Exact Equations of the First Order and of the First Degree

An ordinary differential equation of the first order and of the first degree may be expressed in the form of a total differential equation,

$$Pdx+Qdy=0,$$

where $P$ and $Q$ are functions of $x$ and $y$ and do not involve $p$. If the differential $Pdx+Qdy$ is immediately, that is without multiplication by any factor, expressible in the form $du$, where $u$ is a function of $x$ and $y$, it is said to be exact.

If the equation

$$Pdx+Qdy=0$$

is exact and its primitive is¹

$$u=c,$$

the two expressions for $du$, namely,

$$Pdx+Qdy\text{and}\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$$

must be identical, that is,

$$P=\frac{\partial u}{\partial x},Q=\frac{\partial u}{\partial y}.$$

Then

\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}\tag{A}

provided that the equivalent expression ${\displaystyle \frac{{\partial }^{2}u}{\partial x\partial y}}$ is continuous. The condition of integrability (A) is therefore necessary. It remains to show that the condition is sufficient, that is to say, if it is satisfied the equation is exact and its primitive can be found by a quadrature.

Let $u(x,y)$ be defined by

$$u={\int }_{x_0}^{x}P(x,y)dx+\varphi (y),$$

where $x_0$ is an arbitrary constant, and $\varphi (y)$ is a function of $y$ alone which, for the moment, is also arbitrary. Then $u=c$ will be a primitive of

$$Pdx+Qdy=0$$

if

$$\frac{\partial u}{\partial x}=P,\frac{\partial u}{\partial y}=Q.$$

The first condition is satisfied; the second determines $\varphi (y)$ thus:

\begin{aligned} Q(x, y) & =\frac{\partial u}{\partial y} \\ & =\int_{x_{0}}^{x} \frac{\partial P}{\partial y} d x+\phi^{\prime}(y) \\ & =\int_{x_{0}}^{x} \frac{\partial Q} {\partial x} dx+\phi^{\prime}(y) \\ & =Q(x, y)-Q\left(x_{0}, y\right)+\phi^{\prime}(y) \end{aligned}

and therefore

$$\varphi (y)={\int }_{y_0}^{y}Q(x_0,y)dy,$$

where $y_0$ is arbitrary.

The condition is therefore sufficient, for the equation is exact and has the primitive

$${\int }_{x_0}^{x}P(x,y)dx+{\int }_{y_0}^{y}Q(x_0,y)dy=c.$$

The constants $x_0$ and $y_0$ may be chosen as is convenient, there are not, in all, three arbitrary constants but only one, for a change in $x_0$ or in $y_0$ is equivalent to adding a constant to the left-hand member of the primitive. This is obvious as far as $y_0$ is concerned, and as regards $x_0$, it is a consequence of the condition of integrability.

1. Separation of Variables

A particular instance of an exact equation occurs when $P$ is a function of $x$ alone and $Q$ a function of $y$ alone. In this case $X$ may be written for $P$ and $Y$ for $Q$. The equation

$$Xdx+Ydy$$

is then said to have separated variables. Its primitive is

$$\int Xdx+\int Ydy=c.$$

When the equation is such that $P$ can be factorised into a function $X$ of $x$ alone and $Y_1$ a function of $y$ alone, and $Q$ can similarly be factorised into $X_1$ and $Y$, the variables are said to be separable, for the equation

X Y_{1}\, d x+X_1 Y\, dy=0\tag{I}

may be written in the separated form

\frac{X}{X_1}dx+\frac{Y}{Y_1}dy=0.\tag{II}

It must be noticed, however, that a number of solutions are lost in the division of the equation by $X_1{Y}_1$. If, for example, $x=a$ is a root of the equation $X_1=0$, it would furnish a solution of the equation (I) but not necessarily of the equation (II).

