Equations of Higher Order than the First

The simplest of all differential equations of general order $n$ is the following:

$$\frac{d^{n}y}{d{x}^n}=f(x).$$

Its integration is simply the process of $n$-ple integration and may be carried out in successive stages as follows. Let $x_0$ be a constant, chosen at random, then

$$\frac{d^{n-1}y}{d{x}^{n-1}}={\int }_{x_0}^{x}f(x)dx+C_0,$$$$\frac{d^{n-2}y}{d{x}^{n-2}}={\int }_{x_0}^{x}dx{\int }_{x_0}^{x}f(x)dx+C_0(x-x_0)+C_1,$$$$⋮$$$$y={\int }_{x_0}^{x}dx{\int }_{x_0}^{x}dx\dots {\int }_{x_0}^{x}f(x)dx+C_0\frac{(x-x_0{)}^{n-1}}{(n-1)!}+\cdots +C_{n-1},$$

where $C_0,C_1,\dots ,C_{n-1}$ are arbitrary constants.

The multiple integral may, however, be replaced by a single integral. Let

$$Y=\frac{1}{(n-1)!}{\int }_{x_0}^x(x-t{)}^{n-1}f(t)dt.$$

then

$$\frac{dY}{dx}=\frac{1}{(n-2)!}{\int }_{x_0}^x(x-t{)}^{n-2}f(t)dt.$$

...

$$\frac{d^{n-1}Y}{d{x}^{n-1}}={\int }_{x_0}^{x}f(t)dt,$$

whence, finally,

$$\frac{d^{n}Y}{d{x}^n}=f(x).$$

$Y$ is therefore a solution of the equation which, together with its first $(n-1)$ derivatives vanishes when $x=x_0$. It is therefore identified with the multiple integral. The general solution of the equation is therefore

$$y=\frac{1}{(n-1)!}{\int }_{x_0}^x(x-t{)}^{n-1}f(t)dt+C_0\frac{(x-x_0{)}^{n-1}}{(n-1)!}+\cdots +C_{n-1}.$$

Apart from this simple case, and the case of linear equations with constant coefficients, which will be dealt with in Chapter VI, there are but few equations of order higher than the first which yield to an elementary treatment. In a number of very special cases, however, the order of an equation can be lowered by means of a suitable transformation of the variables, combined with one or more quadratures. The main cases of this kind which can arise will be dealt with in the three following sections.

2.6.1 Equations which do not explicitly involve the Dependent Variable

Consider the equation

$$F(x,\frac{d^{k}y}{d{x}^k},\frac{d^{k+1}y}{d{x}^{k+1}},\dots ,\frac{d^{n}y}{d{x}^n})=0,$$

in which $y$ and its first $k-1$ derivatives do not appear. The transformation

$$v=\frac{d^{k}y}{d{x}^k}$$

reduces the equation to an equation in $v$ of order $n-k$. If this equation can be integrated and its solution is $v=v(x)$, it only remains to integrate the equation

$$\frac{d^{k}y}{d{x}^k}=v(x),$$

which is of the type dealt with in the preceding section.

More generally, however, the reduced equation has a solution of the form

$$\varphi (x,v)=0,$$

which is not readily soluble for $v$. For the method to be practicable it is necessary to express $x$ and $v$ in terms of a parameter $t$, thus

$$y^{(k)}=v(t),x=x(t).$$

Then

dy^{(k-1)} = v(t)\,dx = v(t)x'(t)\,dt,

which, on integration, gives $y^{(k-1)}$. The process is repeated, $k$ times in all, until the explicit solution is reached.

An important particular case is that of equations of the form

$$\frac{d^{n}y}{d{x}^n}=f\left(\frac{d^{n-1}y}{d{x}^{n-1}}\right);$$

such equations are integrable by quadratures.

