Solving Simple Partial Differential Equation I can't solve this partial differential equation.
$$x\frac{\partial \phi}{\partial x}+y\frac{\partial \phi}{\partial y}+ (\alpha+1-x)\phi =0$$
The short answer in the book which i read from it ,
after solving the system $$ \frac{dx}{x}=\frac{dy}{y}=\frac{d\phi}{(x-\alpha-1)\phi} $$
is $$ \sigma ( xy^{-1},\frac{e^{x}}{x^{\alpha+1}\phi}) $$
then the author choose the special case 
$$ (xy^{-1})\frac{e^{x}}{x^{\alpha+1}\phi}-1=0 $$
and he obtain finally $$ \phi=x^{-\alpha} y^{-1} e^{x} $$
Please, can anyone help me?
Thanks
 A: First, we search one particular solution easy to find, on the form $\Phi_p(x)$ : $$x\frac{d\Phi_p}{dx}+(\alpha+1-x)\Phi_p(x)=0$$ $$\Phi_p(x)\: e^x x^{-\alpha-1}$$ Then we change of function : $\Phi(x,y)= F(x,y)\Phi_p(x)= F(x,y)e^x x^{-\alpha-1}$
$\frac{\partial \Phi}{\partial x} =\frac{\partial F}{\partial x}e^x x^{-\alpha-1}+F(x,y)e^x x^{-\alpha-1}-(\alpha+1)F(x,y)e^x x^{-\alpha-2}$
$\frac{\partial \Phi}{\partial y}= \frac{\partial F}{\partial y}e^x x^{-\alpha-1}$
We put $\Phi$ , $\frac{\partial \Phi}{\partial y}$ and $\frac{\partial \Phi}{\partial y}$ into : $$x\frac{d\Phi}{dx}+y\frac{d\Phi}{dy}+(\alpha+1-x)\Phi=0$$ and simplify : 
$$\frac{\partial F}{\partial x}e^x x^{-\alpha} +y\frac{\partial F}{\partial y}e^x x^{-\alpha-1} =0$$
$$x\frac{\partial F}{\partial x} +y\frac{\partial F}{\partial y} =0$$
This PDE is easy to solve (change $x=e^X$ and $y=e^Y$) 
$$\frac{\partial F}{\partial X} +\frac{\partial F}{\partial Y} =0$$
which wellknown general solution is  $F(X,Y)=F(X-Y)$ any derivable function $F$.
$$\Phi(x,y)=e^x x^{-\alpha-1}F(\ln|x|-\ln|y|)=e^x x^{-\alpha-1}f \left(\frac{x}{y}\right)$$
any derivable function $f$
A: One of the simple methods is separation of variables which looks for a solution of the form
$$
\phi(x,y) = f(x)\, g(y)
$$
Putting this into the PDE gives
$$
x\, f'(x)\, g(y) + y\, f(x)\, g'(y) + (\alpha + 1 - x)\, f(x)\, g(y) = 0 
$$
Next is assuming $f(x) \ne 0$ and $g(y) \ne 0$ and for these $(x,y)$ sorting the dependencies on $x$ to the left and those on $y$ to the right:
$$
x \frac{f'(x)}{f(x)} + (\alpha + 1 - x) 
= -y \frac{g'(y)}{g(y)}
= C = \mbox{const.} 
$$
This gives two ODEs
$$
f'(x) + \frac{\alpha + 1 - x - C}{x} f(x) = 0  \\
g'(y) + \frac{C}{y} g(y) = 0
$$
which have the solutions
$$
f(x) = q\, e^x\, x^{-(\alpha+1)+C} \\
g(y) = r\, y^{-C}
$$
for some integration constants $q$ and $r$.
Update: Found a mistake (forgot to feed the solver with $C$ for the first ODE). Better check the resulting solution $\phi(x,y) = f(x)\, g(y)$ to be sure.
Update: JJacquelin spotted another problem with the ODE for $f$, oh my.  
