I read somewhere that a non-linear differential equation can have one or more special solutions which are not obtainable by standard methods like separation of variables. For example, if y'=sqrt(1-y² ) then y=±1 are special solutions.
Is there any method to check whether a given non-linear equation has such special soutions? Specifically, can it be proved that there are no special solutions to y'=sqrt(1+y² )?

David Loeffler

Perhaps the problem with separating the variables in the former case is that to get it in the form:

dy/sqrt(1-y² ) = dx

then you are potentially dividing through by 0. Specifically if sqrt(1-y² ) = 0 and y = ± 1 then there are problems.

It is one of those interesting cases when you can know the differential equation, set its initial condition at x = 0, and yet not necessarily be able to know what y is for any x. Specifically I think:

If dy/dx = sqrt(1-y² ), and when x = 0 then y = -1.

y = -1 for all x is clearly one solution
y = -cos x for all x is fine too

Also

y = -1 for (0 < = x < X)
y = -cos(x-X) for x > = X

is also perfectly good (for any positive X).

What's more it can start off doing the cos curve, keep going for an indeterminate number of cycles and then suddenly (upon reaching ±1) stay there for a while. It really just behaves totally unpredictably whenever y² is 1.

For the sqrt(1+y² ) case I think there will be similar problems if and when y = ±i, but no problems if you keep y real.

Another example (the one I found when working out a differential equation modelling a particle rolling down a hill) is:

[d² y/dx² ]² = 144y

Obviously y = 0 (for all x) is good. But so is y = x⁴ . And these both have the same initial conditions - both y and dy/dx are 0 at x = 0.

Michael