# Complex algebra help

1. Apr 16, 2007

### rsnd

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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

2. Apr 16, 2007

### 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 $\psi(x)$?

3. Apr 17, 2007

### HallsofIvy

Staff Emeritus
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!)