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
Best Answer
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 $f$