For the first method. You get two (uniparametric) families of surfaces ($\\u+\dfrac{1}{x}=c_1$ and $\dfrac{u^2}{2}-y=c_2$), the intersection of wich is a (two parameter) family of curves. Consider some, fixed by now, values of $c_1$ and $c_2$. They determine a curve into the surface of a solution brung about by some boundary conditions. So is, $c_1$ is not free, it has to correspond to some $c_2$ to make the curve pass for some point the boundary conditions impose. But it has to be so for every value of $c_1$ or, in other words, there must exist some functional relation between them: $c_1=f(c_2)$.
The second method works almost the same, but in this case we work directly with parametric equations for the surface (two parameters and three coordinates). Your equations are ok, but not the solution. It must be:
$$x(t)=\frac{-1}{t+x_0}\\u(t)=t+u_0\\y(t)=\frac{(t+u_0)^2}{2}+y_0$$
We can set $x_0=0$ and operate, yelding:
$$x(t)=\frac{-1}{t};t=\frac{-1}{x}\\u=t+u_0=-\dfrac{1}{x}+u_0;u_0=u+\dfrac{1}{x}\\y=\frac{(t+u_0)^2}{2}+y_0=\dfrac{u^2}{2}+y_0;y_0=-\left(\dfrac{u^2}{2}-y\right)$$
We have now exactly a family of curves depending on two parameters. The same argument used in the first method works here: we have some functional relation (too brung about by the boundary conditions) between these parameters: $u_0=g(y_0)$. And we recover the same general solution as with the first method: $u+\dfrac{1}{x}=g\left(\dfrac{u^2}{2}-y\right)$
In this case, the boundary conditions are $u(x,1-x)=0$, enough to determine $f$ in $u+\dfrac{1}{x}=f\left(\dfrac{u^2}{2}-y\right)$
$$\dfrac{1}{x}=f(-(1-x))$$
Making $w=-(1-x)$, we get $x=w+1$ and $f(w)=\dfrac{1}{w+1}$
Finally $u+\dfrac{1}{x}=\dfrac{1}{\dfrac{u^2}{2}-y+1}$
We are almost done. We can make $c_1=1$, as it represents a simple displacementfor the parameter driving the characteristics, $s$. Then, get rid of $s$ because we are interested in the relation among $x,y$ and $u$ for the characteristics.
$y/x=c_2$ or $x/y=c'_2$
$z(s)=u(x(s),y(s))=c_3x(s)^a$ or for easy reading $u=c_3x^a$
But the integration constants must be related: $c_3=f(c'_2)$
Substituting, $u=x^af(x/y)$, being the general solution.
Now $u(x,1)=\phi(x)$
$\phi(x)=x^af(x)\to f(x/y)=(x/y)^{-a}\phi(x/y)$
leading to $u(x,y)=x^a(y/x)^a\phi(x/y)$ or
$u(x,y)=y^a\phi(x/y)$
Best Answer
$$\frac{dt}{1}=\frac{dx}{5u}=\frac{du}{u}\qquad\text{is OK.}$$ As you found it, a first characteristic equation comes from $\frac{du}{dx}=\frac{1}{5}$. $$u-\frac{x}{5}=C_1$$ For a second characteristic equation, you chose $\frac{dx}{dt}=5u$
This is not the simplest equation because it cannot be integrated directly. As already pointed out in comments, $\frac{dt}{1}=\frac{du}{u}$ is straightforward. It doesn't mater, any way leads to the same final result. So we will continue the way you chose. even if is is a bit more complicated.
$$\frac{dx}{dt}=5u=5(C_1+\frac{x}{5})=5C_1+x$$ This is a first order linear ODE easy to solve : $x=-5C_1+C_2e^t$
$xe^{-t}+5C_1e^{-t}=C_2$ $$xe^{-t}+5(u-\frac{x}{5})e^{-t}=C_2$$ $$5ue^{-t}=C_2$$
The general solution of the PDE is on the form of implicite equation $\Phi(C_1,C_2)=0$ or $C_2=F(C_1)$ or $C_1=G(C_2)$ where $F$ and $G$ are arbitrary functions (in fact one inverse from the other). $$5ue^{-t}=F\left(u-\frac{x}{5}\right)$$ $$u=\frac15 e^t F\left(u-\frac{x}{5}\right)$$ $F$ is an arbitrary function to be determined according yo the boundary condition.
$u(0,t)=e^{14t}=\frac15 e^t F\left(e^{14t}-\frac{0}{5}\right)$
$e^{14t}=\frac15 e^t F\left(e^{14t}\right)$
Let $X=e^{14t}$
$5e^{13t}= F\left(X\right)=5X^{13/14}$
Now the function $F(X)$ is determined. We put it into the above general solution where $X=u-\frac{x}{5}$ . The particular solution fitting to the boundary condition is : $$u=e^t \left(u-\frac{x}{5}\right)^{13/14}$$