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Equations of the first order and higher degree, Clairaut’s equation. Singular solutions and extraneous loci. Discriminant of a differential equation. p-discriminant, c-discriminant.

General first order equation of degree n. The general first order equation of degree n is an equation of the form

1) a_{0}(x, y)(y')^{n} + a_{1}(x, y)(y')^{n -1} + .... + a_{n-1}(x, y)y' + a_{n}(x, y) = 0

or, equivalently,

2) a_{0}(x, y) p^{n} + a_{1}(x, y)p^{n -1} + .... + a_{n-1}(x, y)p + a_{n}(x, y) = 0

where in 2) we have used the usual convention of denoting y' by the letter p.

Examples. The following equations are of the first order and varying degrees:

xy (y')^{2} + (x^{2} + xy + y^{2})y' + x^{2} + xy = 0 degree 2

(x^{2} + 1)(y')^{4} + (x + 3 y)(y')^{2} + 2x^{2} + y = 0 degree 4

(xy + 2)(y')^{3} + (y')^{2} + 5x^{2} = 0 degree 3

Such equations can sometimes be solved by one or more of the following methods. In each case the problem is reduced to that of solving one or more equations of the first order and first degree.

Methods of solution

I Equations solvable for p. Here the left member of 2), viewed as a polynomial in p, can be resolved into n linear real factors i.e. 2) can be put into the form

(p - F_{1})(p - F_{2}) ...... (p - F_{n}) = 0

where the F’s are functions of x and y.

Procedure. Factor into n linear real factors, set each factor equal to zero and solve the resulting n differential equations of the first order and first degree

to obtain

3) f_{1}(x, y, C) = 0, f_{2}(x, y, C) = 0, .......... , f_{n}(x, y, C) = 0

The primitive of 2) is the product

f_{1}(x, y, C) · f_{1}(x, y, C) · .................. · f_{1}(x, y, C) = 0

of the n solutions 3).

Example. Solve

xy (y')^{2} + (x^{2} + xy + y^{2})y' + x^{2} + xy = 0

Solution. In terms of p this equation is

xy p^{2} + (x^{2} + xy + y^{2})p + x^{2} + xy = 0

and factored as

(xp + x + y)(yp + x) = 0

The solutions of the component equations

are respectively

2xy + x^{2} - C = 0 and x^{2} + y^{2} - C = 0

The primitive is (2xy + x^{2} - C )(x^{2} + y^{2} - C) = 0

II Equations solvable for y, i.e. equations expressible as y = f(x, p).

Procedure.

1. Differentiate y = f(x, p) with respect to x to obtain

an equation of the first order and first degree.

2. Solve p = F(x, p, dp/dx) to obtain φ(x, p, C) = 0.

3. Obtain the primitive by eliminating p between y = f(x, p) and φ(x, p, C) = 0, when possible, or express x and y separately as functions of the parameter p.

Example. Solve

1) 16x^{2} + 2p^{2}y - p^{3}x = 0

Solution. We can rewrite the equation as

Differentiating with respect to x we get

Clearing of fractions and combining we have

or

This equation is satisfied when p^{3} + 32x = 0 or p - x(dp/dx) = 0. From the later we obtain

Integrating we get

ln p = ln x + ln K = ln Kx

Taking exponentials we get

3) p = Kx

Substituting 3) into 1) we get

16x^{2} + 2K^{2}x^{2}y - K^{3} x^{4} = 0

or, replacing K by 2C,

2 + C^{2}y - C^{3}x^{2} = 0

which is the primitive.

We will not consider the factor p^{3} + 32x of 2) here because it does not contain the derivative
dp/dx. Use of the factor will yield a singular solution, a topic that we will treat shortly.

III Equations solvable for x, i.e. equations expressible as x = f(y, p).

Procedure.

1. Differentiate x = f(y, p) with respect to y to obtain

an equation of the first order and first degree.

