First-Order ODEs of Higher Degree | Ince's Ordinary Differential Equations

An equation of the first order and of degree $m$ may be written

F\left(x, y, \frac{dy}{dx}\right) \equiv \left(\frac{dy}{dx}\right)^m + P_1\left(\frac{dy}{dx}\right)^{m-1} + \dots + P_{m-1}\frac{dy}{dx} + P_m = 0, \tag{A}

where $P_{1}, \dots, P_{m}$ are functions of $x$ and $y$ . Theoretically, the equation may be brought into the factorised form,

(\frac{d y}{d x} - p_{1}) (\frac{d y}{d x} - p_{2}) \dots (\frac{d y}{d x} - p_{m}) = 0,

where $p_{1}, p_{2}, \dots, p_{m}$ are functions of $x$ and $y$ .

Let

ϕ_{r} (x, y, c_{r}) = 0

be the general solution of the equation

\frac{d y}{d x} - p_{r} = 0;

it will also be a solution of the given equation. Conversely if

Φ (x, y, C) = 0

is a solution of the given equation, it must satisfy one or other of the equations

\frac{d y}{d x} - p_{r} = 0 (r = 1, 2, \dots, m) .

It follows that every solution of (A) will be included in the solution

ϕ_{1} (x, y, c) ϕ_{2} (x, y, c) \dots ϕ_{m} (x, y, c) = 0,

which is therefore the general solution. The one arbitrary constant $c$ is sufficient for complete generality, for a particular solution is obtained explicitly by solving one or other of the equations

ϕ_{r} (x, y, c) = 0,

in which $c$ has any numerical value.

Example.

{(\frac{d y}{d x})}^{2} - (1 + y^{2}) = 0.

In the factorised form the equation is

{\frac{d y}{d x} - \sqrt{1 + y^{2}}} {\frac{d y}{d x} + \sqrt{1 + y^{2}}} = 0,

the two factors give rise to solutions

y = \sinh (c \pm x)

respectively, where $c$ is a constant. The general solution therefore is

\begin{align*} y^2 &= \frac{1}{4}(e^{c+x}-e^{-c-x})(e^{c-x}-e^{-c+x}) \\ &= \frac{1}{4}(e^{2c}+e^{-2c}-e^{2x}-e^{-2x}) \\ &= \frac{1}{2}(C-\cosh 2x), \end{align*}

where $C = \cosh 2 c$ .

1. Geometrical Treatment

The theory of the differential equation

F (x, y, \frac{d y}{d x}) = 0

may also be approached from a geometrical point of view. Replace $\frac{d y}{d x}$ by $z$ and regard $z$ as the third rectangular co-ordinate in space. Then the equation

F (x, y, z) = 0

represents a surface $S$ .

Let

y = ϕ (x)

be any solution of the differential equation, then the pair of equations

y=\phi(x), \quad z=\phi'(x)

represents a space-curve $Γ$ which, since

F\{x, \phi(x), \phi'(x)\} = 0

identically, lies upon the surface $S$ . There is not a solution of the differential equation corresponding to every curve which lies on $S$ , but only to those curves at all points of which the differential relation

d y - z d x = 0

is satisfied.

Let

x = x (t), y = y (t), z = z (t)

be the parametric representation of a curve $Γ$ upon $S$ for which the relation

d y - z d x = 0

is satisfied. The projection of $Γ$ upon the $(x, y)$ -plane will be the curve $C$

x = x (t), y = y (t)

y = ϕ (x) .

Since at all points of the curve $Γ$ the equation

F (x, y, z) = 0

becomes

F\{x, \phi(x), \phi'(x)\}=0,

the curve $C$ , or

y = ϕ (x)

is an integral-curve of the equation

F(x, y, y')=0.

Let the parametric representation of the surface $S$ be

x = f (u, v), y = g (u, v), z = h (u, v),

then the relation

d y - z d x = 0

becomes

(\frac{\partial g}{\partial u} - h \frac{\partial f}{\partial u}) d u + (\frac{\partial g}{\partial v} - h \frac{\partial f}{\partial v}) d v = 0

or, say,

\frac{d v}{d u} = k (u, v) .

