<s>[INST] unable to solve Lagrange PDE $U_x +xyU_y +(2x^2z\ln|y|)U_z = 0$ Hi I need to solve $$u_x +xyu_y +(2x^2z\ln|y|)u_z = 0$$
Μy try:
I wrote the equations:
$$ dx = \frac{dy}{xy} = \frac{dz}{2x^2z\ln|y|} $$
so the first surface is $\phi_1 = u$
Second surface:
$$ dx\cdot x = \frac{dy}{y} \Rightarrow  0.5x^2 + C_1= \ln|y| \Rightarrow C = y\cdot e^{-0.5x^2}$$
$$ \phi_2 = ye^{\frac{x^2}{2}}$$
Third surface (I think here lies the problem):
$$ 2x^2\ln|y| = \frac{dz}{z} \Rightarrow \ln|y|(\frac{2x^3}{3} + C) = \ln|z| \Rightarrow \\
\Rightarrow y^{\frac{2x^3}{3}}\cdot y^C = z $$
Now I substitue y:
$$ z \cdot y^{\frac{-2x^3}{3}} = Ce^{0.5x^2} \Rightarrow z \cdot y^{\frac{-2x^3}{3}}e^{-0.5x^2} = C $$
So the third surface is:
$$ \phi_3 = zy^{\frac{-2x^3}{3}}e^{-0.5x^2} $$
Which is a mistake...
according to the solution: $\phi_3 = zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}$
because the solution is $$u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{\frac{x^2}{2}})$$
I can't figure out where I went wrong, any help/hints would be appreciated!
 [/INST] $$ dx = \frac{dy}{xy} = \frac{dz}{2x^2z\ln|y|} \quad\text{is OK.}$$
I agree with $  \quad 0.5x^2 + C_1= \ln|y| \Rightarrow C = y\cdot e^{-0.5x^2}$
$$ \phi_2 = ye^{-\frac{x^2}{2}}$$
$\phi_2 = ye^{\frac{x^2}{2}}$ is not correct.
For the third surface, you forgot $dx$ in $ 2x^2\ln|y| = \frac{dz}{z} $ which should be
$$ 2x^2\ln|y|dx = \frac{dz}{z}$$
with $\ln|y|=0.5x^2 + C_1$
$$ 2x^2(0.5x^2 + C_1)dx = \frac{dz}{z}$$
$$ (x^4 + 2C_1x^2)dx = \frac{dz}{z}$$
Introducing another constant $C\neq C_1$ is a cause of confusion.
I suppose that you can take it from here.
Note : There is a typo in the expected solution $u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{\frac{x^2}{2}})$ which should be :
$$u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{-\frac{x^2}{2}})$$
</s>