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Given an equation in partial derivatives of the form $Af_x+Bf_y=phi(x,y)$, for example $$f_x-f_y=(x+y)^2$$ How do I know which change of coordinates is appropiate to solve the equation? In this example, the change of coordinates is $u=x+y$, $v=x^2-y^2$, why?

lets write $mathbf x = (x,y)^T$. The equation is the linear inhomogeneous transport equation,
$$ mathbf a cdot nabla f =b,$$
where $f=f(x,y)$ and specifically $mathbf a = (a_1,a_2)^T=(1,-1)^T$ and $b = (x+y)^2$, but I will continue almost as if I don't know what $mathbf a$ or $b$ are.

I'll sketch out a theory("The method of characteristics") assuming you are given initial data, that works in a large generality. In particular, with very minor modifications you can also allow the coefficient $mathbf a$ of the derivative to be a function, or even depend on $f$ (but not $nabla f$).

Suppose we are prescribed initial data $f(x,0) = f_0(x)$. (more general initial data along a hyper surface is possible) Consider the family of ODEs that define the "characteristic curves"
$$frac{d}{dt} mathbf X(t;z) = mathbf a, quad mathbf X(0;z) = binom{z}{0} \ frac{d}{dt} F(t;z) = b(mathbf X),quad F(0;z) = f_0(z) $$
These can be solved, by Picard-Lindelof.
In this case, $mathbf X$ is in fact a globally invertible change of coordinates,x since we can explicitly write
$$mathbf X= binom{z}{0} + tmathbf a = mathbf x iff t= y/a_2,quad z = x-ta_1 = x - ya_1/a_2$$
(the condition that $a_2 neq 0$ is related to the fact that the initial data is prescribed on the $x$-axis: the important point is that $mathbf a$ is not tangent to the hypersurface where the initial data is prescribed.)
Let's write the inverse map of $mathbf X$ as $mathbf Z$ with the explicit formula $$mathbf Z(mathbf x):=binom{t(mathbf x)}{z(mathbf x)} = binom{y/a_2}{x - ya_1/a_2}$$

One can now check that if we define
$$ f(mathbf x) := F(mathbf Z(mathbf x))$$
then $f$ solves the original PDE. Indeed, by chain rule and inverse function theorem,
$$ nabla f(mathbf x)^T = (nabla F)^T(mathbf Z(mathbf x)) (nabla mathbf Z)^T(mathbf x)=(nabla F)^T(mathbf Z(mathbf x)) ((nabla mathbf X)^{-1}(mathbf Z(mathbf x)))^T$$
If we hide the point $mathbf x$ where the functions are evaluated, this is written perhaps more legibly,
$$ nabla f^T = (nabla Fcirc mathbf Z)^T ((nabla mathbf X)^{-1}circ mathbf Z)^T $$
Then the dot product $mathbf a cdot nabla f$ is
$$mathbf a cdot nabla f= nabla f^T mathbf a = (nabla Fcirc mathbf Z)^T ((nabla mathbf X)^{-1}circ mathbf Z)^T mathbf a $$
Since $mathbf a$ is precisely the first column of $nabla mathbf Xcirc mathbf Z$ (if we agree to write $t$ derivatives in the first column), $((nabla mathbf X)^{-1}circ mathbf Z)^T mathbf a = binom{1}{0}$. Thus

$$ mathbf a cdot nabla f = (nabla Fcirc mathbf Z)^T binom{1}{0} = (partial_t F)circ mathbf Z = b(mathbf Xcirc mathbf Z) = b$$

as needed.

(See also Method of characteristics (quasilinear pde- nonlinear transport ), Explaining the method of characteristics )

This is a linear PDE then

$$
f = f^h+f^p
$$

with

$$
f^h_x-f^h_y = 0\
f^p_x-f^p_y = (x+y)^2
$$

for the homogeneous solution we have $f^h(x,y) = phi(x+y)$ by characteristics method. Now the particular is obtained proposing for $f^p$ a polynomial form as

$$
f^p = a x(x+y)^2+b y(x+y)^2
$$

and after substitution we have

$$
f^p_x-f^p_y - (x+y)^2=(a-b-1)(x+y)^2=0
$$

so the solution is

$$
f(x,y) = phi(x+y) + a x(x+y)^2+b y(x+y)^2, mbox{such that} a-b=1
$$

NOTE

Regarding the change of coordinates we know one coordinate for sure which is $u = x+y$ but we need two coordinates so we choose the other as $v = x-y$ because $u, v$ form a valid (independent) coordinate system. We could choose instead $x+y, x+ a y$ with $a ne 1$ as well.