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I have the following multivariable function:

$u = x + y + F(xy)$

where F is an arbitrary differentiable function. I would like to construct a first order PDE with u(x,y) as the following solution. Initially I thought of taking $u_x$ and $u_y$ which is easy for the $x+y$ term of $u$. However, what can I do about the $F(xy)$ term? I know that it initially turns out as a solution to characteristic equations, but how do I reverse engineer a first order PDE that has $F(xy)$ as a solution? Would it require me to arbitrarily define a function of $F(xy)$ such as $F(xy) = xy$ or $F(xy) = sin(xy)$, etc. then to take arbitrary derivatives of $u_x$ and $u_y$?

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  • $\begingroup$ Do you know anything about $\dfrac{\partial F}{\partial x}$ or $\dfrac{\partial F}{\partial y}$? If not, we can do a little with the chain rule, but not much. $\endgroup$ – Eric Towers Apr 19 '16 at 3:18
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If you know the "method of characteristics", take it backwards :

We know the explicit solution : $\quad u-x-y=F(xy)\quad $ any differenciable function $F$ of one variable.

The implicit solution is : $\quad \Phi(xy\:,\:u-x-y)=0\quad $ any differentiable function $\Phi$ of two variables.

So, the characteristic equations are : $\begin{cases} xy=c_1 \\ u-x-y=c_2 \end{cases}\quad\to\quad \begin{cases} x\:dy+y\:dx=0 \\ du=dx+dy \end{cases}$

From $x\:dy+y\:dx=0$ : $$\frac{dx}{x}=\frac{dy}{-y}=\frac{dx+dy}{x-y}$$ And with $du=dx+dy$ : $$\frac{dx}{x}=\frac{dy}{-y}=\frac{u}{x-y}$$ The corresponding PDE is : $$x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}=x-y$$

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  • $\begingroup$ Ah, going in reverse makes a lot of sense. Thank you very much! $\endgroup$ – Alvin Nunez Apr 21 '16 at 0:10

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