Solve the PDE: $(xy-x^2)\frac{\partial u}{\partial x} + y^2 \frac{\partial u}{\partial y} + (e^{y/x}+yz)\frac{\partial u}{\partial z} = 0$ As stated in the title, I want to solve the following PDE:
$(xy-x^2)\frac{\partial u}{\partial x} + y^2 \frac{\partial u}{\partial y} + (e^{y/x}+yz)\frac{\partial u}{\partial z} = 0,$
with the condition that $u = \frac{y}{1+y}$ given $z=e^{y/x}$.
Finding two independent integrals $\psi_{1}$ and $\psi_{2}$ of this equation would give me a general solution $u=\Phi(\psi_{1}, \psi_{2})$, $\Phi \in C^{1}$.
The corresponding symmetrical system here is: $$ \frac{dx}{xy-x^2} = \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} = M. $$
By using $y^2dx = (xy-x^2)dy$, I was able to obtain the first integral of this equation as $$\psi_{1} = \frac{y}{x} + \ln{y}.$$
However, finding the second one eludes me. I've tried using $d(e^{y/x}) = -\frac{y}{x^2}e^{y/x}dx + \frac{1}{x}e^{y/x}dy$, to, for example, say $$\frac{-\frac{y}{x^2}e^{y/x}dx + (\frac{1}{x}e^{y/x}+z)dy-ydz}{-\frac{y^2}{x}e^{y/x} + ye^{y/x} +\frac{y^2}{x}e^{y/x} - ye^{y/x} -y^2z + y^2z} = M,$$ therefore $-\frac{y}{x^2}e^{y/x}dx + (\frac{1}{x}e^{y/x}+z)dy-ydz=0$, but I can't integrate this $1$-form to get the second integral.
Is there a good way to find the second integral in this equation?
 A: $$ \frac{dx}{xy-x^2} = \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} \quad \text{is OK.} $$
There is a sign mistake in your first integral.  $\psi_{1} = \frac{y}{x} + \ln|y|$ is false and should be :
$$\psi_{1} = \frac{y}{x} -\ln|y|$$
Or, equivalently with $c_1=e^{\psi_1}$ :
$$c_1=\frac{1}{y}e^{y/x}$$
$ \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} = \frac{dz}{c_1y + yz}$
$$ \frac{dy}{y} = \frac{dz}{c_1 + z}$$
Integrating leads to $\frac{c_1 + z}{y}=c_2$
$$c_2=\frac{\frac{1}{y}e^{y/x} + z}{y}=\frac{1}{y^2}e^{y/x}+\frac{z}{y}$$
The general solution of the PDE is $\qquad u=\Phi(c_1,c_2)$.
Boundary condition: $\quad u=\frac{y}{1+y}\quad$ on $\quad z=e^{y/x}$
$c_1=\frac{1}{y}e^{y/x}=\frac{z}{y}\quad$ and $\quad c_2=\frac{1}{y^2}e^{y/x}+\frac{z}{y} = \frac{z}{y^2}+\frac{z}{y}=\frac{1+y}{y}\frac{z}{y}$
$\frac{y}{1+y}=\Phi\left(\frac{z}{y}\:,\:\frac{1+y}{y}\frac{z}{y}\right)$
An obvious solution is: $\quad u=\Phi(c_1,c_2)=\frac{c_1}{c_2} =\frac{\frac{1}{y}e^{y/x}}{\frac{1}{y^2}e^{y/x}+\frac{z}{y}}$
$$u=\frac{y}{1+y\,z\,e^{-y/x}}$$