2. Homogeneous Equations

If $P$ and $Q$ are homogeneous functions of $x$ and $y$ of the same degree $n$, the equation is reducible by the substitution² $y=vx$ to one whose variables are separable. For

$$P(x,y)=x^{n}P(1,v),P(x,y)=x^{n}Q(1,v),$$

and therefore

$$P(x,y)dx+Q(x,y)dy=0$$

becomes

$$\{P(1,v)+vQ(1,v)\}dx+xQ(1,v)dv=0$$

or

$$\frac{dv}{\varphi (v)}+\frac{dx}{x}=0,$$

where

$$\varphi (v)=v+\frac{P(1,v)}{Q(1,v)}.$$

The solution is

$$\int \frac{dv}{\varphi (v)}=\log \frac{c}{x}.$$

When the equation

$$Pdx+Qdy=0$$

is both homogeneous and exact, it is immediately integrable without the introduction of a quadrature, provided that its degree of homogeneity $n$ is not -1. Its primitive is, in fact,

$$Px+Qy=c.$$

For let $u=Px+Qy$, then

\begin{aligned} \frac{\partial u}{\partial x} & =P+x \frac{\partial P}{\partial x}+y \frac{\partial Q}{\partial x} \\ & =P+x \frac{\partial P}{\partial x}+y \frac{\partial P}{\partial y}=(n+1) P, \end{aligned}

by Euler's theorem (1.2.3.2), and similarly

$$\frac{\partial u}{\partial y}=(n+1)Q.$$

Consequently

\begin{aligned} d u & =\frac{\partial u}{\partial x} d x+\frac{\partial u}{\partial y} d y \\ & =(n+1)(P d x+Q d y) \end{aligned}

and therefore

$$Pdx+Qdy=\frac{d(Px+Qy)}{n+1}.$$

Hence if $n\ne 1$, the primitive is

$$Px+Qy=c.$$

When $n=-1$ the integration in general involves a quadrature. It is a noteworthy fact that the homogeneous equation

$$\frac{Pdx+Qdy}{Px+Qy}=0$$

is exact, for the condition of integrability, namely

$$\frac{\partial }{\partial y}\left(\frac{P}{Px+Qy}\right)=\frac{\partial }{\partial x}\left(\frac{Q}{Px+Qy}\right),$$

reduces to

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

which is true, by Euler's theorem, since $P$ and $Q$ are homogeneous and of the same degree. Thus any homogeneous equation may be made exact by introducing the integrating factor $1/(Px+Qy)$. The degree of homogeneity of this exact equation is, however, -1, so that the integration of a homogeneous equation in general involves a quadrature.

An equation of the type

$$\frac{dy}{dx}=F\left(\frac{Ax+By+C}{ax+by+c}\right),$$

in which $A,B,C,a,b,c$ are constants such that $Ab-aB\ne 0$, may be brought into the homogeneous form by a linear transformation of the variables, for let

$$x=h+\xi ,y=k+\eta ,$$

where $\xi ,\eta$ are new variables and $h,k$ are constants such that

\begin{aligned} A h+B k+C & =0, \\ a h+b k+c & =0 . \end{aligned}

The equation becomes

$$\frac{d\eta }{d\xi }=F\left(\frac{A\xi +B\eta }{a\xi +b\eta }\right),$$

so that $F$ is a homogeneous function of $\xi ,\eta$ of degree zero. The constants $h,k$ are determinate since $Ab-aB\ne 0$.

When $Ab-aB=0$, let $\eta$ be a new dependent variable defined by

$$\eta =x+By/A=x+by/a,$$

then

$$\frac{d\eta }{dx}=1+\frac{b}{a}F\left(\frac{A\eta +C}{a\eta +c}\right).$$

The variables are now separable.

3. Linear Equations of the First Order

The most general linear equation of the first order is of the type

$$\frac{dy}{dx}+\varphi y=\psi ,$$

where $\varphi$ and $\psi$ are functions of $x$ alone. Consider first of all the homogeneous linear equation³

$$\frac{dy}{dx}+\varphi y=0.$$

Its variables are separable, thus:

$$\frac{dy}{y}+\varphi dx=0,$$

and the solution is

$$y=c{e}^{-\int \varphi dx},$$

where $c$ is a constant.
Now substitute in the non-homogeneous equation, the expression

$$y=v{e}^{-\int \varphi dx},$$

in which $v$, a function of $x$, has replaced the constant $c$. The equation becomes

$$\frac{dv}{dx}{e}^{-\int \varphi dx}=\psi ,$$

whence

$$v=C+\int \psi e^{\int \varphi dx}dx.$$

The solution of the general linear equation is therefore

$$y=C{e}^{-\int \varphi dx}+e^{-\int \varphi dx}\int \psi e^{\int \varphi dx}dx,$$

and involves two quadratures.

