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I ∂^2u/∂x∂y = 0

  1. Mar 27, 2017 #1
    I am working my way through a textbook, and whenever this equation is solved (integrated), the answer is given as:

    u = f(x) + f(y)

    I don't understand it. If I integrate it once (with respect to y, say), then I obtain:

    ∂u/∂x = f(x) -----eq.1

    If I integrate again (this time with respect to x), then I obtain:

    u = xf(x) + f(y)

    I know that this can't be correct because the mixed derivative theorem says that if I went the other way (integrating with respect to x and then y), I should get the same answer. But I can't see how integrating eq.1 doesn't produce and 'x' infront of the arbitrary function.
  2. jcsd
  3. Mar 27, 2017 #2


    User Avatar
    Gold Member

    1) ## \int f(x) dx \neq xf(x) ##
    2) ## \frac {\partial u}{\partial x}=F(x) \Rightarrow u=\underbrace{\int F(x) dx}_{f(x)}+g(y)\Rightarrow u=f(x)+g(y) ##.
  4. Mar 27, 2017 #3
    Excellent. Got it now. Not seeing 1) is my fault. Not showing 2) in the working is the textbook's. :P
  5. Mar 27, 2017 #4


    Staff: Mentor

    Wouldn't it be u = f(x) + g(y)? It wouldn't be the same function for both.
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