The Integrating Factor

Let

$ P d x+Q d y=0 $

be a differential equation which is not exact. The theoretical method of integrating such an equation is to find a function $\mu(x, y)$ such that the expression

$ \mu(P d x+Q d y) $

is a total differential $d u$. When $\mu$ has been found the problem reduces to a mere quadrature.

The main question which arises is as to whether or not integrating factors exist. It will be proved that on the assumption that the equation itself has one and only one solution,¹ which depends upon one arbitrary constant, there exists an infinity of integrating factors.

Let the general solution be written in the form

$ \phi(x, y)=c, $

where $c$ is the arbitrary constant. Then, taking the differential,

$ \frac{\partial \phi}{\partial x} d x+\frac{\partial \phi}{\partial y} d y=0 $

or, as it may be written,

$ \phi_{x} d x+\phi_{y} d y=0. $

Since, therefore,

$ \phi(x, y)=c $

is the general solution of

$ P d x+Q d y=0, $

the relation

$ \frac{\phi_{x}}{P}=\frac{\phi_{y}}{Q} $

must hold identically, whence it follows that a function $\mu$ exists such that

$ \phi_{x}=\mu P, \quad \phi_{y}=\mu Q. $

Consequently

$ \mu(P d x+Q d y)=d \phi, $

that is to say an integrating factor $\mu$ exists.

Let $F(\phi)$ be any function of $\phi$, then the expression

$ \mu F(\phi)\{P d x+Q d y\}=F(\phi) d \phi $

is exact. If, therefore, $\mu$ is any integrating factor, giving rise to the solution $\phi=c$, then $\mu F(\phi)$ is an integrating factor. Since $F(\phi)$ is an arbitrary-function of $\phi$, there exists an infinity of integrating factors.

Since the equation

$ \mu(P d x+Q d y)=0 $

is exact, the integrating factor satisfies the relation

$ \frac{\partial(\mu P)}{\partial y}=\frac{\partial(\mu Q)}{\partial x} $

or

$ P \frac{\partial \mu}{\partial y}-Q \frac{\partial \mu}{\partial x}+\mu\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right)=0. $

Thus $\mu$ satisfies a partial differential equation of the first order. In general, therefore, the direct evaluation of $\mu$ depends upon an equation of a more advanced character than the ordinary linear equation under consideration. It is, however, to be noted that any particular solution, and not necessarily the general solution of the partial differential equation is sufficient to furnish an integrating factor. Moreover, in many particular cases, the partial differential equation has an obvious solution which gives the required integrating factor.

As an instance, suppose $\mu$ that is a function of $x$ alone, then

$\frac{1}{\mu}\frac{d\mu}{dx}=\frac{1}{Q}\left(\frac{\partial P}{\partial y}-\frac{\partial R}{\partial x}\right). $

It is therefore necessary that the right-hand member of this equation should be independent of $y$. When this is the case, $\mu$ is at once obtainable by a quadrature. Now suppose also that $Q$ is unity, then $P$ must be a linear function of $y$. The equation is therefore of the form

$ d y+(p y-q) d x=0, $

where $p$ and $q$ are functions of $x$ alone. The equation is therefore linear; the integrating factor, determined by the equation

$\frac{d \mu}{dx}=p\mu$

is

$ \mu=e^{\int p d x}. $

(cf. §2.1.3).

1. The Darboux Equation

A type of equation which was investigated by Darboux is the following:²

$ -L d y+M d x+N(x d y-y d x)=0. $

where $L, M, N$ are polynomials in $x$ and $y$ of maximum degree $m$.
It will be shown that when a certain number of particular solutions of the form

$f(x, y)=0,$

in which $f(x, y)$ is an irreducible polynomial, are known, the equation may be integrated.

Let the general solution be

$u(x, y)=\text{const.}$

then the given equation is equivalent to

$ \frac{\partial u}{\partial x} d x+\frac{\partial u}{\partial y} d y=0, $

and therefore

$ L \frac{\partial u}{\partial x}+M \frac{\partial u}{\partial y}-N\left(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\right)=0 $

Replace $x$ by $\frac{x}{z}, y$, by $\frac{y}{z}$, where $z$ is a third independent variable, then $u\left(\frac{x}{z}, \frac{y}{z}\right)$ is a homogeneous rational function of $x, y$, $z$ of degree zero, and by Euler's Theorem (§ 1.232)

$ x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}+z \frac{\partial u}{\partial z}=0. $

Moreover $u\left(\frac{x}{z}, \frac{y}{z}\right)$ satisfies the relation

$ A(u) \equiv L \frac{\partial u}{\partial x}+M \frac{\partial u}{\partial y}+N \frac{\partial u}{\partial z}=0, $

in which $L, M, N$ are homogeneous polynomials in $x, y, z$ of degree $m$.

