$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z$. How to solve$~z~$?

Question

How to solve $~z~$ from $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z~?$$

I tried to substitude $~x,~y~$ as $~ξ,~η~$, which is $ξ=x+y,~η=x-y$. But it seems useless.

Answer 1

In polar coordinates,

$$x\frac{\partial}{\partial x} + y \frac{\partial}{\partial y} = r\frac{\partial}{\partial r}$$

So the differential equation becomes:

$$r\frac{\partial z}{\partial r} = z$$

Using separation of variables, we get the answer must be of the form

$$z = g(\theta)r \implies z = f\left(\frac{y}{x}\right)\sqrt{x^2+y^2}$$

for an arbitrary function $f$

Answer 2

$$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z\tag1$$ which is a linear partial differential equation of the form $$Pp+Qq=R$$where $$P=x,~~~~Q=y,~~~~R=z,~~~~p=\frac{\partial z}{\partial x},~~~~q=\frac{\partial z}{\partial y}$$ The Lagrange's auxiliary equation for $(1)$ are $$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$$ $$\implies \frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}\tag2$$ Taking the first two equations of $(2)$, we have $$\frac{dx}{x}=\frac{dy}{y}\implies \log x=\log y+\log c\implies \frac{x}{y}=c\qquad \text{where $~c~$ is constant}\tag3$$ Taking the first and third equations of $(2)$, we have $$\frac{dx}{x}=\frac{dz}{z}\implies \log x=\log z-\log d\implies \frac{z}{x}=d\qquad \text{where $~d~$ is constant}\tag4$$ From $(3)$ and $(4)$, the required general solution is $$\frac{z}{x}=\phi\left(\frac{x}{y}\right)\implies z=x~\phi\left(\frac{x}{y}\right)$$ $\phi~$being an arbitrary function.