2. Solve 1/p = F(y, p, dp/dy) to obtain φ(y, p, C) = 0.

3. Obtain the primitive by eliminating p between x = f(y, p) and φ(y, p, C) = 0, when possible, or express x and y separately as functions of the parameter p.

Example. Solve

1) y = 3px + 6p^{2}y^{2}

Solution.

Solving 1) for x we obtain

Differentiating with respect to y we obtain

Clearing of fractions we obtain

or

Setting the second factor equal to zero gives

Integrating we get

2 ln y = - ln p + ln C

which yields

py^{2} = C

or, equivalently,

2) p = C/y^{2}

Substituting 2) into 1) gives the primitive

y^{3} = 3Cx + 6C^{2}

IV Clairaut’s equation. The differential equation

y = px + f(p)

is called Clairaut’s equation. Its primitive is

y = Cx + f(C)

and is obtained simply by replacing p by C in the given equation.

Example. Solve

Solution. The primitive is

Source: Ayres. Differential Equations (Schaum). Chap 9

Singular solutions and extraneous loci

Singular solutions. In addition to the general solution a differential equation may also have a singular solution. A singular solution is a solution not obtainable by assigning particular values to the arbitrary constants of the general solution. It is the equation of an envelope of the family of curves represented by the general solution. This envelope satisfies the differential equation because at every one of its points its slope and the coordinates of the point are the same as those of some member of the family of curves representing the general solution.

James/James. Mathematics Dictionary.

Def. Discriminant of a differential equation. p-discriminant, c-discriminant. For a differential equation of type f(x, y, p) = 0, where p = dy/dx, the p-discriminant is the result of eliminating p between the equations f(x, y, p) = 0 and ∂f(x, y, p)/∂p = 0. If the solution of the differential equation is

u(x, y, c) = 0

the c-discriminant is the result of eliminating c between the equations u(x, y, c) = 0 and ∂u(x, y, c)∂c = 0. The curve whose equation is obtained by setting the p-discriminant equal to zero contains all envelopes of solutions, but also may contain a cusp locus, a tac-locus, or a particular solution (in general, the equation of the tac-locus will be squared and the equation of the particular solution will be cubed). The curve whose equation is obtained by setting the c-discriminant equal to zero contains all envelopes of solutions, but also may contain a cusp locus, a node locus, or a particular solution (in general, the equation of the node-locus will be squared and the equation of the particular solution will be cubed). In general, the cusp locus, node locus and tac-locus are not solutions of the differential equation.

Example. The differential equation (dx/dy)^{2}(2 - 3y)^{2} = 4(1 - y) has the general solution (x -
c)^{2} = y^{2}(1 - y) and the p-discriminant and c-discriminant equations are, respectively,

(2 - 3y)^{2}(1 - y) = 0 and y^{2}(1 - y) = 0 .

The line 1 - y = 0 is an envelope; 2 - 3y = 0 is a tac-locus; y = 0 is a node locus.

James/James. Mathematics Dictionary.

Let us now consider a concrete example of an equation containing a singular solution.. The differential equation

1) y = px + 2p^{2}

has as its general solution the family of straight lines

2) y = Cx + 2C^{2}.

In addition to this general solution it also has the singular solution

3) x^{2} + 8y = 0

which is a parabola. See Fig. 1.

Typical lines of the general solution are shown in Fig. 1 (b). There is a single line for each value
of the arbitrary constant C and each line is tangent to the singular solution i.e. the parabola x^{2} +
8y = 0 . We see that the singular solution corresponds to an envelope of that family of curves
represented by the general solution.

