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!
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 $$ u = f(x-y)+\lambda x + (1-\lambda)y $$