I am trying to find the general solution for the following PDE $$ x {\partial\phi(x,y)\over\partial x} + c y {\partial\phi(x,y)\over\partial y} = - \phi(x,y)$$ where $c = \frac{(\alpha - 1)}{\alpha}$ and $\alpha \in (0,1)$.
The characteritic equations are $$ \frac{d x}{x} = \frac{d y}{cy} = \frac{d \phi}{-1 }. $$ As per my understanding, We now want to find two functions $\Phi(x, y, \phi)$, $\Psi(x, y, \phi)$ ) such that $d \Phi = d \Psi = 0$. The general solution is then given by $F(\Phi, \Psi) = 0$ with $F$ an arbitrary function. Using $$ \frac{d x}{-x} = d \phi \implies d(\phi + log (x)) = 0 $$ and $$ {dx\over dy} = {x\over c y} \implies c y dx - x dy $$ we obtain $$ \Phi = \phi + log (x), $$ but i am not sure how to pick $\Psi$ in this case it appears $x \approx y^c$.
Could i instead work with the ode $$ \frac{d y}{cy} = \frac{d \phi}{-1 } \implies d(c \phi + \log (y)) = 0. $$
The general solution is then given by $$ F(\phi + log (x), c \phi + \log (y)) = 0 $$ but since both terms depend on $\phi$ i am not sure if i can get an expression for $\phi$.
Is there a way to get an general solution expression for $\phi$ ? is there a way to pick a better $\Phi$ ?
Completion of your solution: $$ \frac{d x}{-x} = d \phi \implies d(\phi + log (x)) = 0\Longrightarrow \phi + \log (x)=c_1 \qquad(1) $$ and $$ {dx\over dy} = {x\over c y} \implies c y dx - x dy =0\Rightarrow x-y^c=c_2\qquad (2) $$ Thus the general solution can be written as $c_1=f(c_2)$, or equivalently $$\phi =-\log(x)+f(x-y^c), $$ where $f$ is any differentiable function.