The method here adopted of finding the solution of an equation by regarding the parameter, or constant of integration $c$ of the solution of a simpler equation, as variable, and so determining it that the more general equation is satisfied, is a particular case of what is known as the method of variation of parameters.⁴

It is to be noted that the general solution of the linear equation is linearly dependent upon the constant of integration $C$. Conversely the differential equation obtained by eliminating $C$ between any equation

$$y=Cf(x)+g(x),$$

and the derived equation

y^{\prime}=C f(x)+g^{\prime}(x),

is linear.

If any particular solution of the linear equation is known, the general solution may be obtained by one quadrature. For let $y_1$ be a solution, then the relation

$$\frac{d{y}_1}{dx}+\varphi y_1=\psi$$

is satisfied identically. By means of this relation, $\psi$ can be eliminated from the given equation, which becomes

$$\frac{d}{dx}(y-y_1)+\varphi (y-y_1)=0.$$

The equation is now homogeneous in $y-y_1$, and has the solution

$$y-y_1=C{e}^{-\int \varphi dx},$$

where $C$ is the constant of integration.

If two distinct particular solutions are known, the general solution may be expressed directly in terms of them. For it is known that the general solution has the form

$$y=Cf(x)+g(x),$$

and any two particular solutions $y_1$ and $y_2$ are obtained by assigning definite values $C_1$ and $C_2$ to the arbitrary constant $C$, thus

$$y_1=C_{1}f(x)+g(x),$$$$y_2=C_{2}f(x)+g(x),$$

and therefore

$$\frac{y-y_1}{y_2-y_1}=\frac{C-C_1}{C_2-C_1}.$$

4. The Equations of Bernoulli and Jacobi

The equation

$$\frac{dy}{dx}+\varphi y=\psi y^n,$$

in which $\varphi$ and $\psi$ are functions of $x$ alone, is known as the Bernoulli equation.⁵ It may be brought into the linear form by a change of dependent variable. Let

$$z=y^{1-n},$$

then

$$\frac{dz}{dx}=(1-n)y^{-n}\frac{dy}{dx},$$

and thus if the given equation is written in the form

$$y^{-n}\frac{dy}{dx}+\varphi y^{1-n}=\psi ,$$

it becomes

$$\frac{dz}{dx}+(1-n)\varphi z=(1-n)\psi ,$$

and is linear in $z$.

The Jacobi equation,⁶

$$(a_1+b_{1}x+c_{1}y)(xdy-ydx)-(a_2+b_{2}x+c_{2}y)dy+(a_3+b_{3}x+c_{3}y)dx=0,$$

in which the coefficients $a,b,c$ are constants, is closely connected with the Bernoulli equation. Make the substitution

$$x=X+\alpha ,y=Y+\beta ,$$

where $\alpha ,\beta$ are constants to be determined so as to make the coefficients of $XdY-YdX$, $dY$ and $dX$ separately homogeneous in $X$ and $Y$. When this substitution is made, the equation is so arranged that the coefficient of $XdY-YdX$ is homogeneous and of the first degree, thus

$$\begin{array}{c}(b_{1}X+c_{1}Y)(XdY-YdX)\\ -\{A_2+b_{2}X+c_{2}Y-\alpha (A_1+b_{1}X+c_{1}Y)-A_{1}X\}dY\\ +\{A_3+b_{3}X+c_{3}Y-\beta (A_1+b_{1}X+c_{1}Y)-A_{1}Y\}dX=0\end{array}$$

where

$$A_r=a_r+b_r\alpha +c_r\beta (r=1,2,3).$$

The coefficients of $dY$ and $dX$ also become homogeneous if $\alpha$ and $\beta$ are so chosen that $A_2-\alpha A_1=0,A_3-\beta A_1=0$, or, more symmetrically, if

$$A_1=\lambda ,A_2=\alpha \lambda ,A_3=\beta \lambda ,$$

that is if

a_{1}-\lambda+b_{1} \alpha+c_{1} \beta=a_{2}+\left(b_{2}-\lambda\right) a+c_{2} \beta=a_{3}+b_{3} \alpha+\left(c_{3}-\lambda\right) \beta=0 . \tag{$\Lambda$}

Thus $\lambda$ is determined by the cubic equation

and when $\lambda$ is so determined, $\alpha$ and $\beta$ are then the solutions of any two of the consistent equations ($\mathrm{\Lambda }$).

