# Homework Help: Hyperbolic Equation, Elliptic Equation

1. Sep 22, 2010

### psholtz

1. The problem statement, all variables and given/known data
From the relation:

$$A(x^2+y^2) -2Bxy + C =0$$

derive the differential equation:

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

where $$c^2 = AC(B^2-A^2)$$

3. The attempt at a solution

I'm able to (more or less) do the derivation, but I think the correct relation for the constants is:

$$c^2 = AC/(B^2-A^2)$$

$$u(x,y) = A(x^2+y^2) - 2Bxy + C =0$$

and differentiate:

$$u_x = 2Ax - 2By = 2(Ax-By)$$

$$u_y = 2Ay - 2Bx = 2(Ay-Bx)$$

$$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy = (Ax-By)dx + (Ay-Bx)dy = 0$$

Solving the quadratic expression u(x,y) for x and y respectively, we can write this equation as:

$$\sqrt{(B^2-A^2)y^2 - AC}dx + \sqrt{(B^2- A^2)x^2 - AC}dy=0$$

$$\frac{dx}{\sqrt{(B^2-A^2)x^2-AC}} + \frac{dy}{\sqrt{(B^2-A^2)y^2-AC}} = 0$$

$$\frac{dy}{(B^2-A^2)^{1/2}\sqrt{x^2-\frac{AC}{B^2-A^2}}} + \frac{dy}{(B^2-A^2)^{1/2}\sqrt{y^2-\frac{AC}{B^2-A^2}}} = 0$$

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

and so setting:

$$c^2 = \frac{AC}{B^2-A^2}$$

we can write:

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

My question is: was there a typo in the initial problem statement (i.e., the expression for c^2), or did I make a mistake in my derivation?