots+r_(n))x)\Delta_{n-1}\left[y_{2}\right]=-C_{1}^{3} C_{4} \ldots C_{n} V^{2}\left(r_{1}, r_{1}, r_{1}, r_{4}, \ldots, r_{n}\right) e^{\left(3 r_{1}+r_{4}+\ldots+r_{n}\right) x}Δn−1[y2]=−C13C4…CnV2(R1,R1,R1,R4,…,Rn)it is(3R1+R4+…+Rn)x
With this we consider that we have given sufficient clarifications to be able to finish the proof of formula (18).
4. In summary, the integral (10) of the differential equation (1) was highlighted in which C 1 C 2 … C n ≠ 0 C 1 C 2 … C n ≠ 0 C_(1)C_(2)dotsC_(n)!=0C_{1} C_{2} \ldots C_{n} \neq 0C1C2…Cn≠0, as well as integrals of the form (17), in which r 1 , r 2 , … , r k r 1 , r 2 , … , r k r_(1),r_(2),dots,r_(k)r_{1}, r_{2}, \ldots, r_{k}R1,R2,…,Rkare distinct numbers, k k kkkbeing 1 , or 2 , … 2 , … 2,dots2, \ldots2,…, or n − 1 n − 1 n-1n-1n−1, the polynomials that multiply e r 1 x , e r 2 x , … , e r k x e r 1 x , e r 2 x , … , e r k x e^(r_(1)x),e^(r_(2)x),dots,e^(r_(k)x)e^{r_{1} x}, e^{r_{2} x}, \ldots, e^{r_{k} x}it isR1x,it isR2x,…,it isRkxbeing of effective degrees p 1 − 1 , p 2 − 1 , … , p k − 1 p 1 − 1 , p 2 − 1 , … , p k − 1 p_(1)-1,p_(2)-1,dots,p_(k-1)p_{1}-1, p_{2}-1, \ldots, p_{k-1}p1−1,p2−1,…,pk−1, where p 1 + p 2 + … + p k = n p 1 + p 2 + … + p k = n p_(1)+p_(2)+dots+p_(k)=np_{1}+p_{2}+\ldots+p_{k}=np1+p2+…+pk=n, that is C 1 C 2 … C k ≠ 0 C 1 C 2 … C k ≠ 0 C_(1)C_(2)dotsC_(k)!=0C_{1} C_{2} \ldots C_{k} \neq 0C1C2…Ck≠0For all these integrals Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0.
Let us now prove that any integral of the equation Δ q [ y ] = 0 Δ q [ y ] = 0 Delta_(q)[y]=0\Delta_{q}[y]=0Δq[y]=0for which Δ q − 1 [ y ] ≠ 0 Δ q − 1 [ y ] ≠ 0 Delta_(q-1)[y]!=0\Delta_{q-1}[y] \neq 0Δq−1[y]≠0, where q < n q < n q < nq<nq<n, is the integral of the differential equation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0.
Whether
Δ q [ y ] = | y y ′ … y ( q − 1 ) y ( q ) y ′ y ′ ′ … y ( q ) y ( q + 1 ) ⋅ ⋅ ⋅ ⋅ ⋅ y ( q ) y ( q + 1 ) … y ( 2 q − 1 ) y ( q ) | = 0 Δ q [ y ] = y y ′ … y ( q − 1 ) y ( q ) y ′ y ′ ′ … y ( q ) y ( q + 1 ) ⋅ ⋅ ⋅ ⋅ ⋅ y ( q ) y ( q + 1 ) … y ( 2 q − 1 ) y ( q ) = 0 Delta_(q)[y]=|[y,y^('),dots,y^((q-1)),y^((q))],[y^('),y^(''),dots,y^((q)),y^((q+1))],[*,*,*,*,*],[y^((q)),y^((q+1)),dots,y^((2q-1)),y^((q))]|=0\Delta_{q}[y]=\left|\begin{array}{ccccc} y & y^{\prime} & \ldots & y^{(q-1)} & y^{(q)} \\ y^{\prime} & y^{\prime \prime} & \ldots & y^{(q)} & y^{(q+1)} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ y^{(q)} & y^{(q+1)} & \ldots & y^{(2 q-1)} & y^{(q)} \end{array}\right|=0Δq[y]=|yy′…y(q−1)y(q)y′y′′…y(q)y(q+1)⋅⋅⋅⋅⋅y(q)y(q+1)…y(2q−1)y(q)|=0
the differential equation and y y yyyan integral of it for which Δ q − 1 [ y ] ≠ 0 Δ q − 1 [ y ] ≠ 0 Delta_(q-1)[y]!=0\Delta_{q-1}[y] \neq 0Δq−1[y]≠0. It is proved as in no. 1, that between the elements of the columns of the determinant Δ q [ y ] Δ q [ y ] Delta_(q)[y]\Delta_{q}[y]Δq[y]there is the same linear relationship with constant coefficients, i.e.
B q y + B q − 1 y ′ + … + B 1 y ( q − 1 ) + y ( q ) = 0 (28) B q y ′ + B q − 1 y ′ ′ + … + B 1 y ( q ) + y ( q + 1 ) = 0 B q y ( q ) + B q − 1 y ( q + 1 ) + … + B 1 y ( 2 q − 1 ) + y ( 2 q ) = 0 . B q y + B q − 1 y ′ + … + B 1 y ( q − 1 ) + y ( q ) = 0 (28) B q y ′ + B q − 1 y ′ ′ + … + B 1 y ( q ) + y ( q + 1 ) = 0 B q y ( q ) + B q − 1 y ( q + 1 ) + … + B 1 y ( 2 q − 1 ) + y ( 2 q ) = 0 . {:[B_(q)y+B_(q-1)y^(')+dots+B_(1)y^((q-1))+y^((q))=0],[(28)B_(q)y^(')+B_(q-1)y^('')+dots+B_(1)y^((q))+y^((q+1))=0],[B_(q)y^((q))+B_(q-1)y^((q+1))+dots+B_(1)y^((2q-1))+y^((2q))=0.]:}\begin{align*} & B_{q} y+B_{q-1} y^{\prime}+\ldots+B_{1} y^{(q-1)}+y^{(q)}=0 \\ & B_{q} y^{\prime}+B_{q-1} y^{\prime \prime}+\ldots+B_{1} y^{(q)}+y^{(q+1)}=0 \tag{28}\\ & B_{q} y^{(q)}+B_{q-1} y^{(q+1)}+\ldots+B_{1} y^{(2 q-1)}+y^{(2 q)}=0 . \end{align*}Bqy+Bq−1y′+…+B1y(q−1)+y(q)=0(28)Bqy′+Bq−1y′′+…+B1y(q)+y(q+1)=0Bqy(q)+Bq−1y(q+1)+…+B1y(2q−1)+y(2q)=0.
But the coefficients B 1 , B 2 , … , B q B 1 , B 2 , … , B q B_(1),B_(2),dots,B_(q)B_{1}, B_{2}, \ldots, B_{q}B1,B2,…,Bqbeing constants, we can successively derive the last equation and we will obtain
( ′ ) B q y ( q + 1 ) + B q − 1 y ( q + 2 ) + … + B 1 y ( 2 q ) + y ( 2 q + 1 ) = 0 B q y ( n ) + B q − 1 y ( n + 1 ) + … + B 1 y ( n + q − 1 ) + y ( n + q ) = 0 ( ′ ) B q y ( q + 1 ) + B q − 1 y ( q + 2 ) + … + B 1 y ( 2 q ) + y ( 2 q + 1 ) = 0 B q y ( n ) + B q − 1 y ( n + 1 ) + … + B 1 y ( n + q − 1 ) + y ( n + q ) = 0 {:[('")"B_(q)y^((q+1))+B_(q-1)y^((q+2))+dots+B_(1)y^((2q))+y^((2q+1))=0],[B_(q)y^((n))+B_(q-1)y^((n+1))+dots+B_(1)y^((n+q-1))+y^((n+q))=0]:}\begin{align*} & B_{q} y^{(q+1)}+B_{q-1} y^{(q+2)}+\ldots+B_{1} y^{(2 q)}+y^{(2 q+1)}=0 \tag{$\prime$}\\ & B_{q} y^{(n)}+B_{q-1} y^{(n+1)}+\ldots+B_{1} y^{(n+q-1)}+y^{(n+q)}=0 \end{align*}(′)Bqy(q+1)+Bq−1y(q+2)+…+B1y(2q)+y(2q+1)=0Bqy(n)+Bq−1y(n+1)+…+B1y(n+q−1)+y(n+q)=0
We thus deduce that
Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0
because between the elements of the first q + 1 q + 1 q+1q+1q+1columns of there is the same linear relationship, as shown by formulas (28) and (28').
