# Singular solution

A singular solution of a differential equation is a solution that satisfies the following conditions:

1. It solves the original differential equation.
2. It is tangent to every solution from the family of general solutions of the ODE. 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:

[itex] y(x) = x \cdot y' + (y')^2 \,\![itex]

We write y' = p and then

[itex] y(x) = x \cdot p + (p)^2 \,\![itex]

Now, we shall take the differential according to x:

[itex] p dx = dy = p ( dx ) + x ( dp ) + 2 p ( dp ) \,\![itex]

which by simple algebra yields

[itex] 0 = ( 2 p + x )dp \,\![itex]

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

If dp=0 it means that y' = p = c = Const and the general solution is:

[itex] y_c(x) = c \cdot x + c^2 \,\![itex]

where c is determined by the initial value.

If x + 2p = 0 than we get that p = -(1/2)x and subsituting in the ODE gives

[itex] y_s(x) = -(1/2)x^2 + (-(1/2)x)^2 = -(1/4) \cdot x^2 \,\![itex]

Now we shall check whether this a singular solution.

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

[itex] c \cdot x + c^2 = y_c(x) = y_s(x) = -(1/4) \cdot x^2 \,\![itex]

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

Second condition tangency: y's(x) = y'c(x) .
We calculate the derivatives:

[itex] y_c'(-2 \cdot c) = c \,\![itex]
[itex] y_s'(-2 \cdot c) = -(1/2) \cdot x |_{x = -2 \cdot c} = c \,\![itex]

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

[itex] y_s(x) = -(1/4) \cdot x^2 \,\![itex]

is a singular solution for the family of general solutions

[itex] y_c(x) = c \cdot x + c^2 \,\![itex]

of this Clairaut equation:

[itex] y(x) = x \cdot y' + (y')^2 \,\![itex]

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

[itex] y(x) = x \cdot y' + f(y'). \,\![itex]

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