Show that each of the following equations has a solution of the form $u(x,y) = f(ax+by) $ for a proper choice of constant $a,b$. Find the constant for each example.
(a) $u_x + 3u_y =0$
(b) $3u_x - 7u_y = 0$
(c) $2u_x + \pi u_y =0$
 A: I believe one of the comments to your answer gave a brief version of the answer that I am about to present.  Here is a somewhat more detailed approach.
First I will present the solution to the more general equation
$$
\alpha u_x + \beta u_y = 0,
$$
which I interpret to mean
$$
\alpha \partial_x u(x,y) + \beta \partial_y u(x,y) = 0.
$$
The claim is that all solutions to this equation are of the form $u(x,y)=f(ax+by)=f(g(x,y))$ in which $g(x,y)=ax+by$ for an appropriate choice of $a$ and $b$.  This is equivalent to saying that we can choose $a$ and $b$ to satisfy the following equation.
$$
\alpha \partial_x f(g(x,y)) + \beta \partial_y f(g(x,y)) = 0.
$$
Using the chain rule, I calculate, with $t=g(x,y)$,
$$
\partial_x f(g(x,y)) = \partial_t f(t) \partial_x g(x,y) = a f'(ax+by),\\
\partial_y f(g(x,y)) = \partial_t f(t) \partial_y g(x,y) = b f'(ax+by).
$$
Plugging this into the given equation gives
$$
0 = \alpha u_x + \beta u_y = \alpha a f'(t) + \beta b f'(t) = (\alpha a +\beta b) f'(t)
$$
Hence we need to choose $a$ and $b$ such that $\alpha a + \beta b =0$.  Since this is one equation with two unknowns, we can solve it by fixing $b$ to an arbitrary value.  In this case, I choose $b=1$ so that the dependency on $y$ is maintained, but the equations simplify.  With $b=1$, we must choose $a$ to satisfy $\alpha a + \beta = 0$, or
$$
a = -\frac{\alpha}{\beta}.
$$
You can check that this solves the given equation.  To solve each of the given equations, plug in the appropriate values of $\alpha$ and $\beta$.
