Given the partial differential equation \begin{equation} (1-x)\left[- f(x,y) + \frac{\partial f(x,y)}{\partial x} \right] + (1-y)\left[- f(x,y) + \frac{\partial f(x,y)}{\partial y} \right] = 0 \end{equation} in the interval $0 \leq x < 1$, $0 \leq y < 1$, the following solution can be obtained by separation of variables: \begin{equation} f(x,y) = C \, \exp(x+y) \end{equation}
Is this the general solution?
Thank you.
With the method of characteristics, I obtain the following solution. The characteristic equations are \begin{eqnarray} \frac{d x(s)}{ds} & = & 1-x(s),\\ \frac{d y(s)}{ds} & = & 1-y(s),\\ \frac{d F(x(s),y(s))}{ds} & = & [2-x(s)-y(s)]F(x(s),y(s)). \end{eqnarray} The solution of the third characteristic equation is \begin{equation} F(x(s),y(s)) = F(x(0),y(0)) \, \exp[x(s)-x(0)+y(s)-y(0)]. \end{equation} The solution of the first two characteristic equations is \begin{eqnarray} \frac{1-x(s)}{1-x(0)} &=& \exp(-s),\\ \frac{1-y(s)}{1-y(0)} &=& \exp(-s). \end{eqnarray} Setting $y(0)=0$, the solution of the first two characteristic equations implies that $x(0)$ is a function of $\frac{1-x(s)}{1-y(s)}$: substituting this into the expression above for $F(x(s),y(s))$, we have \begin{equation} F(x,y) = G\left( \frac{1-x}{1-y} \right) \exp(x+y). \end{equation}
This is the 'more general' solution proposed by Robert Israel. My question is: Is this the general solution?
A more general solution is
$$ f(x,y) = G\left( \dfrac{1-y}{1-x}\right) \exp(x+y) $$
for arbitrary differentiable function $G$.
EDIT: To see that this is the general solution, write $f(x,y) = g(x,y) \exp(x+y)$. The equation becomes
$$ (1-x) \dfrac{\partial g}{\partial x} + (1-y) \dfrac{\partial g}{\partial y} = 0$$
With a change of variables $s = (1-y)/(1-x), t = x$ (i.e. $y = st-s+1, x=t$), the equation becomes $$ (1-t) \dfrac{\partial g}{\partial t} = 0 $$ Thus for $x < 1$, $g$ must be a function of $s = (1-y)/(1-x)$ alone.
