I'm trying to solve the following PDE:
$$x^2 y z_x - xy^2 z_y = \frac{1}{z}, \quad on\quad x> 0, y > 0,$$ with the initial value problem, $$z(x,1/x) = \sqrt{2 \ln(x)} \quad x>1,$$ and show that it has infinitely many solutions.
I've found the characteristic curves as $y = \frac{C}{x }, $ and $z^2 (x,y) = \ln(x)^{xy} + h(xy),$ and after plugging the initial value problem, I got $$h(1) = \ln(x),$$ which does not make any sense; h(C) = h(xy) should, in theory, only depend on which characteristic curve that we are, but it clearly is not the case, so normally, I would say this PDE does not have any solution to the given initial value problem, but the question claims that it has infinitely many, in fact, so I'm confused.
Question:
Is there any problem in my analysis ? Why does this PDE have even a solution ? What are those infinitely many solutions ?
$$x^2 y z_x - xy^2 z_y = \frac{1}{z}, \quad on\quad x> 0, y > 0,\qquad z(x,1/x) = \sqrt{2 ln(x)}$$ Fist method : Brut force, solving the equation with condition.
Charpit-Lagrange system of characteristic ODEs : $$\frac{dx}{x^2y}=\frac{dy}{-xy^2}=\frac{dz}{1/z}$$ First characteristic equation from $\frac{dx}{x^2y}=\frac{dy}{-xy^2}$ : $$xy=c_1$$ Second characteristic equation from $\frac{dx}{x^2y}=\frac{dz}{1/z} \quad;\quad \frac{1}{x^2\frac{c_1}{x}}dx=zdz \quad;\quad z^2-\frac{2}{c_1}\ln(x)=c_2$ $$z^2-\frac{2}{xy}\ln(x)=c_2$$ General solution of the PDE in the form of implicit equation : $$z^2-\frac{2}{xy}\ln(x)=F(xy)$$ $F$ is any function to be determined according to some specified condition. $$z(x,y)=\pm\sqrt{\frac{2}{xy}\ln(x)+F(xy)}$$ Condition : $z(x,1/x) = \sqrt{2 ln(x)}$
$$2\ln(x)-\frac{2}{x\frac{1}{x}}\ln(x)=F(x\frac{1}{x})=$$ $$F(1)=0$$ They are an infinity of different functions with the property to have $F(1)=0$.
Thus the problem has an infinity of solutions of this form : $$z(x,y)=\pm\sqrt{\frac{2}{xy}\ln(x)+F(xy)}$$ where $F$ is any function such as $F(1)=0$.
Second method : Inspection of the characteristics equations versus the specified condition.
The condition $z(x,1/x) = \sqrt{2 ln(x)}$ fits one of the characteristic curves $z^2-\frac{2}{x\frac{1}{x}}\ln(x)=c_2=0$
It is well known that they are an infinity of solutions in this case.