In the region above the parabola x^{2} + 8y > 0 and in the region below the parabola x^{2} + 8y < 0.
At each point P:(x, y) in the region above the parabola equation 1) yields a set of two distinct
directions (or slopes) p_{1} and p_{2} and through P passes two distinct lines of the general solution y =
Cx + 2C^{2} , one with the slope p_{1} and the other with the slope p_{2}. For example, consider the point
P:(-2, 4). If we substitute the coordinates (-2, 4) into 1) we get 4 = -2p + 2p^{2} , which when
solved for p, gives the solutions p_{1} = 2 and p_{2} = -1. Using the point-slope formula for straight
lines we compute the equations of the general solution lines passing through P with slopes p_{1} and
p_{2} as y = 2x + 8 and y = -x + 2. See Fig. 1 (a). Thus passing through each point P above the
parabola, there are two straight lines of the general solution, both tangent to the parabola, as
shown in Fig. 1 (a).

Now let us substitute the coordinates pf point P:(-2, 4) into equation 2), the equation of the
general solution. We get 4 = -2C + 2C^{2}, which when solved for C gives the solutions C_{1} = 2 and
C_{2} = -1. These two values of C, when plugged into equation 2), give the two equations y = 2x + 8
and y = -x + 2 that pass through point P.

Let us now consider what happens when we move point P down towards the parabola. As we
move it down the lines adjust, remaining tangent to the parabola. What happens when point P
touches the parabola? The two lines become a single line tangent to the parabola. In the
situation when P lies on the parabola the two slopes p_{1} and p_{2} obtained from equation 1) are equal
i.e. equation 1) has multiple roots (two equal roots). If we solve equation 1) for p using the
quadratic formula we get

In the situation where P lies on the parabola the discriminant x^{2} + 8y inside the radical sign
becomes zero and the formula yields multiple roots i.e. two equal roots.

Points for which x^{2} + 8y < 0 yield imaginary roots for p in equation 1) and for C in equation 2).

Def. p-equation. The differential equation, expressed in terms of the letter p. In our example, equation 1) is the p-equation.

Def. C-equation. The general solution (primitive) of the differential equation. In our example, equation 2) is the C-equation.

Singular solutions and multiple roots. The singular solutions of a differential equation are found by expressing the conditions

a) that the p-equation have multiple roots

b) that the C-equation have multiple roots

Theorem 1. An equation *f*(x) = 0 will have multiple roots if and only if its discriminant
vanishes.

Theorem 2. The discriminant of an equation *f*(x) = 0 can be obtained by eliminating x
between *f*(x) = 0 and *f* '(x) = 0.

Discriminants of some special equations

Equation Discriminant

ax^{2} + bx + c = 0 b^{2} - ac

ax^{3} + bx^{2} + cx + d = 0 b^{2}c^{2} + 18abcd - 4ac^{3} - 4b^{3}d - 27a^{2}d^{2}

From Theorem 2, we see that if we are dealing with a p-equation the discriminant can be
obtained by eliminating p between *f*(p) = 0 and d*f* (p)/dp = 0. I f we are dealing with a C-equation the discriminant can be obtained by eliminating C between *f*(C) = 0 and *df* (C)/dC = 0.

Types of equations having singular solutions. In general, an equation of the first order does not have singular solutions. An equation of the first degree cannot have singular solutions. In addition, an equation f(x, y, p) = 0 cannot have singular solutions if f(x, y, p) can be resolved into factors which are linear in p and rational in x and y.

Theorem 3. Let q(x, y) = 0 be a singular solution of the differential equation f(x, y, p) = 0 and let g(x, y, C) = 0 be the general solution. Then q(x, y) is a factor of both discriminants (p-discriminant and C-discriminant). Each discriminant may, however, have other factors which give rise to other loci associated with the general solution. The p-discriminant may have factors corresponding to a tac-locus, cusp locus or a particular solution. The C-discriminant may have factors corresponding to a cusp locus, node locus or a particular solution. Tac-loci, cusp loci and node loci are extraneous i.e. they do not satisfy the differential equation.

References

1. James/James. Mathematics Dictionary.

2. Frank Ayres. Differential Equations (Schaum).

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