Let the ode be given as \(F\left ( x,y,p\right ) =0\) where \(p=y^{\prime }\left ( x\right ) \), and let the solution be given as \(\Psi \left ( x,y,c\right ) =0\) where \(c\) is the constant of integration.
Given first order ode \(F\left ( x,y,y^{\prime }\right ) =0\) the goal is find its singular solutions (if one exists). This method applies to first order ode’s which is not linear in \(y^{\prime }\). By singular solution here we mean the envelope (called \(E\) below).
These are solutions that satisfy the ode which can not be obtained from the general solution for any value of the constant of integration \(c\) (including \(\pm \infty \)).
A first order ode which is linear in \(y^{\prime }\) do not have such singular solutions, in the sense of singular solution being an envelope of the general solution.
Singular solution will be called \(y_{c}\left ( x\right ) \). This singular solution will be the envelope of the family of solutions of the general solution. It will have no constant in it, unlike the general solution.
If the ode is an initial value problem, and if the uniqueness theorem says there is a unique solution in an interval around \(\left ( x_{0},y_{0}\right ) \) then no singular solution exists in that interval as this will violate the uniqueness theorem.
There are two methods for finding the singular solution. Either from the ode itself (without even knowing the solution) or from the general solution. The first is called the p-discriminant method and the second is called the c-discriminant method. Both of these methods are based on elimination.
In the p-discriminant method, \(p\) is eliminated. In the c-discriminant method, the constant \(c\) is eliminated. Clearly the p-discriminant method is preferred as it does not require solving the ode first.
In both elimination methods, we set up two equations and solve for \(y\). This can result in more than one solution for \(y\). Only those solutions that satisfy the ode itself are the envelope singular solutions we want. Others are important but they are not singular solution as they do not satisfy the ode and are not an envelope for the general solution. This diagram below shows the types of solutions that can be found by each elimination method. Only the envelope is a valid solution singular solution which is common.
In the above we see that there are 4 different type of solutions that show up. Only one type satisfies the ode. This is the envelope singular solution. Let us give some letters to refer to these solutions. Let \(E\) be the envelope. Let \(T\) be the Tac locus, let \(N\) be the nodal locus and let \(C\) be the cusp locus.
What happens in this. When doing the elimination process using either method, we can end up with more than one solution \(y\). These we write in factored form.
The general form from the p-elimination will look like \(ET^{2}C=0\) and the general form from the c-elimination will look like \(EN^{2}C^{3}=0\) (ref: Math24.net).
This is only the most general form. \(T\) or \(C\) or \(N\) could be missing.
We see that both \(E\) and \(C\) can show up from both methods. The factor that satisfies the ode is the \(E\) factor only, which is the envelope singular solution. All others do not satisfy the ode but they have interesting geometrical meanings when plotted against the family of general solution. Will give examples below to show this.
The \(E\) solution from the p-elimination is the locus of points satisfied by \(F\left ( x,y,p\right ) =0\) such that at each point the \(p^{\prime }s\) have equal roots. And the \(E\) solution from the c-elimination is the locus of points satisfied by \(\Psi \left ( x,y,c\right ) =0\) such that at each point the \(c^{\prime }s\) have equal roots. Both \(E\) solution from both methods is the same. So any one of these methods can be used to find \(E\).
For the special case, when \(F\left ( x,y,p\right ) =0\) is quadratic in \(p\) or the solution \(\Psi \left ( x,y,c\right ) =0\) is quadratic in \(c\), we can just use the quadratic discriminant \(b^{2}-4ac=0\) to find solutions for either \(p\) or \(c\). This is simpler than using elimination method. But this only works in this special case. See examples below.
In summary, these are the main steps for finding singular solution for first order ode:
The Examples below show how to use these methods. In all the following examples, the plots will show the singular solution(s) as thick red dashed lines.
Given ode \(F\left ( x,y,p\right ) =0\) the necessary and sufficient conditions that a singular solution exist are the following (see E.L.Ince page 88)
The above should be satisfied simultaneously. However, I am not able to verify these now. Ince says that \(\frac {\partial F}{\partial y}\neq 0\) is necessary for singular solution to exist. So will only use this check in all examples below.