Singular Solutions of Differential Equations

Solved Problems

Example 1.

Find the singular solutions of the equation \[1 + {\left( {y'} \right)^2} = \frac{1}{{{y^2}}}.\]

Solution.

We will use \(p\)-discriminant for investigation of the singular points. Differentiating the equation with respect to \(y'\) gives:

\[2y' = 0,\;\; \Rightarrow y' = 0.\]

Putting this into the differential equation yields the equation of the \(p\)-discriminant:

\[1 + 0 = \frac{1}{{{y^2}}}.\]

It follows from here that the equation of the \(p\)-discriminant describes two horizontal lines: \(y = \pm 1.\) It is easy to check that this solution satisfies the given differential equation:

\[y = \pm 1,\;\; \Rightarrow y' = 0,\;\; \Rightarrow 1 + {0^2} = \frac{1}{{{1^2}}},\;\; \Rightarrow 1 = 1.\]

Now we find the general solution of the differential equation. We can write it in the following form:

\[{\left( {y'} \right)^2} = \frac{1}{{{y^2}}} - 1 = \frac{{1 - {y^2}}}{{{y^2}}},\;\; \Rightarrow y' = \pm \frac{{\sqrt {1 - {y^2}} }}{y},\;\; \Rightarrow \frac{{ydy}}{{\sqrt {1 - {y^2}} }} = \pm dx.\]

Make the replacement:

\[1 - {y^2} = t,\;\; \Rightarrow - 2ydy = dt,\;\; \Rightarrow ydy = - \frac{{dt}}{2}.\]

As a result, we get:

\[\frac{{\left( { - \frac{{dt}}{2}} \right)}}{{\sqrt t }} = \pm dx.\]

After integration we obtain the general solution of the differential equation:

\[\int {\frac{{dt}}{{2\sqrt t }}} = \pm \int {dx} ,\;\; \Rightarrow \sqrt t = \pm x + C,\;\; \Rightarrow \sqrt {1 - {y^2}} = \pm \left( {x + C} \right),\]

where \(C\) is an arbitrary constant.

The last expression can be written as follows:

\[{\left( {x + C} \right)^2} + {y^2} = 1.\]

This equation describes the family of circles of radius \(1,\) filling in the band \(-1 \le y \le 1\) (Figure \(5\)).

A family of circles of radius 1 — Figure 5.

As it can be seen from the Figure, the \(p\)-discriminant lines \(y = \pm 1\) are the envelopes for the given circles. However, we must prove formally that the uniqueness of solution is violated on these straight lines.

Take an arbitrary point \({x_0}.\) Write the condition of touching two integral curves at this point:

\[\left\{ \begin{array}{l} {y_1}\left( {{x_0}} \right) = {y_2}\left( {{x_0}} \right)\\ {y'_1}\left( {{x_0}} \right) = {y'_2} \left( {{x_0}} \right) \end{array} \right..\]

Here \({y_1}\left( x \right)\) denotes our general solution that has the form (for the upper semicircle):

\[{y_1}\left( x \right) = \sqrt {1 - {{\left( {x + C} \right)}^2}} .\]

The function \({y_2}\left( x \right)\) corresponds to the horizontal line \(y = 1.\) Both the lines will touch at the point \({x_0}\) if only the following relations are satisfied:

\[\left\{ \begin{array}{l} \sqrt {1 - {{\left( {{x_0} + C} \right)}^2}} = 1\\ \frac{{ - {x_0} - C}}{{\sqrt {1 - {x_0} + C} }} = 0 \end{array} \right..\]

The given conditions are satisfied, if we set \(C = - {x_0}.\)

Thus, we have proved that at each point \({x_0}\) on the straight line \(y = 1\) there exists a touching circle with \(C = - {x_0}.\) Hence, the uniqueness of solution is violated at each point of the straight line. Therefore, the line \(y = 1\) is a singular solution of the given differential equation. Similarly, we can prove that the line \(y = -1\) is also a singular solution.

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 \[y = Cx + {C^2} + {x^2}.\]

Solution.

We will use \(C\)-discriminant to determine the singular solution. Since the general solution of the differential equation is known, we can write:

\[\Phi \left( {x,y,C} \right) = y - Cx - {C^2} - {x^2}.\]

The partial derivative in \(C\) is

\[\frac{{\partial \Phi \left( {x,y,C} \right)}}{{\partial C}} = - x - 2C.\]

We get the following system of equations:

\[\left\{ \begin{array}{l} y = Cx - {C^2} - {x^2}\\ - x - 2C = 0 \end{array} \right..\]

It follows from the second equation that \(C = - \frac{x}{2}.\) Substituting this in the first equation, we find the \(C\)-discriminant curve, which is a parabola:

\[y = \left( { - \frac{x}{2}} \right) \cdot x - {\left( { - \frac{x}{2}} \right)^2} - {x^2} = \frac{3}{4}{x^2}.\]

Make sure that this function is a solution of the original differential equation:

\[y = \frac{3}{4}{x^2},\;\; \Rightarrow y' = \frac{3}{2}x,\;\; \Rightarrow \frac{3}{4}{x^2} = \left( {\frac{3}{2}x} \right)^2 - 3x \cdot \frac{3}{2}x + 3{x^2},\;\; \Rightarrow \frac{3}{4}{x^2} = \frac{9}{4}{x^2} - \frac{9}{2}{x^2} + 3{x^2},\;\; \Rightarrow \frac{3}{4}{x^2} = \frac{3}{4}{x^2}.\]