5. Conclusion. Let yo be the integral of the equation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0If she doesn't check the equation Δ n − 1 [ y ] = 0 Δ n − 1 [ y ] = 0 Delta_(n-1)[y]=0\Delta_{n-1}[y]=0Δn−1[y]=0, has the form (10), where C 1 C 2 … C n ≠ 0 C 1 C 2 … C n ≠ 0 C_(1)C_(2)dotsC_(n)!=0C_{1} C_{2} \ldots C_{n} \neq 0C1C2…Cn≠0, or the form (17), where k = 1 k = 1 k=1k=1k=1, or 2 , … 2 , … 2,dots2, \ldots2,…, or n − 1 n − 1 n-1n-1n−1, and p 1 , p 2 , p 3 , … , p k p 1 , p 2 , p 3 , … , p k p_(1),p_(2),p_(3),dots,p_(k)p_{1}, p_{2}, p_{3}, \ldots, p_{k}p1,p2,p3,…,pkare positive or zero integers such that p 1 + p 2 + … + p k = n p 1 + p 2 + … + p k = n p_(1)+p_(2)+dots+p_(k)=np_{1}+p_{2}+\ldots+p_{k}=np1+p2+…+pk=n, and C 1 C 2 … C k ≠ 0 C 1 C 2 … C k ≠ 0 C_(1)C_(2)dotsC_(k)!=0C_{1} C_{2} \ldots C_{k} \neq 0C1C2…Ck≠0.
But if the integral y y yyyof the equation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0check the equation Δ n − 1 [ y ] = 0 Δ n − 1 [ y ] = 0 Delta_(n-1)[y]=0\Delta_{n-1}[y]=0Δn−1[y]=0but it doesn't check the equation Δ n − 2 [ y = 0 Δ n − 2 [ y = 0 Delta_(n-2)[y=0\Delta_{n-2}[y=0Δn−2[y=0, it also has one of the forms shown in the previous paragraph, but n n nnnchanges to n − 1 n − 1 n-1n-1n−1.
In general, if the integral y y yyyof the equation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0check the equations
Δ n − 1 [ y ] = 0 , Δ n − 2 [ y ] = 0 , … , Δ n − j [ y ] = 0 Δ n − 1 [ y ] = 0 , Δ n − 2 [ y ] = 0 , … , Δ n − j [ y ] = 0 Delta_(n-1)[y]=0,quadDelta_(n-2)[y]=0,dots,Delta_(n-j)[y]=0\Delta_{n-1}[y]=0, \quad \Delta_{n-2}[y]=0, \ldots, \Delta_{n-j}[y]=0Δn−1[y]=0,Δn−2[y]=0,…,Δn−j[y]=0
but
Δ n → j − 1 [ y ] ≠ 0 Δ n → j − 1 [ y ] ≠ 0 Delta_(n rarr j-1)[y]!=0\Delta_{n \rightarrow j-1}[y] \neq 0Δn→j−1[y]≠0
then it has one of the forms shown in paragraph 1, provided that it changes n n nnnin n − j n − j n-jn-jn−j.
Example. Consider the differential equation
Δ 3 [ y ] = 0 . Δ 3 [ y ] = 0 . Delta_(3)[y]=0.\Delta_{3}[y]=0 .Δ3[y]=0.
1 ∘ 1 ∘ 1^(@)1^{\circ}1∘. If an integral of it y y yyyit is such that Δ 2 [ y ] ≠ 0 Δ 2 [ y ] ≠ 0 Delta_(2)[y]!=0\Delta_{2}[y] \neq 0Δ2[y]≠0, then it has one of the following forms
y = C 1 e r 1 x + C 2 e r 2 x + C 3 e r 2 x y = C 1 e r 1 x + C 2 e r 2 x + C 3 e r 2 x y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)+C_(3)e^(r_(2)x)y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+C_{3} e^{r_{2} x}y=C1it isR1x+C2it isR2x+C3it isR2x
where γ 1 , γ 2 , γ 3 γ 1 , γ 2 , γ 3 gamma_(1),gamma_(2),gamma_(3)\gamma_{1}, \gamma_{2}, \gamma_{3}γ1,γ2,γ3are distinct numbers and C 1 C 2 C 3 ≠ 0 C 1 C 2 C 3 ≠ 0 C_(1)C_(2)C_(3)!=0C_{1} C_{2} C_{3} \neq 0C1C2C3≠0, or
y = ( C 1 x + C 2 ) e r 1 x + C 3 e r 3 x y = C 1 x + C 2 e r 1 x + C 3 e r 3 x y=(C_(1)x+C_(2))e^(r_(1)x)+C_(3)e^(r_(3)x)y=\left(C_{1} x+C_{2}\right) e^{r_{1} x}+C_{3} e^{r_{3} x}y=(C1x+C2)it isR1x+C3it isR3x
where γ 1 , γ 3 γ 1 , γ 3 gamma_(1),gamma_(3)\gamma_{1}, \gamma_{3}γ1,γ3are distinct numbers and C 1 C 2 ≠ 0 C 1 C 2 ≠ 0 C_(1)C_(2)!=0C_{1} C_{2} \neq 0C1C2≠0or
y = ( C 1 x 2 2 ! + C 2 x 1 ! + C 3 ) e r 1 x y = C 1 x 2 2 ! + C 2 x 1 ! + C 3 e r 1 x y=(C_(1)(x^(2))/(2!)+C_(2)(x)/(1!)+C_(3))e^(r_(1)x)y=\left(C_{1} \frac{x^{2}}{2!}+C_{2} \frac{x}{1!}+C_{3}\right) e^{r_{1} x}y=(C1x22!+C2x1!+C3)it isR1x
where C 1 ≠ 0 C 1 ≠ 0 C_(1)!=0C_{1} \neq 0C1≠0.
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘If the integral y y yyyof the differential equation Δ 3 [ y ] = 0 Δ 3 [ y ] = 0 Delta_(3)[y]=0\Delta_{3}[y]=0Δ3[y]=0check the equation Δ 2 [ y ] = 0 Δ 2 [ y ] = 0 Delta_(2)[y]=0\Delta_{2}[y]=0Δ2[y]=0, but Δ 1 [ y ] ≠ 0 Δ 1 [ y ] ≠ 0 Delta_(1)[y]!=0\Delta_{1}[y] \neq 0Δ1[y]≠0, then it has one of the following forms
y = C 1 e r 1 x + C 2 e r 2 x y = C 1 e r 1 x + C 2 e r 2 x y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}y=C1it isR1x+C2it isR2x
where r 1 r 1 r_(1)r_{1}R1and r 2 r 2 r_(2)r_{2}R2are distinct numbers and C 1 C 2 ≠ 0 C 1 C 2 ≠ 0 C_(1)C_(2)!=0C_{1} C_{2} \neq 0C1C2≠0, or
y = ( C 1 x + C 2 ) e r 1 x y = C 1 x + C 2 e r 1 x y=(C_(1)x+C_(2))e^(r_(1)x)y=\left(C_{1} x+C_{2}\right) e^{r_{1} x}y=(C1x+C2)it isR1x
where C 1 ≠ 0 C 1 ≠ 0 C_(1)!=0C_{1} \neq 0C1≠0.
