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# Singular solution

A singular solution ys(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 ys(x) = yc(x) and y's(x) = y'c(x) where yc 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:

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

where primes denote derivatives with respect to x. We write y' = p and then

$y\left(x\right) = x cdot p + \left(p\right)^2. ,!$

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

$0 = \left(2 p + x \right)p\text{'}. ,!$

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:

$y_c\left(x\right) = c cdot x + c^2 ,!$

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

$y_s\left(x\right) = -\left(1/2\right)x^2 + \left(-\left(1/2\right)x\right)^2 = -\left(1/4\right) cdot x^2. ,!$

Now we shall check whether this is a singular solution.

First condition of tangency: ys(x) = yc(x). We solve

$c cdot x + c^2 = y_c\left(x\right) = y_s\left(x\right) = -\left(1/4\right) cdot x^2 ,!$

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

Second condition tangency: y's(x) = y'c(x).

We calculate the derivatives:

$y_c\text{'}\left(-2 cdot c\right) = c ,!$
$y_s\text{'}\left(-2 cdot c\right) = -\left(1/2\right) cdot x |_\left\{x = -2 cdot c\right\} = c. ,!$

We see that both requirements are satisfied and therefore ys is tangent to general solution yc. Hence,

$y_s\left(x\right) = -\left(1/4\right) cdot x^2 ,!$

is a singular solution for the family of general solutions

$y_c\left(x\right) = c cdot x + c^2 ,!$

of this Clairaut equation:

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

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

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