# Solve $u_x+u_y=1$

I am asked to solve $$u_x+u_y=1$$ If is was homogeneous i.e., $u_x+u_y$ the answer would be $u(x,y)=f(y-x)$ where $f$ is an arbitrary function. I have found the following set of solutions: $$u(x,y)=\lambda x +(1-\lambda)y$$ where $\lambda$ is an arbitrary constant(real or imaginary). I just have no idea what method other then trial and error would have lead me here. Any ideas? Thanks!

-
This is a linear first order PDE. Googling will provide lots of write-ups on how to solve it! –  Mariano Suárez-Alvarez Feb 14 '13 at 21:04
Method of characteristics is used on general first-order PDEs. en.wikipedia.org/wiki/Method_of_characteristics –  Ron Gordon Feb 14 '13 at 21:50

## 1 Answer

You can observe that the only difference between homogeneous an inhomogeneous equations is $1$. So you can assume that particular solution is linear on both $x$ and $y$, or $u^p = ax + by$. $$u_x^p + u_y^p = a + b = 1$$ In your case $a = \lambda$ and $b = 1 - \lambda$. General solution of inhomogeneous PDE given is the sum of general solution of homogeneous PDE and particular solution of inhomogeneous PDE, so $$u = f(x-y)+\lambda x + (1-\lambda)y$$

-