Any solution of this differential equation is a relation between $u$ and $v$ which defines a curve $Γ$ on the surface $S$ such that the projection of this curve on the $(x, y)$ -plane is an integral-curve of the differential equation.

Consider, as an example, an equation which can be written in the form

y - g (x, p) = 0.

The corresponding surface $S$ is then representable parametrically as

x = x, y = g (x, p), z = p,

and the relation $d y - z d x = 0$ becomes

p = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial p}p'.

This is a differential equation of the form

\frac{d p}{d x} = k (x, p);

let its general solution be

l (x, p, c) = 0.

Then the integral-curves are the projections on the $(x, y)$ -plane of the intersection of the surface

y - g (x, z) = 0

with the family of cylindrical surfaces

l (x, z, c) = 0.

The general solution of the given equation is therefore obtained by eliminating $p$ between the two equations

y = g (x, p), l (x, p, c) = 0.

2. Equations in which x or y does not explicitly occur

When an equation of either of the forms

F (x, p) = 0, F (y, p) = 0

can be solved for $p$ , the equation can be integrated by quadratures. On the other hand it may occur that the equation is more readily soluble for $x$ (or $y$ as the case may be) in terms of $p$ . Let

x = f (p)

be the solution, then, on differentiating with respect to $y$ ,

\frac{1}{p} = f'(p)\frac{dp}{dy},

whence

y = c + \int p f'(p)\,dp = c+g(p).

say. Then the equations

x = f (p), y = c + g (p)

may be regarded as a parametric representation of the solution, which is obtained explicitly by eliminating $p$ between the two equations.

If the equation does not involve $x$ , it is solved for $y$ and then differentiated with respect to $x$ . The solution is then obtained in the parametric form

y = f (p), x = c + g (p),

where

g(p) = \int p^{-1}f'(p)\,dp.

More generally, it may be possible to express the equation

F (x, p) = 0

parametrically in the form

x = u (t), p = v (t),

then, on differentiating the former with respect to $t$ ,

\frac{1}{p}\frac{dy}{dt} = u'(t),

whence

y = c + \int v(t)u'(t)\,dt.

The solution is then obtained by eliminating $t$ between the expressions for $x$ and $y$ . The equation

F (y, p) = 0,

if expressible in the form

y = u (t), p = v (t),

is solved by eliminating $t$ between

y=u(t) \quad \text{and} \quad x=c+\int \frac{u'(t)}{v(t)}\,dt.

Example. Consider the equation

p^{3} - p^{2} + y^{2} = 0.

It may be represented parametrically as

y = t - t^{3}, p = 1 - t^{2} .

Differentiate the first equation with respect to $t$ , then

p \frac{d x}{d t} = 1 - 3 t^{2},

whence

x = c + \int \frac{1 - 3 t^{2}}{1 - t^{2}} d t = c + 3 t + \log \frac{t - 1}{t + 1} .

Thus $x$ and $y$ are expressed in terms of the parameter $t$ .

3. Equations homogeneous in x and y

An equation which is homogeneous and of degree $m$ in $x$ and $y$ may be written

x^{m} F (\frac{y}{x}, p) = 0.

If it is soluble for $p$ , equations of the type

p = f (\frac{y}{x})

already considered (§ 2.1.2) will arise. This case, therefore, presents no new features of interest. Consider, however, the case in which the equation is soluble for $\frac{y}{x}$ ; thus

\frac{y}{x} = f (p)

y = x f (p) .

Differentiate this equation with respect to $x$ , then

p = f(p) + xf'(p)\frac{dp}{dx}.

Let $p$ be taken as dependent variable, then in this equation the variables are separable, and it has the solution

\log cx = \int \frac{f'(p)}{p-f(p)}\,dp

or, say,

c x = g (p) .

The simultaneous equations

y = x f (p), c x = g (p)

furnish the general solution of the equation.

Example.

y = y p^{2} + 2 p x .

Solution
Solve for $x$ , thus

2 x = y (\frac{1}{p} - p);

differentiate with respect to $y$ , then

\frac{2}{p} = (\frac{1}{p} - p) - y (\frac{1}{p^{2}} + 1) \frac{d p}{d y}

\frac{d p}{p} = - \frac{d y}{y},

whence

p y = c .