The equation may now be written in the form

$$XdY-YdX-\mathrm{\Phi }\left(\frac{Y}{X}\right)dY+\mathrm{\Psi }\left(\frac{Y}{X}\right)dX=0.$$

The substitution $Y=Xu$ brings it into the form of a Bernoulli equation,

$$\frac{dX}{du}+U_{1}X+U_2{X}^2=0,$$

where $U_1$ and $U_2$ are functions of $u$ alone.

It will be shown in a later section (§ 2.2.1) that if the three roots of the equation in $\lambda$ are ${\lambda }_1,{\lambda }_2,{\lambda }_3$ and are distinct,⁷ the general solution of the Jacobi equation is

$$U^{{\lambda }_2-{\lambda }_3}{V}^{{\lambda }_3-{\lambda }_1}{W}^{{\lambda }_1-{\lambda }_2}=\text{const.}$$

where $U,V,W$ are linear expressions in $x$ and $y$.

5. The Riccati Equation

The equation

$$\frac{dy}{dx}+\psi y^2+\varphi y+\chi =0,$$

in which $\psi ,\varphi$ and $\chi$ are functions of $x$, is known as the generalised Riccati equation.⁸ It is distinguished from the previous equations of this chapter in that it is not, in general, integrable by quadratures. It therefore defines a family of transcendental functions which are essentially distinct from the elementary transcendents.⁹

When any particular solution $y=y_1$ is known, the general solution may be obtained by means of two successive quadratures. Let

$$y=y_1+z,$$

then the equation becomes

$$\frac{d{y}_1}{dx}+\frac{dz}{dx}+\psi (y_1^2+2y_{1}z+z^2)+\varphi (y_1+z)+\chi =0,$$

and since $y=y_1$ is a solution, it reduces to

$$\frac{dz}{dx}+(2y_1\psi +\varphi )z+\psi z^2=0.$$

This is a case of the Bernoulli equation; it is reduced to the linear form by the substitution

$$z=1/u,$$

from which the theorem stated follows immediately.
Let $y_1,y_2,y_3$ be three distinct particular solutions of the Riccati equation and $y$ its general solution. Then

$$u=\frac{1}{y-y_1},u_1=\frac{1}{y_2-y_1},u_2=\frac{1}{y_3-y_1}$$

satisfy one and the same linear equation, and consequently

$$\frac{u-u_1}{u_2-u_1}=C$$

where $C$ is a constant. When $u,u_1$ and $u_2$ are replaced by their expressions in terms of $y$, $y_1$ and $y_2$ this relation may be written

$$\frac{y-y_2}{y-y_1}=C\frac{y_3-y_2}{y_3-y_1}.$$

This formula shows that the general solution of the Riccati equation is expressible rationally in terms of any three distinct particular solutions, and also that the anharmonic ratio of any four solutions is constant. It also shows that the general solution is a rational function of the constant of integration. Conversely any function of the type

$$y=\frac{C{f}_1+f_2}{C{f}_3+f_4},$$

where $f_1,f_2,f_3,f_4$ are given functions of $x$ and $C$ an arbitrary constant, satisfies a Riccati equation, as may easily be proved by eliminating $C$ between the expressions for $y$ and the derived expression for y^{\prime}.

When $\psi$ is identically zero, the Riccati equation reduces to the linear equation; when $\psi$ is not zero, the equation may be transformed into a linear equation of the second order. Let $v$ be a new dependent variable defined by

$$y=v/\psi ,$$

then the equation becomes

$$\frac{dv}{dx}+v^2+Pv+Q=0$$

where

P=\phi-\frac{\psi^{\prime}}{\psi}, \quad Q=\psi \chi .