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘If the integral y y yyyof the equation Δ 3 [ y ] = 0 Δ 3 [ y ] = 0 Delta_(3)[y]=0\Delta_{3}[y]=0Δ3[y]=0check the equations Δ 2 [ y ] = 0 Δ 2 [ y ] = 0 Delta_(2)[y]=0\Delta_{2}[y]=0Δ2[y]=0, Δ 1 [ y ] = 0 Δ 1 [ y ] = 0 Delta_(1)[y]=0\Delta_{1}[y]=0Δ1[y]=0, but Δ 0 [ y ] = y ≠ 0 Δ 0 [ y ] = y ≠ 0 Delta_(0)[y]=y!=0\Delta_{0}[y]=y \neq 0Δ0[y]=y≠0, then it has the form
y = C 1 e r 1 x y = C 1 e r 1 x y=C_(1)e^(r_(1)x)y=C_{1} e^{r_{1} x}y=C1it isR1x
where C 1 ≠ 0 C 1 ≠ 0 C_(1)!=0C_{1} \neq 0C1≠0.
4 ∘ 4 ∘ 4^(@)4^{\circ}4∘. If, finally, the integral y y yyyof the differential equation Δ 3 [ y ] = 0 Δ 3 [ y ] = 0 Delta_(3)[y]=0\Delta_{3}[y]=0Δ3[y]=0check the equations Δ 2 [ y ] = 0 , Δ 1 [ y ] = 0 , Δ 0 [ y ] = 0 Δ 2 [ y ] = 0 , Δ 1 [ y ] = 0 , Δ 0 [ y ] = 0 Delta_(2)[y]=0,Delta_(1)[y]=0,Delta_(0)[y]=0\Delta_{2}[y]=0, \Delta_{1}[y]=0, \Delta_{0}[y]=0Δ2[y]=0,Δ1[y]=0,Δ0[y]=0, it is obviously identically zero.
6. Let us proceed to the integration of the differential equation
(29) Δ n [ y ] = A e α x (29) Δ n [ y ] = A e α x {:(29)Delta_(n)[y]=Ae^(alpha x):}\begin{equation*} \Delta_{n}[y]=A e^{\alpha x} \tag{29} \end{equation*}(29)Δn[y]=Ait isαx
where A A AAAand α α alpha\alphaαare constant and A ≠ 0 A ≠ 0 A!=0A \neq 0A≠0For this
, we will use the identity
(30) | Δ n [ y ] Δ n ′ [ y ] Δ n ′ [ y ] Δ n ′ ′ [ y ] | = Δ n − 1 [ y ] Δ n + 1 [ y ] (30) Δ n [ y ] Δ n ′ [ y ] Δ n ′ [ y ] Δ n ′ ′ [ y ] = Δ n − 1 [ y ] Δ n + 1 [ y ] {:(30)|[Delta_(n)[y],Delta_(n)^(')[y]],[Delta_(n)^(')[y],Delta_(n)^('')[y]]|=Delta_(n-1)[y]Delta_(n+1)[y]:}\left|\begin{array}{ll} \Delta_{n}[y] & \Delta_{n}^{\prime}[y] \tag{30}\\ \Delta_{n}^{\prime}[y] & \Delta_{n}^{\prime \prime}[y] \end{array}\right|=\Delta_{n-1}[y] \Delta_{n+1}[y](30)|Δn[y]Δn′[y]Δn′[y]Δn′′[y]|=Δn−1[y]Δn+1[y]
which is proven in the following way.
Let us write the determinant Δ n + 1 [ y ] Δ n + 1 [ y ] Delta_(n+1)[y]\Delta_{n+1}[y]Δn+1[y]in the form of
D = | a 11 … a 1 n a 1 , n + 1 ⋮ ⋮ ⋮ ⋮ … ⋮ ⋮ a n 1 … a n n a n , n + 1 a n + 1 , 1 … a n + 1 , n a n + 1 , n + 1 | D = a 11 … a 1 n a 1 , n + 1 ⋮ ⋮ ⋮ ⋮ … ⋮ ⋮ a n 1 … a n n a n , n + 1 a n + 1 , 1 … a n + 1 , n a n + 1 , n + 1 D=|[a_(11),dots,a_(1n),a_(1,n+1)],[vdots,,vdots,vdots],[vdots,dots,vdots,vdots],[a_(n1),dots,a_(nn),a_(n,n+1)],[a_(n+1,1),dots,a_(n+1,n),a_(n+1,n+1)]|D=\left|\begin{array}{cccc} a_{11} & \ldots & a_{1 n} & a_{1, n+1} \\ \vdots & & \vdots & \vdots \\ \vdots & \ldots & \vdots & \vdots \\ a_{n 1} & \ldots & a_{n n} & a_{n, n+1} \\ a_{n+1,1} & \ldots & a_{n+1, n} & a_{n+1, n+1} \end{array}\right|D=|A11…A1nA1,n+1⋮⋮⋮⋮…⋮⋮An1…AnnAn,n+1An+1,1…An+1,nAn+1,n+1|
and let's notice that we have
Δ n [ y ] = ∂ D ∂ a n + 1 , n + 1 , Δ n ′ [ y ] = − ∂ D ∂ a n , n + 1 = − ∂ D ∂ a r + 1 , n , Δ n ′ ′ [ y ] = ∂ D ∂ a n , n . Δ n [ y ] = ∂ D ∂ a n + 1 , n + 1 , Δ n ′ [ y ] = − ∂ D ∂ a n , n + 1 = − ∂ D ∂ a r + 1 , n , Δ n ′ ′ [ y ] = ∂ D ∂ a n , n . Delta_(n)[y]=(del D)/(dela_(n+1,n+1)),quadDelta_(n)^(')[y]=-(del D)/(dela_(n,n+1))=-(del D)/(dela_(r+1,n)),quadDelta_(n)^('')[y]=(del D)/(dela_(n,n)).\Delta_{n}[y]=\frac{\partial D}{\partial a_{n+1, n+1}}, \quad \Delta_{n}^{\prime}[y]=-\frac{\partial D}{\partial a_{n, n+1}}=-\frac{\partial D}{\partial a_{r+1, n}}, \quad \Delta_{n}^{\prime \prime}[y]=\frac{\partial D}{\partial a_{n, n}} .Δn[y]=∂D∂An+1,n+1,Δn′[y]=−∂D∂An,n+1=−∂D∂AR+1,n,Δn′′[y]=∂D∂An,n.
It is known that we have the identity [4]
| ∂ D ∂ a n , n ∂ D ∂ a n , n + 1 ∂ D ∂ a n + 1 , n ∂ D ∂ a n + 1 , n + 1 | = D 2 D 1 ∂ D ∂ a n , n ∂ D ∂ a n , n + 1 ∂ D ∂ a n + 1 , n ∂ D ∂ a n + 1 , n + 1 = D 2 D 1 |[(del D)/(dela_(n,n)),(del D)/(dela_(n,n+1))],[(del D)/(dela_(n+1,n)),(del D)/(dela_(n+1,n+1))]|=D_(2)D_(1)\left|\begin{array}{cc} \frac{\partial D}{\partial a_{n, n}} & \frac{\partial D}{\partial a_{n, n+1}} \\ \frac{\partial D}{\partial a_{n+1, n}} & \frac{\partial D}{\partial a_{n+1, n+1}} \end{array}\right|=D_{2} D_{1}|∂D∂An,n∂D∂An,n+1∂D∂An+1,n∂D∂An+1,n+1|=D2D1
where D 2 D 2 D_(2)D_{2}D2is the determinant obtained from D D DDDby deleting the last two rows and columns. This identity leads to identity (30).
