Find the general solution of the part differential equation $$yu_x-xu_y=2xyu$$
So $$\frac{dx}{y}=-\frac{dy}{x}=\frac{du}{2xy}$$
Which in turn implies $$\int xdx =- \int ydy \Rightarrow \frac{x^{2}}{2}=-\frac{y^{2}}{2}+C_1 \Rightarrow C_1=\frac{1}{2}(x^{2}+y^{2})$$
Now I don't know how to get second solution. I try in this way: $$\frac{du}{2xy}=\frac{xdx+ydy}{xy-xy} \Rightarrow \frac{du}{2xy}=\frac{xdx+ydy}{0}$$
But I don't know does this equation make sense. And how to solve this? Like this?: $$xdx+ydy=0/ \cdot 2$$ $$2xdx+2ydy=0 \Rightarrow d(x^{2}+y^{2})=0 \Rightarrow C_{2}=x^{2}+y^{2} \Rightarrow u(x,y)=F(\frac{1}{2}(x^{2}+y^{2}),x^{2}+y^{2})$$ I don't think it makes sense, and when I put this code to wolfram alpha
y*D[u[x,y],x]-x*D[u[x,y],y]=2x*y*u -> https://www.wolframalpha.com/input?i=y*D%5Bu%5Bx%2Cy%5D%2Cx%5D-x*D%5Bu%5Bx%2Cy%5D%2Cy%5D%3D2x*y*u
I get this general solution $$u(x,y)=e^{x^{2}} \cdot C_1(\frac{1}{2}(x^{2}+y^{2}))$$ It has first component calculated by me, but I don't know how we get $e^{x^{2}}\cdot .. $
Please help me solve this PDE or give me a hint.
$$yu_x-xu_y=2xyu$$
There is a mistake in your Charpit-Lagrange charasteristic ODEs. After correction of the third term :
$$\frac{dx}{y}=-\frac{dy}{x}=\frac{du}{2xyu}$$ A first characteristic equation comes for solving $\frac{dx}{y}=-\frac{dy}{x}$ : $$x^2+y^2=c_1$$ A second characteristic equation comes from solving $\frac{dx}{y}=\frac{du}{2xyu}$ : $$u\,e^{-x^2}=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is : $$u\,e^{-x^2}=F(x^2+y^2)$$ $$\boxed{u(x,y)=e^{x^2}F(x^2+y^2)}$$ $F$ is an arbitrary function to be determined according to some boundary condition.
LATTER ADDITION. Answer to one of the comments.
An alternative method to find the first characteristic equation (what ever the third term be $\frac{du}{2xyu}$ or $\frac{du}{2xyu}$ or other :
$$\frac{dx}{y}=-\frac{dy}{x}$$ from elementary property of equal fractions with any $A$ and $B$ not both nul : $$\frac{dx}{y}=\frac{-dy}{x}=\frac{A\,dx+B(-dy)}{A\,y+B\,x}$$ With $A=x$ and $B=-y$ $$\frac{dx}{y}=\frac{-dy}{x}=\frac{x\,dx+(-y)(-dy)}{x\,y+(-y)\,x}=\frac{\frac12d(x^2+y^2)}{0}$$ This implies $\quad d(x^2+y^2)=0\quad$ thus $\quad x^2+y^2=C$.
NOTE :
If instead of $yu_x-xu_y=2xyu$ the PDE was
$$yu_x-xu_y=2xy$$
$$\frac{dx}{y}=-\frac{dy}{x}=\frac{du}{2xy}$$ First characteristic equation : $$x^2+y^2=c_1$$ Second characteristic equation from solving $\frac{dx}{y}=\frac{du}{2xy}$ : $$du-2xdx=0\quad\implies\quad u-x^2=c_2$$ General solution on implicit form $c_2=F(c_1)$ leading to : $$u(x,y)=x^2+F(x^2+y^2)$$ $F$ is an arbitrary function. The general solution of the second PDE without exponential term is very different from the general solution of the first PDE which involves an exponential term.