Finding a general solution to an equation

AI Thread Summary
The discussion centers on finding two linearly independent solutions to the equation -c^2x^2 + y^2 + 2By + A = 0, which represents a hyperbola. Participants clarify that linear independence applies to vectors or functions, not directly to points on a curve. A general solution can be derived by completing the square for y, leading to expressions for x in terms of y and vice versa. The user is also seeking to relate this equation to the telegraph equation, a second-order linear differential equation, and requests more information about the original differential equation and any initial conditions. The conversation emphasizes the need for clarity in defining the problem and the solutions sought.
Niles
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Homework Statement


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:

<br /> -c^2x^2+y^2+2By+A=0.<br />

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:

<br /> (-c^2x^2)+(y^2+2By+A)=0,<br />

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?
 
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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?
 
I agree with Defennder :smile:

Your equation is obviously a hyperbola.

Is this part of a larger problem, like a differential equation? :confused:
 
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?
 
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'):

<br /> <br /> -c^2a^2+y^2+2Bb+A=0.<br /> <br />
I wish to solve this.

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

Homework Statement


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:

<br /> -c^2x^2+y^2+2By+A=0.<br />

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:

<br /> (-c^2x^2)+(y^2+2By+A)=0,<br />

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?

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}
 
Niles said:
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'):

<br /> <br /> -c^2a^2+y^2+2Bb+A=0.<br /> <br />



I wish to solve this.

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

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