Find general and particular solution for $yp-xq=0$

Question

I am asked to find $2$ things:

1. Find $z$ such that

$$y \dfrac{\partial z}{\partial x} -x \dfrac{\partial z}{\partial y}=0$$

2. Find a particular solution if $z$ is specified along a circle $$x^2 + y^2 = r^2$$ in the $(x,y)-$plane

I tried to start with Lagrange's auxillary equations, and obtained $$\frac{\mathrm{d}x}{y} = \frac{\mathrm{d}y}{x} = \frac{\mathrm{d}z}{0}$$

Then I tried to go according to this method, but I couldn't understand it.

Can you please show me how to start or give a hint to carry on?

Answer 1

$$y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y}=0$$ System of characteristic ODEs : $$\frac{\mathrm{d}x}{y} = \frac{\mathrm{d}y}{-x} = \frac{\mathrm{d}z}{0}$$ Take care that it isn't $\frac{\mathrm{d}y}{x}$ but $\frac{\mathrm{d}y}{-x}$

A first family of characteristic curves comes from $\frac{\mathrm{d}z}{0}=$finite $\implies \mathrm{d}z=0 \implies z=c_1$.

A second family of characteristic curves comes from $\frac{\mathrm{d}x}{y} = -\frac{\mathrm{d}y}{x}\implies x^2+y^2=c_2$.

The general solution of the PDE expressed on the form of implicit equation is : $$\Phi(z\:,\:x^2+y^2)=0$$ where $\Phi$ is an arbitrary function of two variables.

Or equivalently on explicit form : $$z=F(x^2+y^2)$$ where $F$ is an arbitrary function.

The arbitrary function has to be determined according to the boundary condition.

In the wording of the question, it is said that the boundary is a curve which equation is $x^2+y^2=r^2$, that is a circle. So the boundary is defined, but the condition is not given :

Citation : 2. Particular solution if $z$ is specified along a circle $x^2 + y^2 = r^2$ in the $(x,y)-$plane.

What exactly is the specification ?

IN ADDITION :

In comments, the condition is specified as : $z(x,1)=e^x$ with $x>0$.

Note that the condition is not along a circle $x^2+y^2=r^2$ but is along the straight line $y=1$.

Putting this condition into the general solution leads to : $$F(x^2+1)=e^x$$ Let : $\quad X=x^2+1 \quad\to\quad x=\sqrt{X-1}$ $$F(X)=e^{\sqrt{X-1}}$$ So, the function $F$ is determined. We put it into the above general solution where $X=x^2+y^2$. The particular solution fitting the boundary condition is : $$z(x,y)=\exp\left(\sqrt{x^2+y^2-1}\right)$$

Answer 2

From your equation we know that $(y, -x)^T$ is perpendicular to $\nabla z$. This means that it must be in a plane sharing the normal vector of the x,y-plane and that $\nabla z =\alpha (x, y)^T$. Which means the solution is $z = \alpha(x^2 + y^2 - r^2)$ with $\alpha \in \mathbb{R}$