Usually, singular solutions appear in differential equations when there is a need to divide in a term that might be equal to zero. Therefore, when one is solving a differential equation and using division one must check what happens if the term is equal to zero, and whether it leads to a singular solution.
where primes denote derivatives with respect to x. We write y' = p and then
Now, we shall take the differential according to x:
which by simple algebra yields
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 is:
where c is determined by the initial value.
If x + 2p = 0 than we get that p = −(1/2)x and substituting in the ODE gives
Now we shall check whether this is a singular solution.
First condition of tangency: ys(x) = yc(x). We solve
to find the intersection point, which is ().
Second condition tangency: y's(x) = y'c(x).
We calculate the derivatives:
We see that both requirements are satisfied and therefore ys is tangent to general solution yc. Hence,
is a singular solution for the family of general solutions
of this Clairaut equation: