Consider the homogeneous linear ordinary differential equation

${\displaystyle xy'(x)+2y(x)=0,\,\!}$ 

where primes denote derivatives with respect to x. The general solution to this equation is

${\displaystyle y(x)=Cx^{-2}.\,\!}$ 

For a given ${\displaystyle C}$ , this solution is smooth except at ${\displaystyle x=0}$  where the solution is divergent. Furthermore, for a given ${\displaystyle x\not =0}$ , this is the unique solution going through ${\displaystyle (x,y(x))}$ .

Consider the differential equation

${\displaystyle y'(x)^{2}=4y(x).\,\!}$ 

A one-parameter family of solutions to this equation is given by

${\displaystyle y_{c}(x)=(x-c)^{2}.\,\!}$ 

Another solution is given by

${\displaystyle y_{s}(x)=0.\,\!}$ 

Since the equation being studied is a first-order equation, the initial conditions are the initial x and y values. By considering the two sets of solutions above, one can see that the solution fails to be unique when ${\displaystyle y=0}$ . (It can be shown that for ${\displaystyle y>0}$  if a single branch of the square root is chosen, then there is a local solution which is unique using the Picard–Lindelöf theorem.) Thus, the solutions above are all singular solutions, in the sense that solution fails to be unique in a neighbourhood of one or more points. (Commonly, we say "uniqueness fails" at these points.) For the first set of solutions, uniqueness fails at one point, ${\displaystyle x=c}$ , and for the second solution, uniqueness fails at every value of ${\displaystyle x}$ . Thus, the solution ${\displaystyle y_{s}}$  is a singular solution in the stronger sense that uniqueness fails at every value of x. However, it is not a singular function since it and all its derivatives are continuous.

In this example, the solution ${\displaystyle y_{s}(x)=0}$  is the envelope of the family of solutions ${\displaystyle y_{c}(x)=(x-c)^{2}}$ . The solution ${\displaystyle y_{s}}$  is tangent to every curve ${\displaystyle y_{c}(x)}$  at the point ${\displaystyle (c,0)}$ .

The failure of uniqueness can be used to construct more solutions. These can be found by taking two constant ${\displaystyle c_{1}<c_{2}}$  and defining a solution ${\displaystyle y(x)}$  to be ${\displaystyle (x-c_{1})^{2}}$  when ${\displaystyle x<c_{1}}$ , to be ${\displaystyle 0}$  when ${\displaystyle c_{1}\leq x\leq c_{2}}$ , and to be ${\displaystyle (x-c_{2})^{2}}$  when ${\displaystyle x>c_{2}}$ . Direct calculation shows that this is a solution of the differential equation at every point, including ${\displaystyle x=c_{1}}$  and ${\displaystyle x=c_{2}}$ . Uniqueness fails for these solutions on the interval ${\displaystyle c_{1}\leq x\leq c_{2}}$ , and the solutions are singular, in the sense that the second derivative fails to exist, at ${\displaystyle x=c_{1}}$  and ${\displaystyle x=c_{2}}$ .

The previous example might give the erroneous impression that failure of uniqueness is directly related to ${\displaystyle y(x)=0}$ . Failure of uniqueness can also be seen in the following example of a Clairaut's equation:

${\displaystyle y(x)=x\cdot y'+(y')^{2}\,\!}$ 

We write y' = p and then

${\displaystyle y(x)=x\cdot p+(p)^{2}.}$ 

Now, we shall take the differential according to x:

${\displaystyle p=y'=p+xp'+2pp'}$ 

which by simple algebra yields

${\displaystyle 0=(2p+x)p'.}$ 

This condition is solved if 2p+x=0 or if p′=0.

If p' = 0 it means that y' = p = c = constant, and the general solution of this new equation is:

${\displaystyle y_{c}(x)=c\cdot x+c^{2}}$ 

where c is determined by the initial value.

If x + 2p = 0 then we get that p = −½x and substituting in the ODE gives

${\displaystyle y_{s}(x)=-{\tfrac {1}{2}}x^{2}+(-{\tfrac {1}{2}}x)^{2}=-{\tfrac {1}{4}}x^{2}.}$ 

Now we shall check when these solutions are singular solutions. If two solutions intersect each other, that is, they both go through the same point (x,y), then there is a failure of uniqueness for a first-order ordinary differential equation. Thus, there will be a failure of uniqueness if a solution of the first form intersects the second solution.

The condition of intersection is : y_s(x) = y_c(x). We solve

${\displaystyle c\cdot x+c^{2}=y_{c}(x)=y_{s}(x)=-{\tfrac {1}{4}}x^{2}}$ 

to find the intersection point, which is ${\displaystyle (-2c,-c^{2})}$ .

We can verify that the curves are tangent at this point y'_s(x) = y'_c(x). We calculate the derivatives:

${\displaystyle y_{c}'(-2c)=c\,\!}$ 

${\displaystyle y_{s}'(-2c)=-{\tfrac {1}{2}}x|_{x=-2c}=c.\,\!}$ 

Hence,

${\displaystyle y_{s}(x)=-{\tfrac {1}{4}}\cdot x^{2}\,\!}$ 

is tangent to every member of the one-parameter family of solutions

${\displaystyle y_{c}(x)=c\cdot x+c^{2}\,\!}$ 

of this Clairaut equation:

${\displaystyle y(x)=x\cdot y'+(y')^{2}.\,\!}$ 

• Chandrasekhar equation
• Chrystal's equation
• Caustic (mathematics)
• Envelope (mathematics)
• Initial value problem
• Picard–Lindelöf theorem

• Rozov, N.Kh. (2001) [1994], "Singular solution", Encyclopedia of Mathematics, EMS Press