I would like to solve the first order nonlinear PDE for the function $u(t,y_{1},y_{2})$ over $t\in[0,1]$ given by
$$ \frac{\partial u}{\partial t} + y_{2}\frac{\partial u}{\partial y_{1}} + \frac{1}{2}\left(\frac{\partial u}{\partial y_{2}}\right)^{2} = 0, \quad \text{terminal condition:} \, u(1,y_{1},y_{2}) = v(y_{1},y_{2}).$$
Letting $(x_{1}, x_{2}, x_{3}) := (t, y_{1}, y_{2})$ and $p_{i} := u_{x_{i}}$, we write the PDE as $F:= p_{1} + x_{3}p_{2} + \frac{1}{2}p_{3}^{2}=0$, and standard application of method-of-characteristics gives
$\frac{dx_{1}}{ds}=1, \frac{dx_{2}}{ds}=x_{3}, \frac{dx_{3}}{ds} = p_{3}, \frac{du}{ds} = p_{1} + p_{2}x_{3} + p_{3}^{2}, \frac{dp_{1}}{ds} = 0, \frac{dp_{2}}{ds} = 0, \frac{dp_{3}}{ds} = -p_{2}.$
From here, noting that $p_{1},p_{2}$ are constants along the characteristics, one can get $p_{3}$ as linear function of $s$, $x_{3}$ as quadratic in $s$, and $x_{2}$ as cubic in $s$. My question is: how do I recover the form of the general solution $u(x_{1},x_{2},x_{3})$?