Given an equation, $$u_{xx} + u_{xt} -20u_{tt}= 0$$ it can be shown that there exists $\xi = t - 5x$ and a $\eta= t+4x$ such that
$$u_{\xi \eta} = 0 \space\space\space\space\space\space\space\space \text{ (first canonical form of a hyperbolic)}$$
Thus, we have
$$u(\xi,\eta) = F(\xi)+G(\eta)$$$$u(x,t) = F(t-5x)+G(t+4x)$$
Now lets say we have initial conditions of: $$u(x,0) = f(x) $$ $$ u_t(x,0) = g(x)$$ How do initial conditions change the solution? What exactly are we trying to solve for beyond this?
We have a system from the Cauchy conditions: $$ F(-5 x)+G(4 x) = f(x),\: F'(-5 x)+G'(4 x) = g(x), \: x \in \mathbb{R}. $$ From the first equation we get $F(z) = f\left(-\frac{z}{5}\right)-G\left(-\frac{4 z}{5}\right)$. Now the solution has a form $u(t,x)=f\left(x-\frac{t}{5}\right)-G \left(4 x-\frac{4}{5}t\right)+G(t+4 x)$.
In order to define $G$ we should use the second Cauchy condition $$ \partial_t u(x, 0) = -\frac{1}{5} f'(x)+\frac{9}{5} G'(4 x) = g(x). $$ So, we have ODE, which defines the function $G$: $-\frac{1}{5} f'\left(\frac{x}{4}\right)+ \frac{9}{5} G'(x) =g\left(\frac{x}{4}\right)$. The solution of this ODE is $G(z) = C + \frac{1}{9} \int_0^z \left(5 g\left(\frac{\tau}{4}\right)+f'\left(\frac{\tau}{4}\right)\right) d\tau$, $z \in \mathbb{R}$, $C$ is some constant.
So, the function $$u(x, t) = \frac{1}{9} \left(-\int_0^{4 x-\frac{4 t}{5}} \left(f'\left(\frac{\tau }{4}\right)+5 g\left(\frac{\tau }{4}\right)\right) \, d\tau +\int_0^{t+4 x} \left(f'\left(\frac{\tau }{4}\right)+5 g\left(\frac{\tau }{4}\right)\right) \, d\tau +9 f\left(x-\frac{t}{5}\right) \right)$$ is the solution of your Cauchy problem.