The substitution

v=u^{\prime} / u

now brings the equation into the proposed form, namely,

$$\frac{d^{2}u}{d{x}^2}+P\frac{du}{dx}+Qu=0.$$

In particular, the original equation of Riccati, namely,

\begin{aligned} & d y \\ & d x\end{aligned}+a y^{2}=b x^{m},

where $a$ and $b$ are constants, becomes¹⁰

$$\frac{d^{2}u}{d{x}^2}-ab{x}^{m}u=0.$$

6. The Euler Equation

An important type of equation with separated variables is the following: ¹¹

$$\frac{dx}{X^{\frac{1}{2}}}+\frac{dy}{Y^{\frac{1}{2}}}=0$$

in which

\begin{aligned} X & =a_{0} x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}, \\ Y & =a_{0} y^{4}+a_{1} y^{3}+a_{2} y^{2}+a_{3} y+a_{4} . \end{aligned}

Consider first of all the particular equation

$$\frac{dx}{\sqrt{1-x^2}}+\frac{dy}{\sqrt{1-y^2}}=0,$$

one solution is¹²

$$\arcsin x+\arcsin y=c,$$

but the equation has also the solution

$$x\sqrt{1-y^2}+y\sqrt{1-x^2}=C.$$

Since, as will be proved in Chapter III., the differential equation has but one distinct solution, the two solutions must be related to one another in a definite way. This relation is expressed by the equation

$$C=f(c)$$

Now let

$$x=\sin u,y=\sin v,$$

then

\begin{aligned} u+v&=c\\ \sin u \cos v+\sin v \cos u&=f(c)\\ &=f(u+v) \end{aligned}

Let $v=0$, then

$$\sin u=f(u)$$

and therefore

$$\sin u\cos v+\sin v\cos u=\sin (u+v).$$

Thus the addition formula for the sine-function is established.

In the same way, the differential equation

$$\frac{dx}{(1-x^2{)}^{\frac{1}{2}}(1-k^2{x}^2{)}^{\frac{1}{2}}}+\frac{dy}{(1-y^2{)}^{\frac{1}{2}}(1-k^2{y}^2{)}^{\frac{1}{2}}}=0$$

has the solution

$$\arg \mathrm{sn}x+\arg \mathrm{sn}y=c,$$

where $\arg \mathrm{sn}x$ is the inverse Jacobian elliptic function¹³ defined by

$$\arg \mathrm{sn}x={\int }_0^x\frac{dt}{{(1-t^2)}^{\frac{1}{2}}{(1-k^2{t}^2)}^{\frac{1}{2}}}.$$

Let

$$x=\sin u,y=\sin v,$$

then

$$u+v=c.$$

A second and equivalent solution may be found as follows. By definition

$$\frac{dx}{du}={(1-x^2)}^{\frac{1}{2}}{(1-k^2{x}^2)}^{\frac{1}{2}},$$

and therefore

$$\frac{d^{2}x}{d{u}^2}=-(1+k^2)x+2k^2{x}^3.$$

Similarly

$$\frac{dy}{du}=-\frac{dy}{dv}=-{(1-y^2)}^{\frac{1}{2}}{(1-k^2{y}^2)}^{\frac{1}{2}},$$$$\frac{d^{2}y}{d{u}^2}=\frac{d^{2}y}{d{v}^2}=-(1+k^2)y+2k^2{y}^3,$$

from which it follows that

\begin{aligned} x \frac{d^{2} y}{d u^{2}}-y \frac{d^{2} x}{d u^{2}} & =2 k^{2} x y\left(y^{2}-x^{2}\right), \\ x^{2}\left(\frac{d y}{d u}\right)^{2}-y^{2}\left(\frac{d x}{d u}\right)^{2} & =\left(x^{2}-y^{2}\right)\left(1-k^{2} x^{2} y^{2}\right) . \end{aligned}

Hence

$$\frac{x{\displaystyle \frac{d^{2}y}{d{u}^2}}-y{\displaystyle \frac{d^{2}x}{d{u}^2}}}{x{\displaystyle \frac{dy}{du}}-y{\displaystyle \frac{dx}{du}}}=-(x\frac{dy}{du}+y\frac{dx}{du})\frac{2k^{2}xy}{1-k^2{x}^2{y}^2}.$$

This equation is immediately integrable; the solution is

$$\log (x\frac{dy}{du}-y\frac{dx}{du})=\text{ const. }+\log (1-k^2{x}^2{y}^2)$$

or

$$x\frac{dy}{dv}+y\frac{dx}{du}=C(1-k^2{x}^2{y}^2),$$

that is

\operatorname{sn} u \operatorname{sn}^{\prime} v+\operatorname{sn} v \operatorname{sn}^{\prime} u=f(c)\left(1-k^{2} \operatorname{sn}^{2} u \operatorname{sn}^{2} v\right).

