For a system of PDE's given by $$ \begin{cases} \frac{\partial u}{\partial x} + 6 \frac{\partial u}{\partial y} = 0 \\ \frac{\partial^2 u}{\partial x^2} - 6 \frac{\partial^2 u}{\partial y^2} = 0 \end{cases} $$ I have been told that the general solution is given by $$ u = f(6x-y) \hspace{8mm} \text{and} \hspace{8mm} u' = g(6x+y) $$
Can anyone please tell me why this is?
(Note: This is a more concise wording - and a more specific example - of the question asked here, which recieved no answers)
Claim:
Proof: Start by considering the first PDE in this new system. This has characteristic equation $$ \frac{dy}{dx} = \frac{b}{a} $$ which has solution $$ y = \frac{bx}{a} + c \hspace{6mm} bx - ay = d \hspace{14mm} \left( c \in \mathbb{R}, \hspace{3mm} d = -ac \right) $$ So, we can solve this PDE by using the substitution $$ \epsilon = x \hspace{10mm} \eta = bx - ay \hspace{10mm} \omega = u $$ which, by the chain rule gives $$ \frac{\partial \omega}{\partial x} = \frac{\partial \omega}{\partial \epsilon}\frac{\partial \epsilon}{\partial x} + \frac{\partial \omega}{\partial \eta}\frac{\partial \eta}{\partial x} = \frac{\partial \omega}{\partial \epsilon} + b \frac{\partial \omega}{\partial \eta} \\ \frac{\partial \omega}{\partial y} = \frac{\partial \omega}{\partial \epsilon}\frac{\partial \epsilon}{\partial y} + \frac{\partial \omega}{\partial \eta}\frac{\partial \eta}{\partial y} = -a \frac{\partial \omega}{\partial \eta} $$
Substituting the above into the origional PDE and simplifying, we get $$ \frac{\partial \omega}{\partial \eta} = 0 $$ Taking the partial anti-derivative of the above, we get $$ \omega = f(\eta) $$ where $f(\eta)$ is some arbitrary function. Finally, substitution back in our original variables, we get $$ u = u(x,y) = f(bx - ay) $$
The second result follows in the same way.