2.6.2 Equations which do not explicitly involve the Independent Variable

When an equation has the form

$$F(y,\frac{dy}{dx},\frac{d^{2}y}{d{x}^2},\dots ,\frac{d^{n}y}{d{x}^n})=0,$$

its order may be reduced to $n-1$ by a change of variables. Let $y$ be taken as a new independent variable, and $p$ as the dependent variable. The formulæ by means of which this transformation is effected are

$$\frac{dy}{dx}=p,\frac{d^{2}y}{d{x}^2}=p\frac{dp}{dy},\frac{d^{3}y}{d{x}^3}=p\frac{d}{dy}\left(p\frac{dp}{dy}\right),\dots$$

The given equation is thus reduced to one of the form

$$\mathrm{\Phi }(y,p,\frac{dp}{dy},\dots ,\frac{d^{n-1}p}{d{y}^{n-1}})=0.$$

Let it be supposed that this equation can be integrated, and that its solution is expressible in the parametric form

$$y=f(t),p=g(t),$$

where $f$ and $g$ are functions of the auxiliary variable $t$, and depend also on $n-1$ constants of integration. Then $x$ is obtained, in terms of $t$, by a quadrature, thus:

x = \int \frac{dy}{p} = \int \frac{f'(t)dt}{g(t)}.

In particular, an equation of the second order, which does not explicitly involve $x$, namely

$$F(y,\frac{dy}{dx},\frac{d^{2}y}{d{x}^2})=0,$$

is transformed into the equation

$$F(y,p,p\frac{dp}{dy})=0,$$

which is of the first order.

An equation of the form

$$\frac{d^{n}y}{d{x}^n}=f\left(\frac{d^{n-2}y}{d{x}^{n-2}}\right)$$

is reduced, by the substitution

$$\frac{d^{n-2}y}{d{x}^{n-2}}=v,$$

to

$$\frac{d^{2}v}{d{x}^2}=f(v).$$

If $\frac{dv}{dx}=p$, this last equation becomes

$$p\frac{dp}{dv}=f(v).$$

whence

$$p^2=c+\int f(v)dv,$$

and therefore

$$x=\int \{c+\int f(v)dv{\}}^{-\frac{1}{2}}dv.$$

In order that $y$ may be obtained, $v$ must be expressed in terms of $x$; the solution is then completed by $n-2$ quadratures.

2.6.3 Equations exhibiting a Homogeneity of Form

Two classes of equations will be discussed, the first class being that of equations which are homogeneous in y, y', y'', \dots, y^{(n)}, and which may also involve $x$ explicitly. An equation of this class may, if $m$ is the degree of homogeneity, be written

y^m F\left(x, \frac{y'}{y}, \frac{y''}{y}, \dots, \frac{y^{(m)}}{y}\right) = 0.

Let $u$ be a new dependent variable, defined by the relation

$$y=e^{\int udx},$$

then

y' = ue^{\int u\,dx}, \quad y'' = (u'+u^2)e^{\int u\,dx}, \dots,

and in general

$$y^{(n)}=U_n{e}^{\int udx},$$

where $U_n$ is a polynomial in u, u', \dots, u^{(n-1)}. The change of dependent variable from $y$ to $u$ therefore reduces the order of the equation from $n$ to $n-1$.

The second class includes those equations which are homogeneous in y, xy', x^2y'', \dots, x^ny^{(n)} and do not otherwise involve $x$. Let

F(y, xy', x^2y'', \dots, x^ny^{(n)}) = 0

be the typical equation. Change the independent variable by the substitution

$$x=e^t,$$

then

$$\frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt},x^2\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt},\dots ,$$

and, in general,

$$x^r\frac{d^{r}y}{d{x}^r}=\frac{d}{dt}(\frac{d}{dt}-1)\dots (\frac{d}{dt}-r+1)y.$$

Thus the transformed equation is of the form

$$\mathrm{\Phi }(y,\frac{dy}{dt},\frac{d^{2}y}{d{t}^2},\dots ,\frac{d^{n}y}{d{t}^n})=0$$

and does not explicitly involve $t$. It thus comes under the heading of § 2.6.2.

An equation which comes under the last class, but which can be integrated by a simpler method is the following:¹

F(y'', y'-xy'', y-xy'+\frac{1}{2}x^2y'') = 0.

The derived equation is simply

y'''(F_1-xF_2+\frac{1}{2}x^2F_3)=0,

where $F_1,F_2,F_3$ are the partial derivatives of $F$ with respect to its first, second, and third arguments respectively. It is satisfied by y'''=0, or

$$y=A+Bx+C{x}^2,$$

where $A,B,C$ are arbitrary constants. This will be the general solution of the original equation provided that

$$F(C,B,A)=0.$$

Footnotes

1. Dixon, Phil. Trans. R. S. (A) 186 (1894), p. 563. The generalisation to any order is obvious. See also Raffy, Bull. Soc. Math. France, 25 (1897), p. 71. ↩