Solve the pde $xu_x-yu_y+u=x$ Solve the pde $xu_x-yu_y+u=x$
by solving characterstic method
$\frac{dx}{x}=\frac{dy}{-y}=\frac{dz}{x-u}$
now from $\frac{dx}{x}=\frac{dy}{-y}\implies xy=c_1$ im not getting any idea how to get other constant $c_2$
any help?
 A: $$xu_x-yu_y+u=x$$
You wrote the Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{-y}=\frac{dz}{x-u}$$
The mistake is to introduce a new symbol $z$ which doesn't exist in the PDE. The correct writting is :
$$\frac{dx}{x}=\frac{dy}{-y}=\frac{du}{x-u}$$
The second characteristic equation comes from solving $$\frac{dx}{x}=\frac{du}{x-u}$$
This is a first order linear ODE : $\quad u'+\frac{u}{x}=1\quad$ leading to :
$$xu-\frac{x^2}{2}=c_2$$
With the first characteristic equation that you correctly found, the general solution of the PDE is :
$$u(x,y)=\frac{x}{2}+\frac{1}{x}F(xy)$$
where $F$ is an arbitrary function (to be determined according to some boundary condition).
NOTE. One can introduce a new function $z(x,y)$ but with a specific definition of this function. For example $z$ is considered as equivalent to $u$ on the characteristic curve only but not elsewhere. Then instead of $\frac{dz}{x-u}$ one write $\frac{dz}{x-z}$. The calculus follows as above. This is even more thorough.
