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.

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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) a₀(x, y)(y')ⁿ + a₁(x, y)(y')^{n -1} + .... + a_n-1(x, y)y' + a_n(x, y) = 0

or, equivalently,

2) a₀(x, y) pⁿ + a₁(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 + (x² + xy + y²)y' + x² + xy = 0 degree 2

(x² + 1)(y')⁴ + (x + 3 y)(y')² + 2x² + y = 0 degree 4

(xy + 2)(y')³ + (y')² + 5x² = 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 - F₁)(p - F₂) ...... (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) f₁(x, y, C) = 0, f₂(x, y, C) = 0, .......... , f_n(x, y, C) = 0

The primitive of 2) is the product

f₁(x, y, C) · f₁(x, y, C) · .................. · f₁(x, y, C) = 0

of the n solutions 3).

Example. Solve

xy (y')² + (x² + xy + y²)y' + x² + xy = 0

Solution. In terms of p this equation is

xy p² + (x² + xy + y²)p + x² + xy = 0

and factored as

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

The solutions of the component equations

are respectively

2xy + x² - C = 0 and x² + y² - C = 0

The primitive is (2xy + x² - C )(x² + y² - 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) 16x² + 2p²y - p³x = 0

Solution. We can rewrite the equation as

Differentiating with respect to x we get

Clearing of fractions and combining we have

This equation is satisfied when p³ + 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

16x² + 2K²x²y - K³ x⁴ = 0

or, replacing K by 2C,

2 + C²y - C³x² = 0

which is the primitive.

We will not consider the factor p³ + 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 + 6p²y²

Solution.

Solving 1) for x we obtain

Differentiating with respect to y we obtain

Clearing of fractions we obtain

Setting the second factor equal to zero gives

Integrating we get

2 ln y = - ln p + ln C

which yields

py² = C

or, equivalently,

2) p = C/y²

Substituting 2) into 1) gives the primitive

y³ = 3Cx + 6C²

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 = y²(1 - y) and the p-discriminant and c-discriminant equations are, respectively,

(2 - 3y)²(1 - y) = 0 and y²(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 + 2p²

has as its general solution the family of straight lines

2) y = Cx + 2C².

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

3) x² + 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 x² + 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 x² + 8y > 0 and in the region below the parabola x² + 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) p₁ and p₂ and through P passes two distinct lines of the general solution y = Cx + 2C² , one with the slope p₁ and the other with the slope p₂. For example, consider the point P:(-2, 4). If we substitute the coordinates (-2, 4) into 1) we get 4 = -2p + 2p² , which when solved for p, gives the solutions p₁ = 2 and p₂ = -1. Using the point-slope formula for straight lines we compute the equations of the general solution lines passing through P with slopes p₁ and p₂ 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 + 2C², which when solved for C gives the solutions C₁ = 2 and C₂ = -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 p₁ and p₂ 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 x² + 8y inside the radical sign becomes zero and the formula yields multiple roots i.e. two equal roots.

Points for which x² + 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

ax² + bx + c = 0 b² - ac

ax³ + bx² + cx + d = 0 b²c² + 18abcd - 4ac³ - 4b³d - 27a²d²

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 df (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).