Whether y y yyyan integral of the differential equation (29). It is obvious that u = A e a x u = A e a x u=Ae^(ax)u=A e^{a x}you=Ait isAxbeing an integral of the differential equation Δ 1 [ u ] = 0 Δ 1 [ u ] = 0 Delta_(1)[u]=0\Delta_{1}[u]=0Δ1[you]=0, we will have applying identity (30)
Δ n − 1 [ y ] Δ n + 1 [ y ] = 0 . Δ n − 1 [ y ] Δ n + 1 [ y ] = 0 . Delta_(n-1)[y]Delta_(n+1)[y]=0.\Delta_{n-1}[y] \Delta_{n+1}[y]=0 .Δn−1[y]Δn+1[y]=0.
But it can't be that y y yyyto check the equation Δ n − 1 [ y ] = 0 Δ n − 1 [ y ] = 0 Delta_(n-1)[y]=0\Delta_{n-1}[y]=0Δn−1[y]=0, because we would also have Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0, which is impossible because A ≠ 0 A ≠ 0 A_(!=0)A_{\neq 0}A≠0. So any integral of the differential equation (29) is an integral of the differential equation
(31) Δ n + 1 [ y ] = 0 (31) Δ n + 1 [ y ] = 0 {:(31)Delta_(n+1)[y]=0:}\begin{equation*} \Delta_{n+1}[y]=0 \tag{31} \end{equation*}(31)Δn+1[y]=0
and therefore the integrals of the differential equation (29) must be sought among the integrals of the differential equation (31).
Let us first consider the integral of the differential equation (31)
y = C 1 e r 1 x + C 2 e r 2 x + … + C n e r n x + C n + 1 e r n + 1 x y = C 1 e r 1 x + C 2 e r 2 x + … + C n e r n x + C n + 1 e r n + 1 x y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)+dots+C_(n)e^(r_(n)x)+C_(n+1)e^(r_(n+1)x)y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+\ldots+C_{n} e^{r_{n} x}+C_{n+1} e^{r_{n+1} x}y=C1it isR1x+C2it isR2x+…+Cnit isRnx+Cn+1it isRn+1x
for which Δ n [ y ] ≠ 0 Δ n [ y ] ≠ 0 Delta_(n)[y]!=0\Delta_{n}[y] \neq 0Δn[y]≠0, that is r 1 , r 2 , … , r n , r n + 1 r 1 , r 2 , … , r n , r n + 1 r_(1),r_(2),dots,r_(n),r_(n+1)r_{1}, r_{2}, \ldots, r_{n}, r_{n+1}R1,R2,…,Rn,Rn+1are distinct numbers and C 1 C 2 … C n C n + 1 ≠ 0 C 1 C 2 … C n C n + 1 ≠ 0 C_(1)C_(2)dotsC_(n)C_(n+1)!=0C_{1} C_{2} \ldots C_{n} C_{n+1} \neq 0C1C2…CnCn+1≠0According to formula (12), we have
Δ n [ y ] = C 1 C 2 … C n − C n + 1 V 2 ( r 1 , r 2 , … , r n , r n + 1 ) e ( r 1 + r 2 + … + r n + r n + 1 ) x Δ n [ y ] = C 1 C 2 … C n − C n + 1 V 2 r 1 , r 2 , … , r n , r n + 1 e r 1 + r 2 + … + r n + r n + 1 x Delta_(n)[y]=C_(1)C_(2)dotsC_(n)^(-)C_(n+1)V^(2)(r_(1),r_(2),dots,r_(n),r_(n+1))e^((r_(1)+r_(2)+dots+r_(n)+r_(n+1))x)\Delta_{n}[y]=C_{1} C_{2} \ldots C_{n}^{-} C_{n+1} V^{2}\left(r_{1}, r_{2}, \ldots, r_{n}, r_{n+1}\right) e^{\left(r_{1}+r_{2}+\ldots+r_{n}+r_{n+1}\right) x}Δn[y]=C1C2…Cn−Cn+1V2(R1,R2,…,Rn,Rn+1)it is(R1+R2+…+Rn+Rn+1)x
For equation (29) to be satisfied, we will choose
r 1 + r 2 + … + r n + r n + 1 = α C 1 C 2 … C n C n + 1 V 2 ( r 1 , r 2 , … , r n , r n + 1 ) = A r 1 + r 2 + … + r n + r n + 1 = α C 1 C 2 … C n C n + 1 V 2 r 1 , r 2 , … , r n , r n + 1 = A {:[r_(1)+r_(2)+dots+r_(n)+r_(n+1)=alpha],[C_(1)C_(2)dotsC_(n)C_(n+1)V^(2)(r_(1),r_(2),dots,r_(n),r_(n+1))=A]:}\begin{aligned} & r_{1}+r_{2}+\ldots+r_{n}+r_{n+1}=\alpha \\ & C_{1} C_{2} \ldots C_{n} C_{n+1} V^{2}\left(r_{1}, r_{2}, \ldots, r_{n}, r_{n+1}\right)=A \end{aligned}R1+R2+…+Rn+Rn+1=αC1C2…CnCn+1V2(R1,R2,…,Rn,Rn+1)=A
The first equation determines the r n + 1 r n + 1 r_(n+1)r_{n+1}Rn+1, that is
r n + 1 = α − ( r 1 + r 2 + … + r n ) r n + 1 = α − r 1 + r 2 + … + r n r_(n+1)=alpha-(r_(1)+r_(2)+dots+r_(n))r_{n+1}=\alpha-\left(r_{1}+r_{2}+\ldots+r_{n}\right)Rn+1=α−(R1+R2+…+Rn)
and we will choose r 1 , r 2 , … , r n r 1 , r 2 , … , r n r_(1),r_(2),dots,r_(n)r_{1}, r_{2}, \ldots, r_{n}R1,R2,…,Rnso that r 1 , r 2 , … , r n , r n + 1 r 1 , r 2 , … , r n , r n + 1 r_(1),r_(2),dots,r_(n),r_(n+1)r_{1}, r_{2}, \ldots, r_{n}, r_{n+1}R1,R2,…,Rn,Rn+1to be distinct numbers. Then the second equation determines C n + 1 C n + 1 C_(n+1)C_{n+1}Cn+1and we will have
C n + 1 = A C 1 C 2 … C n V 2 [ r 1 , r 2 … r n , α − ( r 1 + r 2 + … + r n ) ] C n + 1 = A C 1 C 2 … C n V 2 r 1 , r 2 … r n , α − r 1 + r 2 + … + r n C_(n+1)=(A)/(C_(1)C_(2)dotsC_(n)V^(2)[r_(1),r_(2)dotsr_(n),alpha-(r_(1)+r_(2)+dots+r_(n))])C_{n+1}=\frac{A}{C_{1} C_{2} \ldots C_{n} V^{2}\left[r_{1}, r_{2} \ldots r_{n}, \alpha-\left(r_{1}+r_{2}+\ldots+r_{n}\right)\right]}Cn+1=AC1C2…CnV2[R1,R2…Rn,α−(R1+R2+…+Rn)]
The integral of equation (29) is presented in the form
y = C 1 e r 1 x + C 2 e r 2 x + … + C n e r n x + A e [ α − ( r 1 + r 2 + … + r n ) ] x C 1 C 2 … C n V 2 [ r 1 , r 2 , … , r n , α − ( r 1 + r 2 + … + r n ) ] y = C 1 e r 1 x + C 2 e r 2 x + … + C n e r n x + A e α − r 1 + r 2 + … + r n x C 1 C 2 … C n V 2 r 1 , r 2 , … , r n , α − r 1 + r 2 + … + r n y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)+dots+C_(n)e^(r_(n)x)+(Ae^([alpha-(r_(1)+r_(2)+dots+r_(n))]x))/(C_(1)C_(2)dotsC_(n)V^(2)[r_(1),r_(2),dots,r_(n),alpha-(r_(1)+r_(2)+dots+r_(n))])y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+\ldots+C_{n} e^{r_{n} x}+\frac{A e^{\left[\alpha-\left(r_{1}+r_{2}+\ldots+r_{n}\right)\right] x}}{C_{1} C_{2} \ldots C_{n} V^{2}\left[r_{1}, r_{2}, \ldots, r_{n}, \alpha-\left(r_{1}+r_{2}+\ldots+r_{n}\right)\right]}y=C1it isR1x+C2it isR2x+…+Cnit isRnx+Ait is[α−(R1+R2+…+Rn)]xC1C2…CnV2[R1,R2,…,Rn,α−(R1+R2+…+Rn)]
