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A 'singular solution' ''y_s''(''x'') of an ordinary differential equation is a solution that is tangent to every solution from the family of general solutions. By ''tangent'' we mean that there is a point ''x'' where ''y_s''(''x'') = ''y_c''(''x'') and ''y'_s''(''x'') = ''y'_c''(''x'') where ''y_c'' is any general solution.
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.

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Example

Example

Consider the following Clairaut's equation:
:$y(x)\; =\; x\; cdot\; y\text{'}\; +\; (y\text{'})^2\; ,!$
where primes denote derivatives with respect to ''x''. We write ''y' = p'' and then
:$y(x)\; =\; x\; cdot\; p\; +\; (p)^2\; ,!$
Now, we shall take the differential according to ''x'':
:$p\; =\; y\text{'}\; =\; p\; +\; x\; p\text{'}\; +\; 2\; p\; p\text{'}\; ,!$
which by simple algebra yields
:$0\; =\; (\; 2\; p\; +\; x\; )p\text{'}\; ,!$
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:
:$y\_c(x)\; =\; c\; cdot\; x\; +\; c^2\; ,!$
where ''c'' is determined by the initial value.
If ''x'' + 2''p'' = 0 than we get that ''p'' = −(1/2)''x'' and substituting in the ODE gives
:$y\_s(x)\; =\; -(1/2)x^2\; +\; (-(1/2)x)^2\; =\; -(1/4)\; cdot\; x^2\; ,!$
Now we shall check whether this is a singular solution.
First condition of tangency: ''y_s''(''x'') = ''y_c''(''x''). We solve
:$c\; cdot\; x\; +\; c^2\; =\; y\_c(x)\; =\; y\_s(x)\; =\; -(1/4)\; cdot\; x^2\; ,!$
to find the intersection point, which is ($-2c\; ,\; -c^2$).
Second condition tangency: ''y'_s''(''x'') = ''y'_c''(''x'').
We calculate the derivatives:
:$y\_c\text{'}(-2\; cdot\; c)\; =\; c\; ,!$
:$y\_s\text{'}(-2\; cdot\; c)\; =\; -(1/2)\; cdot\; x\; |\_\{x\; =\; -2\; cdot\; c\}\; =\; c\; ,!$
We see that both requirements are satisfied and therefore ''y_s'' is tangent to general solution ''y_c''. Hence,
:$y\_s(x)\; =\; -(1/4)\; cdot\; x^2\; ,!$
is a singular solution for the family of general solutions
:$y\_c(x)\; =\; c\; cdot\; x\; +\; c^2\; ,!$
of this Clairaut equation:
:$y(x)\; =\; x\; cdot\; y\text{'}\; +\; (y\text{'})^2\; ,!$
'Note:' The method shown here can be used as general algorithm to solve any Clairaut's equation, i.e. first order ODE of the form
:$y(x)\; =\; x\; cdot\; y\text{'}\; +\; f(y\text{'}).\; ,!$
See also caustic (mathematics).