I have some conceptual questions regarding a solution technique for second order linear PDEs. The example I have been considering is $u_{xx} + 2u_{xy} + u_{yy} = 0$.
The technique is to use the guess $u = f(mx + y)$, where $f$ is an arbitrary differentiable function, and $m$ is said to be a constant to be determined. I make the substitution $w = mx + y$ and then plug $f(w)$ into the the PDE, yielding $(\frac{d^2 f}{dw^2} (\frac{\partial w}{\partial x})^2 + \frac{df}{dw} \frac{\partial^2 w}{\partial^2 x}) + 2(\frac{d^2 f}{dw^2} \frac{\partial w}{\partial x} \frac{\partial w}{\partial y} + \frac{df}{dw} \frac{\partial^2 w}{\partial y \partial x}) + (\frac{d^2 f}{dw^2} (\frac{\partial w}{\partial y})^2 + \frac{df}{dw} \frac{\partial^2 w}{\partial^2 y}) = 0$, which simplifies to $(m^2 + 2m + 1)\frac{d^2 f}{dw^2} = 0$.
I know how to get a solution to the PDE from the $m^2 + 2m + 1 = 0$ branch, and have been instructed that the $\frac{d^2 f}{dw^2} = 0$ will not produce any additional solutions and thus can be ignored, but I wanted to work this out myself.
It is unclear to me whether $m$ can still be treated as a constant to be determined in the context of this branch. In this branch, the characteristic equation does not have to equal $0$, so I cannot import $m$'s solutions from the other branch, and I do not think $m$ will be solvable here. Should I think of $m$ as a variable instead of a constant when considering this branch?
When I solve $f = \int\int 0\ dw\ dw$, I can see algebraically that this is ordinary antidifferentiation, since $f$ is, in some sense, a function only of the variable $w$, leading to the solution $f = A * (mx + y) + B$. However, $f$ is ultimately a function of $mx + y$, an input that lives in $\mathbb{R}^2$, and geometrically I have a strong intuition that this should be partial antidifferentiation. For an $\mathbb{R}^2$ input-space, when antiderivatives are taken with respect to one dimension, the other, orthogonal dimension is held constant, and the constants of integrations become functions of that orthogonal dimension; antidifferentiating some $g(x,\ y)$ with respect to $x$ yields a constant that is a function of $y$. But this seems to be the same geometry at play here, up to a rotation transformation. If I antidifferentiate with respect to $w$, which is (though not often written as such) antidifferentiating with respect to $mx + y$, then the orthogonal dimension is $y - \frac{x}{m}$. Thus, from this perspective, the solution to $f = \int\int 0\ dw\ dw$ should be $f = A(y - \frac{x}{m}) * (mx + y) + B(y - \frac{x}{m})$. And all of this is treating $m$ as a constant, which I'm not sure is appropriate in this branch. Depending on the answer to the first question, I may have to account for $m$ being a variable.