Here is what I did.
I think the method is very lame, but the idea should be this way. Could any one correct it for me please ?
1. Because of the regularity conditions on f (that is, it is a continuously differentiable function), you can use Clairaut's Theorem to equate the mixed partial derivatives.
2. Your g(y) and h(x) are not unique. Almost any ol' functions will do there! But this does not hurt your final result, since you're not asked to find unique functions.
Other than these comments, it looks fine to me.
1. A previous problem in my book said:
Let f: R^n --> R , differentiable on the open ball centering at a = (a1, ...,an) and radius r and f'_n = 0 Then there is a UNIQUE g: R^n-1 centering at (a1,...,an-1) and radius r such that f(x1,...,xn) = g(x1,...,xn-1).
2. I am mostly concerned with the bottom line beginning with
The integrals? Is it correct that Df(x,y) = f(x,y) ?
Is it correct to just put in front of h(x)dx ??
I think integral must come with
3. Clairaut's Theorem ??? Never heard it before. Is it the same as Young Theorem ?
Thanks a lot
How could you explain this step? I cannot see the reason why came from.
and, is there a way without using partial integration ? The book I am studying does not mention it. (I try doing problems without resorting to other definitions outside the book.)
Thank you very much.
Not sure there's a way to do this problem without partial integration. In any case, when you partially integrate w.r.t. one independent variable, the "constant" of integration must be allowed to be a function of the other independent variables. You can verify this by taking the partial derivative w.r.t. the variable of partial integration. The "constant" of integration gets killed by the partial differentiation. Remember finding harmonic functions? Or solving exact ODE's? It's the same idea here.