# Euler Equation, Arcsine

Suppose we have the differential equation:

$$\frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} = 0$$

It can be rewritten as:

$$\sqrt{1-y^2}dx + \sqrt{1-x^2}dy = 0$$

One solution of this equation (besides arcsin(x) + arcsin(y) = C), is given by:

$$x\sqrt{1-y^2} + y\sqrt{1-x^2} = C$$

However, I'm wondering how this equation can be derived from the equation immediately preceding it (i.e., how the "solution", equation 3 above, can be derived from equation 2)?

It looks like it was simply integrated, term-by-term, that is the "dx" term was integrated to give "x" and the "dy" term was integrated to give "y", but usually this can only be done if the D.E. is exact, and this equation does not appear to be exact.

So how is one justified in arriving at that answer, if the equation is not exact?

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I am sorry, but your 2nd equation:

$$\sqrt{1-y^2}dx + \sqrt{1-x^2}dy = 0$$

cannot be integrated to get the 3rd equation:

$$x\sqrt{1-y^2} + y\sqrt{1-x^2} = C$$

You cannot integrate term by term, from dx to x, since $$\sqrt{1-y^2}$$ depends on x, and $$\sqrt{1-x^2}$$ depends on y.

In fact, if you have equation 3:

$$d\left(x\sqrt{1-y^2}+y \sqrt{1-x^2}\right) = \left(-\frac{x y}{\sqrt{1-x^2}}+\sqrt{1-y^2}\right) dx+\left(\sqrt{1-x^2}-\frac{x y}{\sqrt{1-y^2}}\right) d y$$

Yes, but the third equation is a solution for the second (and/or first), and indeed you've indicated the method for how to get from the second equation to the third, which was my original question: the "integrating" factor is just (much) more complex than I had thought.

$$\frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} =0$$

which is equivalent to:

$$\sqrt{1-y^2} dx + \sqrt{1-x^2} dy = 0$$

Call these equations 1 and 2, respectively.

Now multiply equation 1 by -xy:

$$\frac{-xydx}{\sqrt{1-x^2}} + \frac{-xydy}{\sqrt{1-y^2}} = 0$$

$$\left(\sqrt{1-y^2} - \frac{xy}{\sqrt{1-x^2}}\right)dx + \left(\sqrt{1-x^2} - \frac{xy}{\sqrt{1-y^2}}\right)dy = 0$$

It's easy to show that this equation is, in fact, exact, and indeed the exact differential is given by the expression you provided above, and which indeed is the expression we were trying to derive. You have now:

$$d\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) = 0$$

from which we get:

$$x\sqrt{1-y^2} + y\sqrt{1-x^2} = C$$

which is what I was trying to prove.

Thanks!

It's interesting to note, in passing, that if we take the "two" solutions for the D.E. given above, and calculate their Jacobian, the answer is 0, which is what we would expect if the "two" solutions were actually the same, and just differed in how they were represented.

Differential equation:

$$\frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} = 0$$

Two solutions:

$$u_1 = \sin^{-1}x + \sin^{-1}y$$

$$u_2 = x\sqrt{1-y^2} + y\sqrt{1-x^2}$$

The Jacobian being:

$$J = \left|\array{cc}\frac{\partial u_1}{\partial x} & \frac{\partial u_1}{\partial y} \\ \frac{\partial u_2}{\partial x} & \frac{\partial u_2}{\partial y} \endarray\right|$$

$$J = \left|\array{cc}\frac{1}{\sqrt{1-x^2}} & \frac{1}{\sqrt{1-y^2}} \\ \sqrt{1-y^2} - \frac{xy}{\sqrt{1-x^2}} & \sqrt{1-x^2} - \frac{xy}{\sqrt{1-y^2}} \endarray\right|$$

$$J = 0$$

Hello. psholtz. I thought you just do the integration without considering their dependence. It is interesting that $$x\sqrt{1-y^2} + y\sqrt{1-x^2} = C$$ is a solution. Thank you for your information in the Jacobian as well.

Yes, it is interesting.

It's also interesting to consider the geometrical implications of each solution.

The first solution is as follows:

$$\sin^{-1}x + \sin^{-1}y = c_1$$

or in other words, if we define two angles u and v such that:

$$u = \sin^{-1}x$$

$$v = \sin^{-1}y$$

then this solution becomes:

$$u + v = c_1$$

Or in other words, the differential equation is simply saying that the "sum" of two angles is constant.

Consider now expressing x and y in terms of u and v:

$$x = \sin u$$

$$y = \sin v$$

The second solution then becomes:

$$x\sqrt{1-y^2} + y\sqrt{1-x^2} = c_2$$

$$\sin u \cos v + \sin v \cos u = c_2$$

$$\sin(u+v) = c_2$$

Or in other words, the "second" solution is really saying the same identical thing: that the same of the two angles is constant, it's simply saying it in a different way.

The relation between the two constants of integration is clearly:

$$\sin(c_1) = c_2$$