Surface from equation I have no experience with this kind of tasks. Could you help?
Find the surface that follows $yu_x + u(x-1)u_y = 0$ and contains the curve $x=2$, $y=t$, $u=2t^2$.
 A: The idea is to cook up a direction on the surface transverse to the curve and evolve it along the PDE to sweep out a surface.  I'll briefly sketch it:
Given $yu_x + (x-1)uu_y=0$, so $u_x = (1-x)u\lambda$, $u_y = y\lambda$ for some $\lambda=\lambda(x,y)$.
So
$$ \begin{align*}
\lambda_x y &= u_{yx} = u_{xy} \\
&= (1-x)u_y\lambda+(1-x)u\lambda_y \\
&= (1-x)y\lambda^2 + (1-x)u\lambda_y
\end{align*}
$$
and hence
$$ (y,x-1)\cdot\nabla\lambda = (1-x)\lambda^2 y $$
Ignoring the problematic case $t=0$ for which $(y,x-1)$ is actually tangential to the given curve, the integral curves of $(y,x-1)$ starting at $(x,y)=(2,t)$ has $\sinh s$, $\cosh s$ and some $t$ lying around.  Plugging back gives
$$
\frac{\partial}{\partial s}\lambda(x(s,t),y(s,t))
=(1-x(s,t))\lambda(x(s,t),y(s,t))^2 y(s,t)
$$
which you can solve for $\lambda(x(s,t),y(s,t))$ (easy to determine the value $\lambda(s=0)=4$), hence
$$ u_s = u_x x_s + u_y y_s, u(s=0)=2t^2 $$
is a simple.integration.
A: The standard way so solve a quasilinear first order PDE is to set up the characteristic equations. Given
$$ a(x,y,u)u_x+b(x,y,u)u_y+c(x,y,u)u = 0 $$
we construct the characteristic curves as solutions to the system
$$ \frac{\text dx}{a} = \frac{\text dy}{b} = \frac{\text du}{c} $$
In our case, we have
$$ \frac{\text dx}{y} = \frac{\text dy}{u(x-1)} = \frac{\text du}{0} $$
We interpret $\frac{\text du}{0}$ as $u$ being a constant (informally, $\frac{\text dx}{y} = \frac{\text du}{0} \Rightarrow \frac{\text du}{\text dx} = 0 \Rightarrow u = c_1$). Then, since $u$ is a constant,
$$ \frac{\text dx}{y} = \frac{\text dy}{u(x-1)} \Rightarrow u(x-1)\text dx = y\text dy \Rightarrow u(x^2-2x) - y^2 = c_2 $$
Then, any surface with $g(c_1(x,y,u),c_2(x,y,u))=0$ for any $g$ will satisfy the original equation. Equivalently, we can write $c_1 = f(c_2)$ for any $f$. Plugging in our values, we get
$$ u = f(u(x^2-2x)-y^2) $$
is a family of surfaces which satisfy the PDE. To get the particular one which contains the curve $(x(t),y(t),u(t)) = (2,t,2t^2)$, we convert this to the condition $u(2,y) = 2y^2$ and plug in $x=2$ in our family of surfaces to solve for $f$.
$$ u(2,y) = f(u(2,y)(2^2-2\cdot2)-y^2) \text{ and } u(2,y) = 2y^2 \\
   \Rightarrow 2y^2 = (-2)(-y^2) = f(-y^2) \Rightarrow f(\zeta) = -2\zeta $$
Finally, substituting this definition of $f$ into our surface equation gives the solution
$$ u = 2y^2-2u(x^2-2x) \Rightarrow u(x,y) = \frac{2y^2}{2x^2-4x+1} $$
We can check that $u(2,y) = 2y^2$ and that $yu_x(x,y) + u(x,y)(x-1)u_y(x,y) = 0$.
