I am stuck on the following problem: Solve for $f(x,y)$, where:
$\frac{\partial f}{\partial y} = y$,
$\frac{\partial f}{\partial x} = \frac{1}{2}xy$
My original strategy was to integrate the first equation giving:
$\frac{y^2}{2} + g(x)$, where $g(x)$ is a constant function, and then differentiating this and setting it equal to the second equation giving:
$g'(x) = \frac{1}{2}xy$, which I integrate again to give:
$g(x) = \frac{1}{4}x^2y + h(y)$, giving:
$f(x,y) = \frac{y^2}{2} + \frac{1}{4}x^2y + h(y)$,
but, where do you go from here? Any suggestions would be appreciated!
Thanks!
Not in any strong sense, suppose your function $f$ is twice times continuously differentiable, you can use Schwarz's theorem, i.e. \begin{equation} \frac{\partial^{2}f(x,y)}{\partial x\partial y}\equiv \frac{\partial^{2}f(x,y)}{\partial y\partial x} \end{equation} the partial derivatives of second order commute, and would end up with the condition \begin{equation}\frac{1}{2}x\equiv 0\end{equation} which is impossible. Notice, that your question is strongly related to the question if there exists a potential with gradient $(\tfrac{1}{2}xy,y)$ and then the condition I gave would be called the integrability condition...