# Weird differential equation

1. Feb 22, 2013

### Bipolarity

1. The problem statement, all variables and given/known data
Solve the following differential equation, with the initial-value y(2)=2.

$$y' = \sqrt{ \frac{1-y^{2} } { 1-x^{2} } }$$

2. Relevant equations

3. The attempt at a solution

This is a strange ODE. It is continuous when either both |x|<1 and |y|<1 or when both |x|>1 and |y|>1. (Both of the regions guarantee the existence and continuity of the derivative)

$$\frac{y'}{ \sqrt{ 1-y^{2} } } = \frac{1 } { \sqrt{1-x^{2} } }$$

However, when you separate the variables (to solve it), the radical splits, causing the derivative to be defined only when both |x|<1 and |y|<1. The other region i.e. |x|>1 and |y|>1 is lost as a region in which a solution exists. This is strange.

By the existence theorem, that the function y'(x,y) is continuous when both |x|>1 and |y|>1 should guarantee the existence of a solution, yet separation of variables forces the loss of a solution in that region.

What then, is the (explicit) solution for that region, or how might I go about finding it?

I did end up finding the explicit solution to the IVP to be $y = x$ but to do so I required the computation $arcsin(2)-arcsin(2) = 0$. This makes no sense.

BiP

Last edited: Feb 22, 2013
2. Feb 22, 2013

### Staff: Mentor

sqrt(a*b)=sqrt(a)*sqrt(b) is true only for a,b>0, otherwise your operation is invalid in the real numbers. For |x|, |y|>1, you cannot split them like that - unless you use complex numbers.

3. Feb 22, 2013

### Ray Vickson

For |x|,|y|>1 you need
$$\frac{y'}{\sqrt{y^2-1}} = \frac{1}{ \sqrt{x^2-1}}.$$

For the first case you need integrals of the form $\int dt/\sqrt{1-t^2},$ while for the second case you need $\int dt/\sqrt{t^2-1}.$

4. Feb 22, 2013

### Bipolarity

OK thanks, so then how would the explicit solution be obtained when |x|,|y|>1 ???

BiP

5. Feb 22, 2013

### Staff: Mentor

Oh, of course, see Ray Vickson's post how to split the sqrt in this case.