Solve the pde $xu_x-yu_y+u=x$

Question

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?

Answer 1

$$xu_x-yu_y+u=x$$

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.