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Math Help - Non homogeneous PDE

  1. #1
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    Non homogeneous PDE

    d^2 f/dx^2 - d^f/dy^2 = x

    This is a non homogeneous PDE. I first tried to find the solution to the homogeneous PDE associated with it.

    d^2 f/dx^2 - d^f/dy^2 = 0

    I equated d^2 f/dx^2 = d^f/dy^2 = k (a constant)

    Now integrated both the sides to get f = k [(x^2)/2 + (y^2)/2]

    Is this correct?
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  2. #2
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    Quote Originally Posted by kushagra452 View Post
    d^2 f/dx^2 - d^f/dy^2 = x

    This is a non homogeneous PDE. I first tried to find the solution to the homogeneous PDE associated with it.

    d^2 f/dx^2 - d^f/dy^2 = 0

    I equated d^2 f/dx^2 = d^f/dy^2 = k (a constant)

    Now integrated both the sides to get f = k [(x^2)/2 + (y^2)/2]

    Is this correct?
    The solution will consist of two parts - a particular solution and the general solution of the homeogenous problem (which is f_{xx} - f_{yy} = 0). For the particular solution, any one will work. Say f_p = \frac{1}{6}x^3. The homogenous equation

    f_{xx} - f_{yy} = 0 is the standard wave equation that has the general solution

    f = F(x+y) + G(x-y) where F and G are arbitrary.

    Thus, the general solution is f = F(x+y) + G(x-y) + \frac{1}{6}x^3 .

    As for your solution is doesn't satisfy the homogenous equation.
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  3. #3
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    Thanks Danny for the reply.

    How did you solve the homogeneous equation f(xx)-f(yy) = 0? I could not figure out how to use the change of variable. x = u+v , y= u-v.

    Thanks again!

    Cheers!
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  4. #4
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    Quote Originally Posted by kushagra452 View Post
    Thanks Danny for the reply.

    How did you solve the homogeneous equation f(xx)-f(yy) = 0? I could not figure out how to use the change of variable. x = u+v , y= u-v.

    Thanks again!

    Cheers!
    Yes, that's it! If you use x = u+v, y = u-v, then

     <br />
f_{xx}-f_{yy} = f_{uv} = 0,<br />

    which you can integrate twice.
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