Complex Algebra Help: Finding u(x,y) & Proving Result

[rsnd]

Homework Statement

I am told: {\frac {{\it du}}{{\it dx}}}=y and {\frac {{\it du}}{{\it dy}}}=x. Need to find u(x,y) which is a real valued function and prove the result.

Homework Equations

The Attempt at a Solution

Well, I think the answer is of the form u(x,y) = xy + c because the answer makes sense but how should I go about proving it?

Thanks

 

[quasar987]

By integrating and reasoning logically.

\frac{\partial u}{\partial x}=y

Therefor, integrating wrt x:

\int\frac{\partial u}{\partial x}dx=\int ydx

u(x,y)= yx+\phi(y)

phi(y) is kind of like the constant of integration, but we reckon that in the most general case, it may actually be a function of y. Make sure you understand why.

We can do the same with the other equation:

\frac{\partial u}{\partial y}=x

u(x,y)=xy+\psi(x)

Now compare the two equations. What do you conclude about the forms of \phi(y) and \psi(x)?

 

[HallsofIvy]

I would have done this slightly differently.
You are given that
\frac{\partial u}{\partial x}= y
so u(x,y)= xy+ \phi (y)
as quasar987 said.

Now differentiate that with respect to y:
\frac{\partial u}{\partial y}= x+ \frac{d\phi}{dy}
and that must be equal to x. What does that tell you about
\frac{d\phi}{dy}?

(I said slightly differently!)