# I ∂^2u/∂x∂y = 0

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1. Mar 27, 2017

### humphreybogart

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. Mar 27, 2017

### ShayanJ

1) $\int f(x) dx \neq xf(x)$
2) $\frac {\partial u}{\partial x}=F(x) \Rightarrow u=\underbrace{\int F(x) dx}_{f(x)}+g(y)\Rightarrow u=f(x)+g(y)$.

3. Mar 27, 2017

### humphreybogart

Excellent. Got it now. Not seeing 1) is my fault. Not showing 2) in the working is the textbook's. :P

4. Mar 27, 2017

### Staff: Mentor

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