The theory depends on the fact that if

$u(x, y)=\text{const.}$

is a solution of the given equation, $u\left(\frac{x}{z}, \frac{y}{z}\right)$ is homogeneous and of degree zero, and satisfies the relation $A(u)=0$. The converse is clearly also true.

Now let

$ f(x, y)=0 $

be any particular solution, where $f(x, y)$ is an irreducible polynomial of degree $h$, and let

$g(x, y, z)=z^{h}f\left(\frac{x}{z}, \frac{y}{z}\right).$

Then, since $g$ is homogeneous and of degree $h$,

$ x \frac{\partial g}{\partial x}+y \frac{\partial g}{\partial y}+z \frac{\partial g}{\partial z}=h g. $

Also

$ \begin{aligned} A(g) & \equiv L \frac{\partial g}{\partial x}+M \frac{\partial g}{\partial y}+N \frac{\partial g}{\partial z} \\ & =z^{h}\left(L \frac{\partial f}{\partial x}+M \frac{\partial f}{\partial y}+N \frac{\partial f}{\partial z}\right)+h z^{h-1} f N \\ & =h z^{-1} N g \end{aligned} $

since $f=0$ is a solution. This relation may be written in the form

$ A(g)=K g, $

since $A(g)$ is a polynomial of degree $m+h-1$ and $g$ is a polynomial of degree $h, K$ is a polynomial of degree $m-1$.

The operator $A$ has the property that if $F$ is any function of $u, v, w, \ldots$, where $u, v, w, \ldots$ are themselves functions of $x, y, z$,

$ A(F)=\frac{\partial F}{\partial u} A(u)+\frac{\partial F}{\partial v} A(v)+\frac{\partial F}{\partial w} A(w)+\cdots $

Let

$ f_{1}(x, y)=0, \quad f_{2}(x, y)=0, \ldots, \quad f_{p}(x, y)=0 $

be particular solutions of the given equation, where $f(x, y)$ is an irreducible polynomial of degree $h_{r}$. Let

$\begin{aligned} g_{r}(x, y, z)=z^{h_{r}} &f\left(\frac{x}{z}, \frac{y}{z}\right)\\ &(r=1,2, \ldots, p) \end{aligned}$

and consider the function

$ u(x, y, z)=\prod_{r=1}^{p}\left(g_{r}\right)^{a_{r}} $

where $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}$ are constants to be determined. Now

$ \begin{aligned} A(u) & =\sum_{r=1}^{p} \frac{\partial u}{\partial g_{r}} A\left(g_{r}\right) \\ & =\sum \alpha_{r} g_{1}^{\alpha_{1}} \ldots g_{r}^{\alpha_{r}-1} \ldots g_{p}^{\alpha_{p}} \ldots . K_{r} g_{r} \\ & =u \sum \alpha_{r} K_{r} \end{aligned} $

where $K_{r}$ is, for every value of $r$, a polynomial of degree $m-1$. Also $u(x, y, z)$ is a polynomial in $x, y, z$ of degree $h_{1} \alpha_{1}+h_{2} \alpha_{2}+\ldots+h_{p} \alpha_{p}$. If $u(x, y, z)$ is to furnish the required solution when $z=1$, it must be a polynomial in $x, y, z$ of degree zero, and must satisfy the relation $A(u)=0$, whence

$ \begin{array}{r} h_{1} a_{1}+h_{2} \alpha_{2}+\ldots+h_{p} \alpha_{p}=0 \\ K_{1} \alpha_{1}+K_{2} \alpha_{2}+\ldots+K_{p} \alpha_{p}=0 \end{array} $

Each polynomial $K_{r}$ contains at most $\frac{1}{2} m(m+1)$ terms, so that the last equation, being an identity in $x, y, z$, is equivalent to not more than $\frac{1}{2} m(m+1)_{\text {relations }}$ between the constants $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{p}$. There are, therefore, in all, at most

$ \frac{1}{2} m(m+1)+1 $

equations between the $p$ unknown constants $\alpha$. Suitable values can therefore be given to these constants if the number $p$ exceeds the number of equations, that is if

$ p \geqslant \frac{1}{2} m(m+1)+2. $

If, therefore, $\frac{1}{2} m(m+1)+2$ particular solutions are known, the general solution can be obtained without quadratures.

