For i) we directly obtain $x(s,\tau) = x_0(s)e^{a\tau}$, $y(s,\tau) = y_0(s)e^{b\tau}$, $u(s,\tau) = u_0(s)e^{c\tau}$ and for ii) we obtain the general solution by intersecting $y - \frac{b}{a}x = const$ and $u - \frac{c}{a}x = const$, say, i.e. $u = \frac{c}{a}x + F(y - \frac{b}{a}x)$ where $F$ is an arbitrary function on a single variable.
But how to show that the forms are equivalent (what should actually be meant)? In fact, both solutions are equivalent to $au_x + bu_y = c$, so they have to be mutually equivalent?!
Any help appreciated!
The equations are
$\dfrac{dx}{d\tau}=a,\;\dfrac{dy}{d\tau}=b$ and $\dfrac{du}{d\tau}=c$ and their solutions are instead
$x(s,\tau)=x_0(s)+a\tau,\;y(s,\tau)=y_0(s)+b\tau$ and $u(s,\tau)=u_0(s)+c\tau$
The equivalence is established in the sense that any solution of the form in (i) can be written in the form in (ii). The idea is to eliminate $s$ and $\tau$:
We can set $x_0(s)=0$, then $\tau=x/a$ and $y-\dfrac{b}{a}x=y_0(s)$ We have too $u=\dfrac{c}{a}x+u_0(s)$
Now, the functions $y_0(s)$ and $u_0(s)$ are related through $s$, so, any value for $y_0$ determines the value for $u_0$ or $u_0=F(y_0)$ Finally,
$u=\dfrac{c}{a}x+F(y_0)$
$u=\dfrac{c}{a}x+F(y-\dfrac{b}{a}x)$