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Partial Differential Equations Help!

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that u=f(2x+y^2)+g(2x-y^2) satisfies the equation y^2 d^2u/dx^2 + (1/y) du/dy - d^2u/dy^2=0 where f and g are arbitrary (twice differentiable) functions.



    2. Relevant equations



    3. The attempt at a solution
    I came up with fxx=0 fyy=2 gxx=0 gyy= 2. But didn't know if this was correct and if so what to do next. I do not want the solution but how to go about solving this problem. Thanks!
     
  2. jcsd
  3. Feb 1, 2009 #2

    Dick

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    You want to find things like uxx, uy and uyy to substitute into your equation. What are those? Oh, and fxx is not 0. E.g. fx=f'(2x+y^2)*2. Use the chain rule.
     
    Last edited: Feb 1, 2009
  4. Feb 1, 2009 #3
    I am still confused, I do not see where the chain rule comes into play, f(x,y) = 2x + y2. So why isn't fx equal to 2? I do not see how you are coming up with (2x + y2)*2. I desperately need to understand this whole concept! Thanks!
     
  5. Feb 1, 2009 #4

    Dick

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    Chain rule. f(x,y) isn't equal to 2x+y^2. It's equal to f(2x+y^2). The x derivative is f'(2x+y^2)*2.
     
  6. Feb 1, 2009 #5
    Sorry to be a pain but I think I am finally getting the idea and want to make sure this is true. ux = f'(2x+y2)*2 + g'(2x-y2)*2
    uy = f'(2x+y2)*2y + g'(2x-y2)*-2y
    uxx = f''(2x+y2)*0 + g''(2x-y2)*0
    uyy = f''(2x+y2)*2 + g''(2x-y2)*-2

    then I just plug these results into the equation y2uxx + (1/y) uy - uyy = 0. Is that correct?

    Thanks!
     
  7. Feb 2, 2009 #6

    Dick

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    ux and uy are ok. Not uxx and uyy. fxx=(f'(2x+y^2)*2)'=2*(f'(2x+y^2))'=4*f''(2x+y^2). The '2' is a multiplicative constant. You need to use the product rule on 2y*f'(2x+y^2), since it's the product of two functions of y. Yes, then just plug them into your equation.
     
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