Solving $u_x^2+u_y^2=n_0^2$ Using Method of characteristics $$
\begin{cases}
u_x^2+u_y^2=n_0^2\\
u(x,2x)=1\\
\end{cases}$$
For constant $n_0$
This is the Eikonal equation which is a non-linear PDE, to use the method of characteristics we "bring it down" to an ode by substituting the partial derivatives.
$$P_1=u_x$$
$$P_2=u_y$$
$$P_1^2+P_2^2-n_0^2=0$$
And the characteristics strips are:
$$\begin{cases}
\frac{dx_i}{dt}=\frac{\partial F}{\partial P_i}\\
\frac{du}{dt}=\sum_{i=1}^{n}P_i\frac{\partial F}{\partial P_i}\\
\frac{dP_i}{dt}=-\frac{\partial F}{\partial x_i}-P_i\frac{\partial F}{\partial u}\\
\end{cases}$$
In our case it is:
$$\begin{cases}
\frac{dx}{dt}=\frac{\partial F}{\partial P_1}\iff x_t=2P_1\\
\frac{dy}{dt}=\frac{\partial F}{\partial P_2}\iff y_t=2P_2\\
\frac{du}{dt}=P_1\frac{\partial F}{\partial P_1}+P_2\frac{\partial F}{\partial P_2}\iff u_t=2P_1^2+2P_2^2=2(P_1^2+P_2^2)=2n_0^2\\
\frac{dP_1}{dt}=-\frac{\partial F}{\partial x}-P_1\frac{\partial F}{\partial u}\iff P_{1_t}=-0-P_1\cdot 0=0\\
\frac{dP_2}{dt}=-\frac{\partial F}{\partial y}-P_2\frac{\partial F}{\partial u}\iff P_{2_t}=-0-P_2\cdot 0=0\\
\end{cases}$$
We have $$P_{1_t},P_{2_t}=0$$ as $x,y,u$ does not appear in $F$?

$$u_x^2+u_y^2-n_0^2=0=p^2+q^2-n_0^2=F(x,y,p,q,u)$$
$$F_x=0\quad;\quad F_y=0\quad;\quad F_p=2p\quad;\quad F_q=2q\quad;\quad F_u=0$$
Charpit-Lagrange system of characteristic ODEs :
$$\dfrac{dp}{F_x+pF_u}=\dfrac{dq}{F_y+qF_u}=\dfrac{du}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}$$
$$\dfrac{dp}{0}=\dfrac{dq}{0}=\dfrac{du}{-2p^2-2q^2}=\dfrac{dx}{-2p}=\dfrac{dy}{-2q}=dt$$
Note that this is the same as :
$$\begin{cases}
\frac{dp}{dt}=0\\
\frac{dq}{dt}=0\\
\frac{du}{dt}=-2p^2-2q^2\\
\frac{dx}{dt}=-2p\\
\frac{dy}{dt}=-2q
\end{cases}$$
A first characteristic equation comes from $dp=0$
$$p=c_1$$
A second characteristic equation comes from $dq=0$
$$q=c_2$$
A third characteristic equation comes from $\dfrac{du}{-2c_1^2-2c_2^2}=\dfrac{dx}{-2c_1}=\dfrac{dy}{-2c_2}=\frac{du-c_1dx-c_2dy}{-2c_1^2-2c_2^2-c_1(-2c_1)-c_2(-c_2)}=\frac{du-c_1dx-c_2dy}{0}$
$du-c_1dx-c_2dy=0\quad\implies\quad u-c_1x-c_2y=c_3$
$$u=c_1x+c_2y+c_3$$
Since we are not looking for the general solution but only for a particular solution which satisfies the boundary condition $u(x,2x)=1$ , there is no need for further calculus about more characteristic equations.
Condition : $\quad u(x,2x)=c_1x+c_2(2x)+c_3=(c_1+2c_2)x+c_3=1\implies\begin{cases}c_1=-2c_2\\c_3=1\end{cases}$
$u=-2c_2x+c_2y+1$
$p^2+q^2=(-2c_2)^2+c_2^2=n_0^2\quad\implies c_2=\pm\frac{n_0}{\sqrt{5}}$
$$\boxed{u(x,y)=\frac{\pm n_0}{\sqrt{5}}(-2x+y)+1}$$
