Solving Simple Partial Differential Equation

Question

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

Answer 1

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 $f$

Answer 2

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.