Singular y=-x for x+y^2y' = 0

[psholtz]

Just so I have the concept of a singular solution down correctly, suppose I have an equation like:

$$\left(x+y\right)^2y' = 0$$

This admits of two solutions:

$$y=-x$$

and, from:

$$y' = 0$$

$$y = C$$

where C is a constant.

So the "two" solutions for the equation would be:

$$y_1=-x, y_2 = C$$

In this case, y=-x would be considered the "singular" solution, correct?

 

[Mark44]

Take a look here: http://en.wikipedia.org/wiki/Singular_solution, especially in the Failure of uniqueness and Further example of failure of uniqueness sections.

 

[psholtz]

Yes, I'm familiar w/ Wikipedia.

If there's a specific example from Wikipedia that you'd like to discuss in more depth, please do so.

Getting back to my example:

$$(x+y)^2y' = 0$$

The equation is solved by "two" equations:

$$y_1 = -x$$

$$y_2 = C$$

Suppose we set initial conditions to something very simple, like:

$$y(0) = 0$$

Then there would be "two" curves (or lines, for precisely), which would satisfy the equation and initial conditions. That is, the line y=-x passes through the origin, as does the line y=0, both of which are solutions to the differential equation, and both of which satisfy the initial conditions.

So it would appear that failure of uniqueness is something that happens here.

My question is just, do we say that the whole system is singular, or just the y=x part, or what?

 

[HallsofIvy]

The system is singular. You cannot say that one solution or the other is a "singular solution".

 

[psholtz]

In well-known case of the Clairaut equation, the singular solution forms an "envelope" or "boundary" around the family of general solutions to the D.E.

Is this always the case (i.e., do singular solutions always form a boundary around the set of general solutions), or does this only happen to occur in the case of the Clairaut equation?