If $p=\frac{1}{2} m(m+1)+1$ and the discriminant of the equations is zero, the same result holds. Let $p=\frac{1}{2} m(m+1)+1$ and let the discriminant be not zero. In this case, let the constants be determined by the equations

$ \begin{aligned} h_{1} a_{1}+h_{2} a_{2}+\cdots+h_{p} a_{p} & =-m-2, \\ K_{1} a_{1}+K_{2} a_{2}+\cdots+K_{p} a_{p} & =-\frac{\partial L}{\partial x}-\frac{\partial M}{\partial y}-\frac{\partial N}{\partial z} . \end{aligned} $

There are now $\frac{1}{2} m(m+1)+1$ non-homogeneous equations which determine the constants $\alpha$. This determination of the constants gives rise to a function $u(x, y, z)$ such that

$ \begin{gathered} x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}+z \frac{\partial u}{\partial z}=-(m+2) u \\ A(u) \equiv L \frac{\partial u}{\partial x}+M \frac{\partial u}{\partial y}+N \frac{\partial u}{\partial z}=-\left(\frac{\partial L}{\partial x}+\frac{\partial M}{\partial y}+\frac{\partial N}{\partial z}\right) u . \end{gathered} $

Eliminate $\frac{\partial u}{\partial z}$ between these equations, then

$ \left\{L \frac{\partial u}{\partial x}+M \frac{\partial u}{\partial y}+\left(\frac{\partial L}{\partial x}+\frac{\partial M}{\partial y}+\frac{\partial N}{\partial z}\right) u\right\} z-\left\{x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}+(m+2) u\right\} N=0. $

But since $N$ is homogeneous and of degree $m$,

$ x \frac{\partial N}{\partial x}+y \frac{\partial N}{\partial y}+z \frac{\partial N}{\partial z}=m N, $

and therefore, eliminating $\frac{\partial N}{\partial z}$,

$ (L z-N x) \frac{\partial u}{\partial x}+(M z-N y) \frac{\partial u}{\partial y}+\left(z \frac{\partial L}{\partial x}+z \frac{\partial M}{\partial y}-x \frac{\partial N}{\partial x}-y \frac{\partial N}{\partial y}-2 N\right) u=0. $

Let $z=1$, then $u(x, y)$ satisfies the equation

$ (L-N x) \frac{\partial u}{\partial x}+(M-N y) \frac{\partial u}{\partial y}+\left\{\frac{\partial(L-N x)}{\partial x}+\frac{\partial(M-N y)}{\partial y}\right\} u=0. $

But this is precisely the condition that $u(x, y)$ should be an integrating factor for the equation

$ -L d y+M d x+N(x d y-y d x)=0. $

If, therefore, $\frac{1}{2} m(m+1)+1$ particular solutions are known, an integrating factor can be obtained.

To return to the Jacobi equation (§ $2 \cdot 14$ ),

$ \left(a_{1}+b_{1} x+c_{1} y\right)(x d y-y d x)-\left(a_{2}+b_{2} x+c_{2} y\right) d y+\left(a_{3}+b_{3} x+c_{3} y\right) d x=0. $

In this case $m=1$. The equation will have a solution of the linear form

$\alpha x+\beta y+\gamma=\text { const. }$$ A(f) \equiv\left(a_{2} z+b_{2} x+c_{2} y\right) \frac{\partial f}{\partial x}+\left(a_{3} z+b_{3} x+c_{3} y\right) \frac{\partial f}{\partial y}+\left(a_{1} z+b_{1} x+c_{1} y\right) \frac{\partial f}{\partial z}=\lambda f, $

where $\lambda$ is a constant and $f=\alpha x+\beta y+\gamma z$. This leads to three equations between $\alpha, \beta, \gamma$, $\lambda$, namely,

$ \gamma\left(a_{1}-\lambda\right)+\alpha a_{2}+\beta a_{3}=0, \gamma b_{1}+\alpha\left(b_{2}-\lambda\right)+\beta b_{3}=0, \gamma c_{1}+\alpha c_{2}+\beta\left(c_{3}-\lambda\right)=0, $

whence

$\left|\begin{array}{lll} a_1-\lambda, & a_2, & a_3 \\ b_1, & b_2-\lambda, & b_3 \\ c_1, & c_2, & c_3-\lambda \end{array}\right|=0 .$

It will be assumed that this equation has three distinct roots, $\lambda_{1}, \lambda_{2}, \lambda_{3}$, to which correspond three values of $f$, namely, $U, V, W$. Then

$U^{i} V^{j} W^{k}=\text{const.}$

will be the general solution, when $z$ is made equal to unity, if

$ \begin{array}{r} i+j+k=0 \\ \lambda_{1} i+\lambda_{2} j+\lambda_{3} k=0 . \end{array} $

It is sufficient to take $i=\lambda_{2}-\lambda_{3}, j=\lambda_{3}-\lambda_{1}, k=\lambda_{1}-\lambda_{2}$. The general solution is therefore

$U^{\lambda_{2}-\lambda_{3}} V^{\lambda_{3}-\lambda_{1}} W^{\lambda_{1}-\lambda_{2}}=\text{const.}$

Footnotes

1. This assumption will be justified in the following chapter. ↩

2. Bull. Sc. Math. (2), 2 (1878), p. 72. ↩