Eliminating $p$ from the original equation gives the required solution

y^{2} = 2 c x + c^{2} .

4. Equations linear in x and y

A general type of equation whose solution can be obtained in a parametric form by differentiation is the following:¹

y = x ϕ (p) + ψ (p) .

The derived equation is

p = \phi(p) + \{x\phi'(p)+\psi'(p)\}\frac{dp}{dx},

if $x$ is regarded as dependent variable, and $p$ as independent variable the equation may, when $p - ϕ (p) \neq 0$ , be written

\frac{dx}{dp} - \frac{\phi'(p)}{p-\phi(p)}x = \frac{\psi'(p)}{p-\phi(p)}

and is then a linear equation in the ordinary sense. Its solution in general involves two quadratures; let it be

x = c f (p) + g (p),

then $x$ may be eliminated from the original equation, giving an expression for $y$ in the form

y = c f_{1} (p) + g_{1} (p) .

The general solution is thus expressed parametrically in terms of $p$ .

Consider now those particular values of $p$ , say $p_{1}, p_{2}, \dots$ for which

p - ϕ (p) = 0;

for those values of $p$ ,

\frac{d p}{d x} = 0.

Thus there arises a certain set of isolated integral curves such as

y = x ϕ (p_{1}) + ψ (p_{1}) .

They are straight lines such that if an integral curve of the general family meets one of them, it will have, in general, an inflexion at the common point. The straight lines furnish an example of singular solutions, that is of solutions of the equation which are not included in the general family of integral curves, and not obtainable from the general solution by attributing a special value to the constant of integration.

Example.

y = 2 p x - p^{2} .

The derived equation is

p = 2 p + 2 (x - p) \frac{d p}{d x},

whence, if $p \neq 0$ ,

\frac{d x}{d p} + \frac{2 x}{p} = 2.

The solution of this linear equation is

x = \frac{c}{p^{2}} + \frac{2}{3} p,

which, combined with the original equation, gives the required solution.
On the other hand, when $p = 0$ , there is a solution

y = 0.

5. The Clairaut Equation

The Clairaut equation,²

y = p x + ψ (p),

is not included in the class of equations studied in the preceding section because, in the notation of that section,

ϕ (p) = p

identically, and therefore the method adopted fails.

The derived equation is

p = p + \{x+\psi'(p)\}\frac{dp}{dx};

it can be satisfied either by $\frac{d p}{d x} = 0$ , i.e. $p = c$ , a constant, or by

x+\psi'(p)=0.

The first possibility, $p = c$ , leads to the general solution

y = c x + ψ (c) .

The second possibility leads to a particular solution obtained by eliminating $p$ between the two equations

y = px + \psi(p), \quad x+\psi'(p)=0.

It contains no arbitrary constant, and is not a particular case of the general solution; it is therefore a singular solution.

Now the envelope of the family of straight lines

y = c x + ψ (c)

is obtained by eliminating $c$ between this equation and

0 = x + \psi'(c),

and is identical with the curve furnished by the singular solution. In the case of the Clairaut equation, therefore, the singular solution represents the envelope of the family of integral-curves.

Conversely, the family of tangents to a curve

y = f (x)

satisfies an equation of the Clairaut form, for if

y = α x + β

is a tangent, then

\alpha x+\beta = f(x), \quad \alpha=f'(x).

The elimination of $x$ between these equations gives rise to a relation

β = ψ (α),

and since, on the tangent, $α = p$ , the tangents satisfy the equation

y = p x + ψ (p) .

Example.

y = p x + 1 / p .

Differentiating,

p = p + (x - 1 / p^{2}) \frac{d p}{d x},

whence either $p = c$ , giving the general solution

y = c x + 1 / c,

or else

p^{2} = 1 / x .

The singular solution is found by eliminating $p$ between

p^{2} = 1 / x and y = p x + 1 / p

and is

y^{2} = 4 x .

Equations of the First Order but not of the First Degree

1. Geometrical Treatment

2. Equations in which x or y does not explicitly occur

3. Equations homogeneous in x and y

4. Equations linear in x and y

5. The Clairaut Equation

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