However, let us also consider the integrals of the differential equation (31) of the form (17), that is
y = [ C 1 ( p 1 − 1 ) ! x p 1 − 1 + C 1 , 1 ( p 1 − 2 ) ! x p 1 − 2 + … + C 1 , p 1 − 1 ] e r 2 x + ⋯ ⋯ + ⋯ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + [ C k − 1 ( p k − 1 − 1 ) ! x p k − 1 − 1 + C k − 1 , 1 ( p k − 1 − 2 ) ! x p k − 1 − 2 + … + C k − 1 , p k − 1 − 1 ] e r k − 1 x (33) + [ C k ( p k − 1 ) ! x p k − 1 + C k − 1 , 1 ( p k − 2 ) ! x p k − 2 + … + C k , p k − 1 ] e r k x y = C 1 p 1 − 1 ! x p 1 − 1 + C 1 , 1 p 1 − 2 ! x p 1 − 2 + … + C 1 , p 1 − 1 e r 2 x + ⋯ ⋯ + ⋯ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + C k − 1 p k − 1 − 1 ! x p k − 1 − 1 + C k − 1 , 1 p k − 1 − 2 ! x p k − 1 − 2 + … + C k − 1 , p k − 1 − 1 e r k − 1 x (33) + C k p k − 1 ! x p k − 1 + C k − 1 , 1 p k − 2 ! x p k − 2 + … + C k , p k − 1 e r k x {:[y=[(C_(1))/((p_(1)-1)!)x^(p_(1)-1)+(C_(1,1))/((p_(1)-2)!)x^(p_(1)-2)+dots+C_(1,p_(1)-1)]e^(r_(2)x)],[+cdots cdots+cdots***********],[+[(C_(k-1))/((p_(k-1)-1)!)x^(p_(k-1)-1)+(C_(k-1,1))/((p_(k-1)-2)!)x^(p_(k-1)-2)+dots+C_(k-1,p_(k-1)-1)]e^(r_(k-1)x)],[(33)+[(C_(k))/((p_(k)-1)!)x^(p_(k)-1)+(C_(k-1,1))/((p_(k)-2)!)x^(p_(k)-2)+dots+C_(k,p_(k)-1)]e^(r_(k)x)]:}\begin{align*} y & =\left[\frac{C_{1}}{\left(p_{1}-1\right)!} x^{p_{1}-1}+\frac{C_{1,1}}{\left(p_{1}-2\right)!} x^{p_{1}-2}+\ldots+C_{1, p_{1}-1}\right] e^{r_{2} x} \\ & +\cdots \cdots+\cdots \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\ & +\left[\frac{C_{k-1}}{\left(p_{k-1}-1\right)!} x^{p_{k-1}-1}+\frac{C_{k-1,1}}{\left(p_{k-1}-2\right)!} x^{p_{k-1}-2}+\ldots+C_{k-1, p_{k-1}-1}\right] e^{r_{k-1} x} \\ & +\left[\frac{C_{k}}{\left(p_{k}-1\right)!} x^{p_{k}-1}+\frac{C_{k-1,1}}{\left(p_{k}-2\right)!} x^{p_{k}-2}+\ldots+C_{k, p_{k}-1}\right] e^{r_{k} x} \tag{33} \end{align*}y=[C1(p1−1)!xp1−1+C1,1(p1−2)!xp1−2+…+C1,p1−1]it isR2x+⋯⋯+⋯⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅+[Ck−1(pk−1−1)!xpk−1−1+Ck−1,1(pk−1−2)!xpk−1−2+…+Ck−1,pk−1−1]it isRk−1x(33)+[Ck(pk−1)!xpk−1+Ck−1,1(pk−2)!xpk−2+…+Ck,pk−1]it isRkx
where
(34) p 1 + p 2 + … + p k = n + 1 (34) p 1 + p 2 + … + p k = n + 1 {:(34)p_(1)+p_(2)+dots+p_(k)=n+1:}\begin{equation*} p_{1}+p_{2}+\ldots+p_{k}=n+1 \tag{34} \end{equation*}(34)p1+p2+…+pk=n+1
and r 1 , r 2 , … , r k r 1 , r 2 , … , r k r_(1),r_(2),dots,r_(k)r_{1}, r_{2}, \ldots, r_{k}R1,R2,…,Rkare distinct. According to formula (18), we have
Δ n [ y ] = ( − 1 ) k C 1 p 1 C 2 p 2 … C k − 1 p k − 1 C k p k V 2 ( r 1 , … , r 1 ⏟ p 1 ori , … , r k , … , r k ⏟ p 1 ori ) e ( p 1 r 1 + … + p k r Λ ) x Δ n [ y ] = ( − 1 ) k C 1 p 1 C 2 p 2 … C k − 1 p k − 1 C k p k V 2 ( r 1 , … , r 1 ⏟ p 1  ori  , … , r k , … , r k ⏟ p 1  ori  ) e p 1 r 1 + … + p k r Λ x Delta_(n)[y]=(-1)^(k)C_(1)^(p_(1))C_(2)^(p_(2))dotsC_(k-1)^(p_(k-1))C_(k)^(p_(k))V^(2)(ubrace(r_(1),dots,r_(1))_(p_(1)" ori "),dots,ubrace(r_(k),dots,r_(k))_(p_(1)" ori "))e^((p_(1)r_(1)+dots+p_(k)r_(Lambda))x)\Delta_{n}[y]=(-1)^{k} C_{1}^{p_{1}} C_{2}^{p_{2}} \ldots C_{k-1}^{p_{k-1}} C_{k}^{p_{k}} V^{2}(\underbrace{r_{1}, \ldots, r_{1}}_{p_{1} \text { ori }}, \ldots, \underbrace{r_{k}, \ldots, r_{k}}_{p_{1} \text { ori }}) e^{\left(p_{1} r_{1}+\ldots+p_{k} r_{\Lambda}\right) x}Δn[y]=(−1)kC1p1C2p2…Ck−1pk−1CkpkV2(R1,…,R1⏟p1 OR ,…,Rk,…,Rk⏟p1 OR )it is(p1R1+…+pkRΛ)x
For the second member to reduce to A e α x A e α x Ae^(alpha x)A e^{\alpha x}Ait isαx, we will choose
p 1 r 1 + … + p k − 1 r k − 1 + p k r k = α ( − 1 ) k C 1 p 1 C 2 p 2 … C k p k V 2 ( r 1 , … , r 1 ⏟ p 1 ori , … , r k , … , r k ⏟ p k ori ) = A p 1 r 1 + … + p k − 1 r k − 1 + p k r k = α ( − 1 ) k C 1 p 1 C 2 p 2 … C k p k V 2 ( r 1 , … , r 1 ⏟ p 1  ori  , … , r k , … , r k ⏟ p k  ori  ) = A {:[p_(1)r_(1)+dots+p_(k-1)r_(k-1)+p_(k)r_(k)=alpha],[(-1)^(k)C_(1)^(p_(1))C_(2)^(p_(2))dotsC_(k)^(p_(k))V^(2)(ubrace(r_(1),dots,r_(1))_(p_(1)" ori ")","dots","ubrace(r_(k),dots,r_(k))_(p_(k)" ori "))=A]:}\begin{gathered} p_{1} r_{1}+\ldots+p_{k-1} r_{k-1}+p_{k} r_{k}=\alpha \\ (-1)^{k} C_{1}^{p_{1}} C_{2}^{p_{2}} \ldots C_{k}^{p_{k}} V^{2}(\underbrace{r_{1}, \ldots, r_{1}}_{p_{1} \text { ori }}, \ldots, \underbrace{r_{k}, \ldots, r_{k}}_{p_{k} \text { ori }})=A \end{gathered}p1R1+…+pk−1Rk−1+pkRk=α(−1)kC1p1C2p2…CkpkV2(R1,…,R1⏟p1 OR ,…,Rk,…,Rk⏟pk OR )=A
where the numbers p 1 , p 2 , … , p k − 1 , p k p 1 , p 2 , … , p k − 1 , p k p_(1),p_(2),dots,p_(k-1),p_(k)p_{1}, p_{2}, \ldots, p_{k-1}, p_{k}p1,p2,…,pk−1,pkare fixed so that we have the relation (34) and that p k ≠ 0 p k ≠ 0 p_(k)!=0p_{k} \neq 0pk≠0.
