PDE to obtain function $(x,y)\mapsto C x^{\alpha} y^{1-\alpha}$ If I want to calculate the function $C\cdot x^{\alpha}$ I have to solve the differential equation $$\frac{dy}{dx}\cdot \frac{x}{y}=\alpha$$
$\frac1y \, dy=\alpha\cdot \frac1x \,dx $
$\int \frac1y \, dy=\int \alpha\cdot \frac1x \,dx $
$\int \frac1y \, dy=\alpha\cdot\int  \frac1x \,dx $
$\ln(y)=\alpha\cdot ln(x)+c$
$e^{\ln(y)}=e^{\alpha\cdot ln(x)+c}$
$y=e^{\alpha\cdot ln(x)}\cdot e^{c}$
$y=\left(e^{ ln(x)}\right)^{\alpha}\cdot e^{c}$
$y=x^{\alpha}\cdot e^{c}$
$y=C\cdot x^{\alpha}$
So far so good. But what kind of PDE (system) and conditions I have to start with to obtain $z(x,y)=C\cdot x^{\alpha}\cdot y^{1-\alpha}$?  The PDE I think I have to use is $$\frac{\partial z}{\partial x}\cdot \frac{x}{z}+\frac{\partial z}{\partial y}\cdot \frac{y}{z}=1$$
And I have the two conditions  $z(0,y)=0$ and  $z(x,0)=0$. My two questions are:


*

*Are these conditions sufficient to obtain $z(x,y)$?

*What method I have to apply to solve the PDE?
Thanks to all who takes time to read the question.
 A: For the function $f: x\mapsto x^\alpha$, we compute the derivative
$$
f'(x) = \alpha x^{\alpha-1} = \alpha\frac{f(x)}{x}\, ,
$$
which gives a differential equation.
For the separable function $u: (x,y)\mapsto x^\alpha y^{1-\alpha}$, we compute the partial derivatives
$u_x(x,y) = \alpha u(x,y)/x$
and
$u_y(x,y) = (1-\alpha) u(x,y)/y$.
Therefore,
$$
\frac{x}{\alpha} u_x - \frac{y}{1-\alpha} u_y = 0\, .
$$
This PDE may be solved by the method of characteristics, which gives $u(x,y) = F\big(y x^{\alpha/(1-\alpha)}\big)$ for arbitrary $F$. The "arbitrary function $F$" is deduced from the boundary conditions. Here, $F: t\mapsto t^{1-\alpha}$ would work. For instance, this is obtained by imposing $u(1,y) = y^{1-\alpha}$.
Alternatively, one can write the PDE
$$
x u_x + y u_y = u\, ,
$$
which solutions obtained by the method of characteristics are $u(x,y) = x F(x/y)$ for arbitrary $F$. Here, imposing e.g. $u(x,1)=x^{\alpha}$ would work.

Note: To get a linear first-order PDE, one could consider any linear combination of $u_x$ and $u_y$: $$a u_x+b u_y=\left(\alpha \frac{a}{x} + (1-\alpha)\frac{b}{y}\right) u \, ,$$ where the coefficients $a, b \neq 0$ are functions of $(x,y)$. The first PDE corresponds to $a = x/\alpha$ and $b = -y/(1-\alpha)$, while the second PDE corresponds to $a = x$ and $b = y$.
