Use Charpits to solve this PDE

Use Charpits Equations to Solve the following given that u=1 on the line x+2y=2.

Is the solution unique?

$\displaystyle

4u \frac{\partial {u}}{\partial x}=(\frac{\partial {u}}{\partial y})^2$

$\displaystyle

4up-q^2=0, F(x,y,u,p,q)=0

$

$\displaystyle \frac{\partial F}{\partial x}=0, \frac{\partial {F}}{\partial y}=0,\frac{\partial {F}}{\partial u}=4p$

$\displaystyle \frac{dx}{dt}=F_p, \frac{dy}{dt}=F_q, \frac{dp}{dt}=-F_x-pF_u, \frac{dq}{dt}=-F_y-qF_u, \frac{du}{dt}=pF_P+qF_q$

Initial Condition are

$\displaystyle t=0,u=1, x=s,y=1-\frac{s}{2}, \frac{d}{ds} (u(s,1-\frac{s}{2}))=\frac{d}{ds}(1)$

Therefore

$\displaystyle \frac{\partial u}{\partial x}+ \frac{\partial u}{\partial y}( \frac{-1}{2})=0$

Therefore

$\displaystyle p - \frac{q}{2}=0, q=2p, @t=0, u=1 \implies 4p-4p^2=0$

2 solutions p=0, p=1

$\displaystyle \frac{dp}{dt}=-4p^2 => p=\frac{1}{4t-A}$

No solution for p=0, therefore continuing with p=1 at t=0 to find A

$\displaystyle \frac{dp}{dt}=-4p^2 => p=\frac{1}{4t-A}$

$\displaystyle

\implies A=\frac{-1}{4}

$.

Therefore $\displaystyle p=\frac{1}{4t+1}$. Now q=2p therefore $\displaystyle q=\frac{2}{4t+1}$.

Since 4up=q^2 implies u is $\displaystyle \frac{4t+1}{16t^2+1}$

How am I doing so far? (Thinking)