Then we will take
(35) r k = α − ( p 1 r 1 + … + p k − 1 r k − 1 ) p k (35) r k = α − p 1 r 1 + … + p k − 1 r k − 1 p k {:(35)r_(k)=(alpha-(p_(1)r_(1)+dots+p_(k-1)r_(k-1)))/(p_(k)):}\begin{equation*} r_{k}=\frac{\alpha-\left(p_{1} r_{1}+\ldots+p_{k-1} r_{k-1}\right)}{p_{k}} \tag{35} \end{equation*}(35)Rk=α−(p1R1+…+pk−1Rk−1)pk
the numbers r 1 , r 2 , … , r k − 1 r 1 , r 2 , … , r k − 1 r_(1),r_(2),dots,r_(k-1)r_{1}, r_{2}, \ldots, r_{k-1}R1,R2,…,Rk−1being chosen so that all numbers r 1 , r 2 , … , r k r 1 , r 2 , … , r k r_(1),r_(2),dots,r_(k)r_{1}, r_{2}, \ldots, r_{k}R1,R2,…,Rkto be distinct.
Next we will take
C k = A ( − 1 ) k C 1 p 1 C 2 p 2 … C k − 1 p k − 1 V 2 ( r 1 , … , r 1 ⏟ p 1 ori , … , r p k , … , r k r k ) p k C k = A ( − 1 ) k C 1 p 1 C 2 p 2 … C k − 1 p k − 1 V 2 ( r 1 , … , r 1 ⏟ p 1  ori  , … , r p k , … , r k r k ) p k C_(k)=root(p_(k))((A)/((-1)^(k)C_(1)^(p_(1))C_(2)^(p_(2))dotsC_(k-1)^(p_(k-1))V^(2)(ubrace(r_(1),dots,r_(1))_(p_(1)" ori "),dots,r_(p_(k),dots,r_(k))^(r_(k)))))C_{k}=\sqrt[p_{k}]{\frac{A}{(-1)^{k} C_{1}^{p_{1}} C_{2}^{p_{2}} \ldots C_{k-1}^{p_{k-1}} V^{2}(\underbrace{r_{1}, \ldots, r_{1}}_{p_{1} \text { ori }}, \ldots, r_{p_{k}, \ldots, r_{k}}^{r_{k}})}}Ck=A(−1)kC1p1C2p2…Ck−1pk−1V2(R1,…,R1⏟p1 OR ,…,Rpk,…,RkRk)pk
and then the differential equation (29) still has the integrals given by formula (33), where r k r k r_(k)r_{k}Rkand C k C k C_(k)C_{k}Ckare given by formulas (35) and (36).

Examples.

1 ∘ 1 ∘ 1^(@)1^{\circ}1∘. The equation of the chain. This is
Δ [ y ] = | y y ′ y ′ y ′ ′ | = 1 Δ [ y ] = y y ′ y ′ y ′ ′ = 1 Delta[y]=|[y,y^(')],[y^('),y^('')]|=1\Delta[y]=\left|\begin{array}{ll} y & y^{\prime} \\ y^{\prime} & y^{\prime \prime} \end{array}\right|=1Δ[y]=|yy′y′y′′|=1
Its integral is found among the integrals of the differential equation
Δ 2 [ y ] = 0 Δ 2 [ y ] = 0 Delta_(2)[y]=0\Delta_{2}[y]=0Δ2[y]=0
This equation has integrals of the form
(37) y = C 1 e r 1 x + C 2 e r 2 x (37) y = C 1 e r 1 x + C 2 e r 2 x {:(37)y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x):}\begin{equation*} y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x} \tag{37} \end{equation*}(37)y=C1it isR1x+C2it isR2x
and
(38) y = ( C 1 x + C 2 ) e r 1 x (38) y = C 1 x + C 2 e r 1 x {:(38)y=(C_(1)x+C_(2))e^(r_(1)x):}\begin{equation*} y=\left(C_{1} x+C_{2}\right) e^{r_{1} x} \tag{38} \end{equation*}(38)y=(C1x+C2)it isR1x
For these integrals we have
Δ 1 [ y ] = C 1 C 2 ( r 1 − r 2 ) 2 e ( r 1 + r 2 ) x Δ 1 [ y ] = − C 1 2 e 2 r 1 x . Δ 1 [ y ] = C 1 C 2 r 1 − r 2 2 e r 1 + r 2 x Δ 1 [ y ] = − C 1 2 e 2 r 1 x . {:[Delta_(1)[y]=C_(1)C_(2)(r_(1)-r_(2))^(2)e^((r_(1)+r_(2))x)],[Delta_(1)[y]=-C_(1)^(2)e^(2r_(1)x).]:}\begin{aligned} & \Delta_{1}[y]=C_{1} C_{2}\left(r_{1}-r_{2}\right)^{2} e^{\left(r_{1}+r_{2}\right) x} \\ & \Delta_{1}[y]=-C_{1}^{2} e^{2 r_{1} x} . \end{aligned}Δ1[y]=C1C2(R1−R2)2it is(R1+R2)xΔ1[y]=−C12it is2R1x.
For integrals of the form (37), so that we have Δ 1 [ y ] = 1 Δ 1 [ y ] = 1 Delta_(1)[y]=1\Delta_{1}[y]=1Δ1[y]=1we will choose
r 1 + r 2 = 0 , r 1 − r 2 = 1 C 1 C 2 r 1 + r 2 = 0 , r 1 − r 2 = 1 C 1 C 2 r_(1)+r_(2)=0,quadr_(1)-r_(2)=(1)/(sqrt(C_(1)C_(2)))r_{1}+r_{2}=0, \quad r_{1}-r_{2}=\frac{1}{\sqrt{C_{1} C_{2}}}R1+R2=0,R1−R2=1C1C2
which leads us to
( ′ ) y = C 1 e x 2 C 1 C 2 + C 2 e − x 2 C 1 C 2 ( ′ ) y = C 1 e x 2 C 1 C 2 + C 2 e − x 2 C 1 C 2 {:('")"y=C_(1)e^((x)/(2sqrt(C_(1)C_(2))))+C_(2)e^(-(x)/(2sqrt(C_(1)C_(2)))):}\begin{equation*} y=C_{1} e^{\frac{x}{2 \sqrt{C_{1} C_{2}}}}+C_{2} e^{-\frac{x}{2 \sqrt{C_{1} C_{2}}}} \tag{$\prime$} \end{equation*}(′)y=C1it isx2C1C2+C2it is−x2C1C2
For integrals of the form (38), so that we have Δ 1 [ y ] = 1 Δ 1 [ y ] = 1 Delta_(1)[y]=1\Delta_{1}[y]=1Δ1[y]=1we will choose r 1 = 0 , C 1 = i r 1 = 0 , C 1 = i r_(1)=0,C_(1)=ir_{1}=0, \mathrm{C}_{1}=iR1=0,C1=and, which leads us to
(38') y = i x + C 2 (38') y = i x + C 2 {:(38')y=ix+C_(2):}\begin{equation*} y=i x+C_{2} \tag{38'} \end{equation*}(38')y=andx+C2
The integrals of the differential equation of the chain are ( 37 ′ ) 37 ′ (37^('))\left(37^{\prime}\right)(37′)and ( 38 ′ ) 38 ′ (38^('))\left(38^{\prime}\right)(38′).
