Different solutions for PDE $u_x+u_y=1$ I try to solve a very basic PDE problem "$u_x+u_y=1$"
Here is my approach.
I try to use change coordinates method. notice $(u_x, u_y)$ $\cdot$ (1, 1)=1
Let $x_1=x+y, y_1=x-y$, so $x = (x_1+y_2)/2, y = (x_1 - y_2)/2$
Then $"u_x = u_{x_1} + u_{y_1}, u_y= u_{x_1} - u_{y_1}$
Then we get $U_x+u_y=2u_{x_1}=1, i.e. u_{x_1}=1/2.$ 
Then general solution for such ODE is $u=(x_1)/2 + f(y_1)$
So we get solution to the ODE =(x+y)/2 + f(x-y)
However, I search online found there are solutions y+ f(x - y) =1 or = x+ f(x - y) which still work. Are all of them(including mine) the correct solutions? Why there are such huge differences?
 A: The three forms of general solutions that you mention are the same. But don't use the same symbol $f$ for different functions. Better write :
$$u(x,y)=\frac{x+y}{2}+f(x-y) \tag 1$$
$$u(x,y)=x+g(x-y) \tag 2$$
$$u(x,y)=y+h(x-y)\tag 3$$
$f$ , $g$ , $h$ are any arbitrary functions, but not simultaneously equal one to another.
Explication :
For example, consider $u(x,y)=\frac{x+y}{2}+f(x-y)$ . 
$f$ which is any function can be written $f(x-y)=g(x-y)+\phi(x-y)$ where $g$ and $\phi$ are any functions.
$$u(x,y)=\frac{x+y}{2}+g(x-y)+\phi(x-y)$$
Let $\phi(x-y)=\frac{x-y}{2}\quad\to\quad u(x,y)=\frac{x+y}{2}+g(x-y)+\frac{x-y}{2}$
After simplification $\quad u(x,y)=x+g(x-y)$
So, you got Eq.(2) . This shows that Eq.(1) and Eq.(2)  are equivalent, which means that both express the general solution of the PDE on two different forms.
Also, $f$  can be written $f(x-y)=h(x-y)+\varphi(x-y)$ where $h$ and $\varphi$ are any functions.
$$u(x,y)=\frac{x+y}{2}+h(x-y)+\varphi(x-y)$$
Let $\varphi(x-y)=-\frac{x-y}{2}\quad\to\quad u(x,y)=\frac{x+y}{2}+h(x-y)-\frac{x-y}{2}$
After simplification $\quad u(x,y)=y+h(x-y)$
So, you got Eq.(3) . This shows that Eq.(1) and Eq.(3)  are equivalent, which means that both express the general solution of the PDE on two different forms.
You can also write $h(x-y)=F(x-y)+c\quad$ where $c$ is any constant, since $F(x-y)+c$ is a function of $(x-y)$. This leads to another form of general solution $u(x,y)=y+c+F(x,y)$ , again equivalent to the other forms.
Of course $f(x-y)$ can be replaced by a lot of different sum of functions of $(x-y)$ or of $(y-x)$. So, you see that they are an infinity of equivalent forms, apparently different from one to the others, but which all express the same general solution of the PDE.