Now we check that uniqueness of solution is violated on this curve. Denote:

\[{y_1} = Cx + {C^2} + {x^2},\;\; {y_2} = \frac{3}{4}{x^2}.\]

Write the conditions of touching the two curves at a certain arbitrary point \({x_0}\) as

\[\left\{ \begin{array}{l} {y_1}\left( {{x_0}} \right) = {y_2}\left( {{x_0}} \right)\\ {y'_1}\left( {{x_0}} \right) = {y'_2} \left( {{x_0}} \right) \end{array} \right..\]

As a result, we have:

\[\left\{ \begin{array}{l} C{x_0} + {C^2} + x_0^2 = \frac{3}{4}x_0^2\\ C + 2{x_0} = \frac{3}{2}{x_0} \end{array} \right..\]

The given system of equations is consistent when the constant \(C\) at each point \({x_0}\) is equal to

\[C = - \frac{{{x_0}}}{2}.\]

Thus, we have proved that the \(C\)-discriminant curve \(y = \frac{3}{4} {x^2}\) is the envelope (that is the singular solution) for the family of parabolas \(y = Cx + {C^2} + {x^2}\) representing the general solution of the differential equation.

Example 3.

Investigate the singular solutions of the differential equation \[{\left( {y'} \right)^2}{\left( {1 - y} \right)^2} = 2 - y.\]

Solution.

First we find \(p\)-discriminant of the given equation. Differentiating with respect to \(x\) gives:

\[2y'{\left( {1 - y} \right)^2} = 0.\]

Eliminate \(y'\) from the system of the equations:

\[\left\{ \begin{array}{l} {\left( {y'} \right)^2}{\left( {1 - y} \right)^2} = 2 - y\\ y'{\left( {1 - y} \right)^2} = 0 \end{array} \right..\]

Then we obtain:

\[{\left( {y'} \right)^2} = \frac{{2 - y}}{{{{\left( {1 - y} \right)}^2}}},\;\; \Rightarrow \frac{{2 - y}}{{{{\left( {1 - y} \right)}^2}}} \cdot {\left( {1 - y} \right)^4} = 0,\;\; \Rightarrow {\left( {1 - y} \right)^2}\left( {2 - y} \right) = 0.\]

Now we determine \(C\)-discriminant. Unfortunately, to find it, we need to solve the differential equation and find its general solution :(. Rewrite the equation in the following form:

\[\left( {\frac{{dy}}{{dx}}} \right)^2 = \frac{{2 - y}}{{{{\left( {1 - y} \right)}^2}}},\;\; \Rightarrow \frac{{dy}}{{dx}} = \pm \frac{{\sqrt {2 - y} }}{{1 - y}},\;\; \Rightarrow \frac{{\left( {1 - y} \right)dy}}{{\sqrt {2 - y} }} = \pm dx.\]

By integrating both sides we have:

\[\int {\frac{{\left( {1 - y} \right)dy}}{{\sqrt {2 - y} }}} = \pm \int {dx} + C.\]

We change the variable in the left integral:

\[2 - y = t,\;\; \Rightarrow dy = - dt,\;\; \Rightarrow 1 - y = t - 1.\]

This yields:

\[ \int {\frac{{\left( {t - 1} \right)\left( { - dt} \right)}}{{\sqrt t }}} = \pm x + C,\;\; \Rightarrow \int {\left( {\sqrt t - \frac{1}{{\sqrt t }}} \right)dt} = \mp x - C,\;\; \Rightarrow \frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}} = \mp x - C,\;\; \Rightarrow \frac{2}{3}{t^{\frac{3}{2}}} - 2{t^{\frac{1}{2}}} = \mp x - C,\;\; \Rightarrow \frac{2}{3}{\left( {2 - y} \right)^{\frac{3}{2}}} - 2{\left( {2 - y} \right)^{\frac{1}{2}}} = \mp x - C,\;\; \Rightarrow \frac{2}{3}\sqrt {2 - y} \left( {2 - y - 3} \right) = \mp x - C,\;\; \Rightarrow \frac{4}{9}\left( {2 - y} \right){\left( {y + 1} \right)^2} = {\left( {x + C} \right)^2},\;\; \Rightarrow 4\left( {2 - y} \right){\left( {y + 1} \right)^2} = 9{\left( {x + C} \right)^2}.\]

Differentiate the general solution with respect to \(C:\)

\[0 = 18\left( {x + C} \right).\]

Putting \(x + C = 0\) back into the general solution, we obtain the equation of the \(C\)-discriminant:

\[{\left( {y + 1} \right)^2}\left( {2 - y} \right) = 0.\]

Now we can write both discriminants together:

\[{\psi _p}\left( y \right) = {\left( {1 - y} \right)^2}\left( {2 - y} \right) = 0,\]

\[{\psi _C}\left( y \right) = {\left( {y + 1} \right)^2}\left( {2 - y} \right) = 0.\]

As it follows from the structure of the discriminants, the equation \(2 - y = 0\) is the equation of the envelope because it's contained in the both discriminants as the first degree factor. We can also find the equation of the Tac locus from the expression for \(p\)-discriminant:

\[{\left( {1 - y} \right)^2} = 0,\;\; \Rightarrow y = 1.\]

And similarly, as it follows from the expression for \(C\)-discriminant, the equation of the Node locus is given by

\[{\left( {y + 1} \right)^2} = 0,\;\; \Rightarrow y = - 1.\]

In the given example, only the envelope \(y = 2\) is the singular solution of the differential equation.