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘. Darboux equation. It is the equation
Δ 2 [ y ] = | y y ′ y ′ ′ y ′ y ′ ′ y ′ ′ ′ y ′ ′ y ′ ′ ′ y ′ v | = 1 Δ 2 [ y ] = y y ′ y ′ ′ y ′ y ′ ′ y ′ ′ ′ y ′ ′ y ′ ′ ′ y ′ v = 1 Delta_(2)[y]=|[y,y^('),y^('')],[y^('),y^(''),y^(''')],[y^(''),y^('''),y^('v)]|=1\Delta_{2}[y]=\left|\begin{array}{ccc} y & y^{\prime} & y^{\prime \prime} \\ y^{\prime} & y^{\prime \prime} & y^{\prime \prime \prime} \\ y^{\prime \prime} & y^{\prime \prime \prime} & y^{\prime v} \end{array}\right|=1Δ2[y]=|yy′y′′y′y′′y′′′y′′y′′′y′V|=1
Its integrals are found among the integrals of the equation
which are of the form
Δ 3 [ y ] = 0 Δ 3 [ y ] = 0 Delta_(3)[y]=0\Delta_{3}[y]=0Δ3[y]=0
(39) y = C 1 e r 1 x + C 2 e r 2 x + C 3 e r 3 x (40) y = ( C 1 x + C 2 ) e r 1 x + C 3 e r 3 x (41) y = ( C 1 x 2 2 + C 2 x + C 3 ) e r 1 x (39) y = C 1 e r 1 x + C 2 e r 2 x + C 3 e r 3 x (40) y = C 1 x + C 2 e r 1 x + C 3 e r 3 x (41) y = C 1 x 2 2 + C 2 x + C 3 e r 1 x {:[(39)y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)+C_(3)e^(r_(3)x)],[(40)y=(C_(1)x+C_(2))e^(r_(1)x)+C_(3)e^(r_(3)x)],[(41)y=(C_(1)(x^(2))/(2)+C_(2)x+C_(3))e^(r_(1)x)]:}\begin{align*} & y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+C_{3} e^{r_{3} x} \tag{39}\\ & y=\left(C_{1} x+C_{2}\right) e^{r_{1} x}+C_{3} e^{r_{3} x} \tag{40}\\ & y=\left(C_{1} \frac{x^{2}}{2}+C_{2} x+C_{3}\right) e^{r_{1} x} \tag{41} \end{align*}(39)y=C1it isR1x+C2it isR2x+C3it isR3x(40)y=(C1x+C2)it isR1x+C3it isR3x(41)y=(C1x22+C2x+C3)it isR1x
For these types of integrals, we have
Δ 2 [ y ] = C 1 C 2 C 3 V 2 ( r 1 , r 2 , r 3 ) e ( r 1 + r 2 + r 3 ) x Δ 2 [ y ] = − C 1 2 C 3 V 2 ( r 1 , r 1 , r 3 ) e ( 2 r 1 + r 3 ) x Δ 2 [ y ] = − C 1 3 e 3 r 1 x Δ 2 [ y ] = C 1 C 2 C 3 V 2 r 1 , r 2 , r 3 e r 1 + r 2 + r 3 x Δ 2 [ y ] = − C 1 2 C 3 V 2 r 1 , r 1 , r 3 e 2 r 1 + r 3 x Δ 2 [ y ] = − C 1 3 e 3 r 1 x {:[Delta_(2)[y]=C_(1)C_(2)C_(3)V^(2)(r_(1),r_(2),r_(3))e^((r_(1)+r_(2)+r_(3))x)],[Delta_(2)[y]=-C_(1)^(2)C_(3)V^(2)(r_(1),r_(1),r_(3))e^((2r_(1)+r_(3))x)],[Delta_(2)[y]=-C_(1)^(3)e^(3r_(1)x)]:}\begin{aligned} & \Delta_{2}[y]=C_{1} C_{2} C_{3} V^{2}\left(r_{1}, r_{2}, r_{3}\right) e^{\left(r_{1}+r_{2}+r_{3}\right) x} \\ & \Delta_{2}[y]=-C_{1}^{2} C_{3} V^{2}\left(r_{1}, r_{1}, r_{3}\right) e^{\left(2 r_{1}+r_{3}\right) x} \\ & \Delta_{2}[y]=-C_{1}^{3} e^{3 r_{1} x} \end{aligned}Δ2[y]=C1C2C3V2(R1,R2,R3)it is(R1+R2+R3)xΔ2[y]=−C12C3V2(R1,R1,R3)it is(2R1+R3)xΔ2[y]=−C13it is3R1x
Because Δ 2 [ y ] Δ 2 [ y ] Delta_(2)[y]\Delta_{2}[y]Δ2[y]to be equal to 1, we will choose in the first case
r 3 = − ( r 1 + r 2 ) , C 3 = 1 C 1 C 2 V 2 [ r 1 , r 2 , − ( r 1 + r 2 ) ] r 3 = − r 1 + r 2 , C 3 = 1 C 1 C 2 V 2 r 1 , r 2 , − r 1 + r 2 r_(3)=-(r_(1)+r_(2)),C_(3)=(1)/(C_(1)C_(2)V^(2)[r_(1),r_(2),-(r_(1)+r_(2))])r_{3}=-\left(r_{1}+r_{2}\right), C_{3}=\frac{1}{C_{1} C_{2} V^{2}\left[r_{1}, r_{2},-\left(r_{1}+r_{2}\right)\right]}R3=−(R1+R2),C3=1C1C2V2[R1,R2,−(R1+R2)]
in the second case
r 3 = − 2 r 1 , C 3 = − 1 C 1 2 V 2 ( r 1 , r 1 , − 2 r 1 ) r 3 = − 2 r 1 , C 3 = − 1 C 1 2 V 2 r 1 , r 1 , − 2 r 1 r_(3)=-2r_(1),quadC_(3)=-(1)/(C_(1)^(2)V^(2)(r_(1),r_(1),-2r_(1)))r_{3}=-2 r_{1}, \quad C_{3}=-\frac{1}{C_{1}^{2} V^{2}\left(r_{1}, r_{1},-2 r_{1}\right)}R3=−2R1,C3=−1C12V2(R1,R1,−2R1)
and in the third case
r 1 = 0 C 1 = − 1 r 1 = 0 C 1 = − 1 r_(1)=0quadC_(1)=-1r_{1}=0 \quad C_{1}=-1R1=0C1=−1
This leads us to the following integrals of Darboux's equation:
(39) y = C 1 e r 1 x + C 2 e r 2 x + e − ( r 1 + r 2 ) x C 1 C 2 ( r 2 − r 1 ) 2 ( 2 r 1 + r 2 ) 2 ( r 1 + 2 r 2 ) 2 ( ′ ) y = ( C 1 x + C 2 ) e r 1 x − e − 2 r 1 x 81 C 1 2 r 1 4 (41́) y = − x 2 2 + C 2 x + C 3 (39) y = C 1 e r 1 x + C 2 e r 2 x + e − r 1 + r 2 x C 1 C 2 r 2 − r 1 2 2 r 1 + r 2 2 r 1 + 2 r 2 2 ( ′ ) y = C 1 x + C 2 e r 1 x − e − 2 r 1 x 81 C 1 2 r 1 4 (41́) y = − x 2 2 + C 2 x + C 3 {:[(39)y=C_(1)e^(r_(1)x)+C_(2)e^(r_(2)x)+(e^(-(r_(1)+r_(2))x))/(C_(1)C_(2)(r_(2)-r_(1))^(2)(2r_(1)+r_(2))^(2)(r_(1)+2r_(2))^(2))],[('")"y=(C_(1)x+C_(2))e^(r_(1)x)-(e^(-2r_(1)x))/(81C_(1)^(2)r_(1)^(4))],[(41́)y=-(x^(2))/(2)+C_(2)x+C_(3)]:}\begin{align*} & y=C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+\frac{e^{-\left(r_{1}+r_{2}\right) x}}{C_{1} C_{2}\left(r_{2}-r_{1}\right)^{2}\left(2 r_{1}+r_{2}\right)^{2}\left(r_{1}+2 r_{2}\right)^{2}} \tag{39}\\ & y=\left(C_{1} x+C_{2}\right) e^{r_{1} x}-\frac{e^{-2 r_{1} x}}{81 C_{1}^{2} r_{1}^{4}} \tag{$\prime$}\\ & y=-\frac{x^{2}}{2}+C_{2} x+C_{3} \tag{41́} \end{align*}(39)y=C1it isR1x+C2it isR2x+it is−(R1+R2)xC1C2(R2−R1)2(2R1+R2)2(R1+2R2)2(′)y=(C1x+C2)it isR1x−it is−2R1x81C12R14(41́)y=−x22+C2x+C3

BIBLIOGRAPHY

