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Finding a general solution to an equation

  1. Sep 19, 2008 #1
    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:


    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:


    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. jcsd
  3. Sep 19, 2008 #2


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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?
  4. Sep 19, 2008 #3


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    I agree with Defennder :smile:

    Your equation is obviously a hyperbola.

    Is this part of a larger problem, like a differential equation? :confused:
  5. Sep 19, 2008 #4
    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?
  6. Sep 19, 2008 #5
    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'):




    I wish to solve this.

    - I should probably have written this from the beginning.. sorry.
  7. Sep 19, 2008 #6


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    [tex]c^2x^2- (y^2+ 2By )= A[/tex]
    Complete the square: [itex]y^2+ 2By= y^2+ 2By+ B^2- B^2= (y+ B)^2- B^2[/itex]
    [tex]c^2x^2- (y- B)^2= A+ B^2[/tex]
    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,
    [tex]c^2x^2= (y- B)^2+ A+ B^2[/tex]
    [tex]x^2= \frac{(y-B)^2+ A+ B^2}{c^2}[/tex]
    [tex]x= \frac{\pm\sqrt{(y-b)^2+ A+ B^2}}{c}[/tex]
    [tex](y- B)^2= c^2x^2- A- B^2[/tex]
    [tex]y- B= \pm\sqrt{c^2x^2- A- B^2}[/tex]
    [tex]y= B\pm\sqrt{c^2x^2- A- B^2}[/tex]
  8. Sep 19, 2008 #7


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