By putting $v=0$ it is found that $f(u)=\mathrm{sn}u$, and therefore

\operatorname{sn}(u+v)=\frac{\operatorname{sn} u \operatorname{sn}^{\prime} v+\operatorname{sn} v \operatorname{sn}^{\prime} u}{1-k^{2} \operatorname{sn}^{2} u \operatorname{sn}^{2} v}.

This is the addition formula for the Jacobian elliptic function $\mathrm{sn}u$.

The same process of integration may be applied to the general Euler equation.¹⁴ In particular it may be noted that when $a_0=0$ a linear transformation brings the equation into the form

$$\frac{dx}{\sqrt{4x^3-g_{2}x-g_3}}+\frac{dy}{\sqrt{4y^3-g_{2}y-g_3}}=0.$$

If $\mathrm{\wp }(z)$ is the Weierstrassian elliptic function defined by

$$z={\int }_{\mathrm{\wp }(z)}^{\mathrm{\infty }}{(4t^3-g_{2}t-g_3)}^{-t}dt,$$

and $x=\mathrm{\wp }(u),y=\mathrm{\wp }(v)$, the general solution of the equation is $u+v=c$.

An equivalent general solution is

$${\{{(4x^3-g_{2}x-g_3)}^{\frac{1}{2}}-{(4y^3-g_{2}y-g_3)}^{\frac{1}{2}}\}}^2=(x-y{)}^2(C+4x+4y)$$

It may thus be shown that the addition-formula for the $\mathrm{\wp }$-function is

\wp(u+v)=-\wp(u)-\wp(v)+\frac{1}{4}\left\{\frac{\wp^{\prime}(u)-\wp^{\prime}(v)}{\wp(u)-\wp(v)}\right\}^{2}.

Footnotes

1. Throughout this Chapter the letter $c$ or $C$ generally denotes a constant of integration. Any other use of these letters will be evident from the context. ↩

2. This device was first used by Leibniz in 1691. ↩

3. The term homogeneous is applied to a linear equation when it contains no term independent of $y$ and the derivatives of $y$. This usage of the term is to be distinguished from that of the preceding section in which an equation (in general non-linear) was said to be homogeneous when $P$ and $Q$ were homogeneous functions of $x$ and $y$ of the same degree. There should be no confusion between the two usages of the term. ↩

4. Vide § 5.2.3. The application of the method to the linear equation of the first order is due to John Bernoulli, Acta Erud., 1697, p. 113, but the solution by quadratures was known to Leibniz several years earlier. ↩

5. James Bernoulli, Acta Erud. 1695, p. 553 [Opera 1, p. 663]. The method of solution was discovered by Leibniz, Acta Erud. 1696, p. 145 [Math. Werke 5, p. 329]. ↩

6. J. für Math. 24 (1842), p. 1 [Ges. Werke, 4, p. 256]. See also the Darboux equation, § 2.2.1, infra. ↩

7. The case in which they are not distinct is discussed by Serret, Cale. Diff. et Int. 2, p. 431. ↩

8. Riccati, Acta Erud. Suppl., VIII. (1724), p. 73, investigated the equation y^{\prime}+a y^{2}=b x^{m}, with which his name is usually associated. The generalised equation was studied by d'Alembert, vide infra, § 12.5.1. ↩

9. The elementary transcendents are functions which can be derived from algebraic functions by integration, and the inverses of such functions. Thus the logarithmic function is defined as ${\int }_1^x{x}^{-1}dx$; ; its inverse is the exponential function. From the exponential function the trigonometrical and the hyperbolic functions are derived by rational processes, and such functions as the error-function by integration. ↩

10. The solution of this equation may be expressed in terms of Bessel functions (§7.3.1). ↩

11. Euler, Inst. Calc. Int., 1, Chaps. V., VI. ↩

12. The function $\arcsin x$ is defined as. ${\int }_0^x{(1-t^2)}^{-\frac{1}{2}}dt;\sin x$ is defined as the inverse of are $\sin x$, so that $\sin 0=0$; and $\cos x$ is defined as ${(1-{\sin }^{2}x)}^{\frac{2}{2}}$ with the condition that $\cos 0=1$. No further properties of the trigonometrical functions are assumed. ↩

13. Whittaker and Watson, Modern Analysis, Chap. XXII. ↩

14. Cayley, Elliptic Functions, Chap. XIV. ↩