Finding a general solution to an equation

1. Sep 19, 2008

Niles

1. The problem statement, all variables and given/known data
Hi all.

I have the following equation containing the variables x and y, where A, c > 0 and B is a constant, which I am not told anything about:

$$-c^2x^2+y^2+2By+A=0.$$

I wish to find two solutions {x_1, y_1} and {x_2, y_2}, to this problem and they must be linearly independant. I thought of doing something like this:

$$(-c^2x^2)+(y^2+2By+A)=0,$$

where we now have two second-order polynomials.

My question is: Is there any way to find a general solution to this problem as a function of either x or y?

2. Sep 19, 2008

Defennder

I'm not following you here. You have an equation in terms of x and y. The solutions of that equation are the collection of (x,y) points which satisfy that equation. So what does it mean for a collection of points to be "linearly independent"? Linear independence should apply only to vectors or functions, don't they?

3. Sep 19, 2008

tiny-tim

I agree with Defennder

Your equation is obviously a hyperbola.

Is this part of a larger problem, like a differential equation?

4. Sep 19, 2008

Niles

But I need to distinct solution - so if one solution is a multple of the other, then I wouldn't be able to use it.

Is there a way to find a general solution to this problem?

5. Sep 19, 2008

Niles

Ok, here's the story:

I wish to find two solutions to the telegraph equation (a differential equation, second order, linear and homogeneous). I know that it is on the form of u(x,y) = e^(ax) * e^(by), where the 'a' and 'b' in this equation corresponds to the 'x' and 'y' I wrote in my first post. The 'a' and 'b' can by found using the above equation (inserting 'a' and 'b'):

$$-c^2a^2+y^2+2Bb+A=0.$$

I wish to solve this.

- I should probably have written this from the beginning.. sorry.

6. Sep 19, 2008

HallsofIvy

$$c^2x^2- (y^2+ 2By )= A$$
Complete the square: $y^2+ 2By= y^2+ 2By+ B^2- B^2= (y+ B)^2- B^2$
so
$$c^2x^2- (y- B)^2= A+ B^2$$
That is, as tiny-tim said, a hyperbola. I don't understand what you want as a "general solution". That is the general solution: choose any value of y and solve for the corresponding value of x or vice versa.

For example,
$$c^2x^2= (y- B)^2+ A+ B^2$$
$$x^2= \frac{(y-B)^2+ A+ B^2}{c^2}$$
$$x= \frac{\pm\sqrt{(y-b)^2+ A+ B^2}}{c}$$
or
$$(y- B)^2= c^2x^2- A- B^2$$
$$y- B= \pm\sqrt{c^2x^2- A- B^2}$$
$$y= B\pm\sqrt{c^2x^2- A- B^2}$$

7. Sep 19, 2008

gabbagabbahey

Can you post the original differential equation, along with any initial conditions and additional information you are given?