Consider the following first order linear PDE $$a u_x + b u_y + c u = 0.$$
I have solved this PDE by changing my coordinates to $x' = ax + by $ and $y' = bx - ay$, and the general solution was $\tilde u(x', y') = f(y') \exp{-c/(a^2 + b^2) x'}$.
But then, I wanted to solve it by other means. For example, let's consider the parametrised curve $\alpha$ s.t $dx / ds = a$ & $dy / ds = b$. This gives us, $$\frac{du(x,y)}{ ds} = -c u(x,y),$$ and after solving this we get $x(x,y) = x(x(s_0), y(s_0)) \exp{(-c s)}.$
However this latter analysis only valid on the curve $\alpha$ whose tangent vector at any point is $(a,b)$, and we can see that, on this curve, both results agree with each other.
Question: However, I found the latter solution much more intuitive, so there any way to obtain the general solution using the second result ? I mean, in the latter analysis, we found the solution of this PDE on $\alpha$, so using this information, can we generalise this solution to whole $\mathbb{R}^2$ ?
$$au_x+bu_y=-cu$$ The Charpit-Legendre system of characteristic ODEs is : $$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{-cu}=ds $$ A first characteristic equation comes from $\frac{dx}{a}=\frac{dy}{b}$ : $$bx-ay=c_1$$ A second characteristic equation comes from : $\frac{dx}{a}=\frac{du}{-cu}$ : $$u\,e^{\frac{c}{a}x}=c_2$$ The general solution is $\quad u\,e^{\frac{c}{a}x}=F(bx-ay)\quad$ where $F$ is any function (to be determined according to some boundary condition). $$u(x,y)=e^{-\frac{c}{a}x}F(bx-ay)$$