  1. H. Löwner, Über monotone Matrixfunctionen. Math. Zeit., 38 (1934) 177-216.
  2. G. Darboux, Sur une équation différenceielle du quatriòme ordre. CR de l'Ac. often Sci. de Paris, vol. CXLI, pp. 415-417.
    • Sur une équation differential du quatrième ordre, CR de l'Ac. often Sci. de Paris, vol. CXLI, pp. 483-484.
  4. AG Kuros, Course in Higher Algebra. Technical Publishing House, Buc., 1955, p. 133.
  5. Th. J. Stieltjes, Quelques recherches sur la théorie des quadratures dites mécaniques. Ann. of the Ecole Normale Supérieure, 1884, pp. 409-426.
  6. DV Ionescu, Numerical Quadratures. Technical Publishing House, Buc., 1957, p. 262.

Integration of one differential equation

(Brief content)
In this work, the differential equation is integrated, (1) Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0and differential equation (2) Δ n [ y ] = A e α x Δ n [ y ] = A e α x Delta_(n)[y]=Ae^(alpha x)\Delta_{n}[y]=A e^{\alpha x}Δn[y]=Ait isαx, where A and α α alpha\alphaαconstants. Equation Δ 1 [ y ] = 1 Δ 1 [ y ] = 1 Delta_(1)[y]=1\Delta_{1}[y]=1Δ1[y]=1it is the differential equation of the chain line, and the equation Δ 2 [ y ] = 1 Δ 2 [ y ] = 1 Delta_(2)[y]=1\Delta_{2}[y]=1Δ2[y]=1was studied by G. Darbu [ 1 , 2 ] [ 1 , 2 ] [1,2][1,2][1,2].
Integral equations Δ n F [ y ] = A Δ n F [ y ] = A Delta_(n)^(F)[y]=A\Delta_{n}^{F}[y]=AΔnF[y]=Afor which Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0, given by formulas (10) and (17) and they are true in the entire interval ( − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞).
For these integrals, we have formulas (12) and (18), which are true in the interval ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞).
It is proved that any integral equation Δ q [ y ] = 0 Δ q [ y ] = 0 Delta_(q)[y]=0\Delta_{q}[y]=0Δq[y]=0with Δ q − 1 [ y ] ≠ 0 Δ q − 1 [ y ] ≠ 0 Delta_(q-1)[y]!=0\Delta_{q-1}[y] \neq 0Δq−1[y]≠0where q < n q < n q < nq<nq<n, is also integral Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0.
With the help of these results, all the integrals of the differential equation are obtained Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0.
As an application, the differential equation (2) is integrated, preliminarily proving the identity (30), which reduces the finding of the integrals of equation (2) to the integration of the equation Δ n + 1 [ y ] = 0 Δ n + 1 [ y ] = 0 Delta_(n+1)[y]=0\Delta_{n+1}[y]=0Δn+1[y]=0. The results are given by formulas (32) and (33), where Υ k Υ k Υ_(k)\Upsilon_{k}Ykand C k C k C_(k)C_{k}Ckgiven by formulas (35) and (36).
The integrals of Darbou's differential equation (4) are given by formulas (39'), (40'), (41').

The integration of a differential equation

(Résumé)
Dans ce mémoire on integré l'équation différenceielle(1), Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0and the differential equation (2) Δ n [ y ] = A ϵ α r Δ n [ y ] = A ϵ α r Delta_(n)[y]=Aepsilon^(alpha r)\Delta_{n}[y]=A \epsilon^{\alpha r}Δn[y]=AεαR, where A A AAAand α α alpha\alphaαsont des constantes L'équation Δ 1 [ y ] = 1 Δ 1 [ y ] = 1 Delta_(1)[y]=1\Delta_{1}[y]=1Δ1[y]=1est l'équation différenceielle de la chaînette et l'équation Δ 2 [ y ] = 1 Δ 2 [ y ] = 1 Delta_(2)[y]=1\Delta_{2}[y]=1Δ2[y]=1was studied by G. Darboux [ 1 , 2 ] [ 1 , 2 ] [1,2][1,2][1,2].
Les intégrales de l'équation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0for which Δ n − 1 [ y ] ≠ 0 Δ n − 1 [ y ] ≠ 0 Delta_(n-1)[y]!=0\Delta_{n-1}[y] \neq 0Δn−1[y]≠0sont données par les formulas (10) et (17) valid dans l'intervale ( − ∞ − ∞ -oo-\infty−∞, + ∞ ) + ∞ ) +oo)+\infty)+∞). Pour ces intégrales on a les formulas (12) et (18), valid dans 1'intervale ( − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞).
On demune aussi que toute intégrale de l'équation Δ q [ y ] = 0 Δ q [ y ] = 0 Delta_(q)[y]=0\Delta_{q}[y]=0Δq[y]=0, with Δ q − 1 [ y ] ≠ 0 Δ q − 1 [ y ] ≠ 0 Delta_(q-1)[y]!=0\Delta_{q-1}[y] \neq 0Δq−1[y]≠0, where q < n q < n q < nq<nq<n, est également intégrale de l'équation Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0.
A l'aide de ces résultats we obtain toutes les integrales de l'équation différenceielle Δ n [ y ] = 0 Δ n [ y ] = 0 Delta_(n)[y]=0\Delta_{n}[y]=0Δn[y]=0.
Comme application on integré l'équation différenceielle (2), en démontrant au préalé l'identité (30) qui amène la recherche des integrales de l'équation (2) à l'intégration de l'équation Δ n + 1 [ y ] = 0 Δ n + 1 [ y ] = 0 Delta_(n+1)[y]=0\Delta_{n+1}[y]=0Δn+1[y]=0. The results are given by formulas (32) and (33) where Υ k Υ k Υ_(k)\Upsilon_{k}Ykand C k C k C_(k)C_{k}Ckare given by formulas (35) and (36).
The integrals of the differential equation (4) of Darboux are given by the formulas ( 39 ′ ) , ( 40 ′ ) , ( 41 ′ ) 39 ′ , 40 ′ , 41 ′ (39^(')),(40^(')),(41^('))\left(39^{\prime}\right),\left(40^{\prime}\right),\left(41^{\prime}\right)(39′),(40′),(41′).
1957

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