Singular solution

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.

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

:$y(x)\; =\; x\; cdot\; y\text{'}\; +\; f(y\text{'}).\; ,!$

ee also

* caustic (mathematics).