general solution

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.


Consider the following Clairaut's equation:

y(x) = x cdot y' + (y')^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' = p + x p' + 2 p p' ,!

which by simple algebra yields

0 = (2 p + x )p'. ,!

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 + 2p = 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: ys(x) = yc(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'(-2 cdot c) = c ,!
y_s'(-2 cdot c) = -(1/2) cdot x |_{x = -2 cdot c} = c. ,!

We see that both requirements are satisfied and therefore ys is tangent to general solution yc. 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' + (y')^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' + f(y'). ,!

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