# Differential Equations

## First Order Equations # Singular Solutions of Differential Equations

### Definition of Singular Solution

A function $$\varphi \left( x \right)$$ is called the singular solution of the differential equation $$F\left( {x,y,y’} \right) = 0,$$ if uniqueness of solution is violated at each point of the domain of the equation. Geometrically this means that more than one integral curve with the common tangent line passes through each point $$\left( {{x_0},{y_0}} \right).$$

#### Note.

Sometimes the weaker definition of the singular solution is used, when the uniqueness of solution of differential equation may be violated only at some points.

A singular solution of a differential equation is not described by the general integral, that is it can not be derived from the general solution for any particular value of the constant $$C.$$ We illustrate this by the following example:

Suppose that the following equation is required to be solved: $${\left( {y’} \right)^2} – 4y = 0.$$

It is easy to see that the general solution of the equation is given by the function $$y = {\left( {x + C} \right)^2}.$$ Graphically, it is represented by the family of parabolas (Figure $$1$$).

Besides this, the function $$y = 0$$ also satisfies the differential equation. However, this function is not contained in the general solution! Since more than one integral curve passes through each point of the straight line $$y = 0,$$ the uniqueness of solution is violated on this line, and hence it is a singular solution of the differential equation.

### $$p$$-discriminant

One of the ways to find a singular solution is investigation of the so-called $$p$$-discriminant of the differential equation. If the function $$F\left( {x,y,y’} \right)$$ and its partial derivatives $${\large\frac{{\partial F}}{{\partial y}}\normalsize}, {\large\frac{{\partial F}}{{\partial y’}}\normalsize}$$ are continuous in the domain of the differential equation, the singular solution can be found from the system of equations:

$\left\{ \begin{array}{l} F\left( {x,y,y’} \right) = 0\\ \frac{{\partial F\left( {x,y,y’} \right)}}{{\partial y’}} = 0 \end{array} \right..$

The equation $$\psi \left( {x,y} \right) = 0$$ obtained by solving the given system of equations is called the $$p$$-discriminant of the differential equation. The corresponding curve determined by this equation is called a $$p$$-discriminant curve.

Upon finding the $$p$$-discriminant curve, one should check the following:

1. Whether it is a solution of the differential equation?
2. Whether it is a singular solution, that is are there any other integral curves of the differential equation that touch the $$p$$-discriminant curve at each point?

This can be done as follows:

• Find the general solution of the differential equation (denote it by $${y_1}$$);
• Write the conditions of touching the singular solution (denote it by $${y_2}$$) and the general solution $${y_1}$$ at an arbitrary point $${x_0}:$$
$\left\{ \begin{array}{l} {y_1}\left( {{x_0}} \right) = {y_2}\left( {{x_0}} \right)\\ {y’_1}\left( {{x_0}} \right) = {y’_2} \left( {{x_0}} \right) \end{array} \right.;$

If the system has a solution at the arbitrary point $${x_0},$$ the function $${y_2}$$ is a singular solution. The singular solution usually corresponds to the envelope of the family of integral curves of the general solution of the differential equation.

### Envelope of the Family of Integral Curves and $$C$$-discriminant

Another way to find a singular solution as the envelope of the family of integral curves is based on using $$C$$-discriminant.

Let $$\Phi \left( {x,y,C} \right)$$ be the general solution of a differential equation $$F\left( {x,y,y’} \right) = 0.$$ Graphically the equation $$\Phi \left( {x,y,C} \right) = 0$$ corresponds to the family of integral curves in the $$xy$$-plane. If the function $$\Phi \left( {x,y,C} \right)$$ and its partial derivatives are continuous, the envelope of the family of integral curves of the general solution is defined by the system of equations:

$\left\{ \begin{array}{l} \Phi \left( {x,y,C} \right) = 0\\ \frac{{\partial \Phi \left( {x,y,C} \right)}}{{\partial C}} = 0 \end{array} \right..$

To make sure whether a solution of the system of equations is really the envelope, one can use the method mentioned in the previous section.

### General Algorithm of Finding Singular Points

A more common way of finding singular points of a differential equation is based on the simultaneous using $$p$$-discriminant and $$C$$-discriminant.

Here first we find the equations of the $$p$$-discriminant and $$C$$-discriminant:

• $${\psi_p}\left( {x,y} \right) = 0$$ is the equation of the $$p$$-discriminant;
• $${\psi_C}\left( {x,y} \right) = 0$$ is the equation of the $$C$$-discriminant.

It turns out that these equations have a certain structure. In general case, the equation of the $$p$$-discriminant can be factored into the product of three functions:

${{\psi _p}\left( {x,y} \right) }={ E \times {T^2} \times C }={ 0,}$

where$$E$$ means the equation of the Envelope, $$T$$ is the equation of the Tac locus, and $$C$$ is the equation of the Cusp locus.

Similarly, the equation of the $$C$$-discriminant can be also factored into the product of three functions:

${{\psi _C}\left( {x,y} \right) }={ E \times {N^2} \times {C^3} }={ 0,}$

where $$E$$ is the equation of the Envelope, $$N$$ is the equation of the Node locus, and $$C$$ is the equation of the Cusp locus.

Here we meet with new kinds of singular points: $$C$$ – Cusp loci, $$T$$- Tac loci, and $$N$$ – Node loci. Their view in the $$xy$$-plane is shown schematically in Figures $$2-4.$$ Figure 2 Figure 3 Figure 4

Three of the four types of points, namely, the Tac loci, Cusp loci and Node loci are extraneous points, i.e. they do not satisfy the differential equation and, therefore, they are not singular solutions of the differential equation. Only the envelope of the considered points is the singular solution. Since the envelope is presented in the equations of the both discriminants as a first degree factor, this allows to find the equation of the envelope.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the singular solutions of the equation $$1 + {\left( {y’} \right)^2}$$ $$= {\large\frac{1}{{{y^2}}}\normalsize}.$$

### Example 2

Find the singular solution of the differential equation $$y = {\left( {y’} \right)^2}$$ $$- 3xy’$$ $$+ 3{x^2}.$$ The general solution of the equation is known and given by the function $$y = Cx + {C^2}$$ $$+ {x^2}.$$

### Example 3

Investigate the singular solutions of the differential equation $${\left( {y’} \right)^2}{\left( {1 – y} \right)^2}$$ $$= 2 – y.$$
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Problems 1-3