Is there a particular way to solve this type of PDE? $u_pf+u_v=g$
$g_pf+g_v=0$
$\frac{dp}{dv}=f(p(v),v)$
$g$ cannot equal 0
$g=du$
Consider $p(v_0)=p_0$
Given u,f, and g is all function of p and v.
I checked this condition with a tutor before as mentioned this type of PDE is too general and have no simpler way to solve it? g and f can be any known function
My 2nd proposition is if above condition is true, then the "$p_2$" that satisfy this formulae 
$u(p_0,v_0)+g(p_0,v_0)=u(p_2,v_0+h)$
will also be the exact solution to the differential equation f but my tutor think this is true only for some special f?
As an example a DE $\frac{dp}{dv}=-1.4\frac{p}{v}$ so the function of g that satisfy the condition $g_pf+g_v=0$ will be $g=pv^{1.4}$
For second condition, $u_pf+u_v=g$ , the function of u that satisfy this equation $u_p(-1.4\frac{p}{v})+u_v=pv^{1.4}$ will be $u=pv^{2.4}$
The exact solution of the DE $\frac{dp}{dv}=-1.4{p}{v}$ assuming $p(v_0)=p_0$ will be $p_0v_0^{1.4}$= constant hence $p(v_0+h)=p_0[\frac{v_0}{v_0+h}]^{1.4}$
The solution of $p_2$ that satisfy this equation $u(p_0,v_0)+g(p_0,v_0)h=u(p_2,v)$ for this example will be
$p_0v_0^{2.4}+p_0v_0^{1.4}h=(p_2)(v_0+h)^{2.4}$
$p_2=\frac{p_0v_0^{1.4}}{(v_0+h)^{1.4}}$
while if consier $v_0$ take a -h stepsize
$p_0v_0^{2.4}+p_0v_0^{1.4}(-h)=(p_2)(v_0-h)^{2.4}$
$p_2=\frac{p_0v_0^{1.4}}{(v_0-h)^{1.4}}$
If it does not satisfy the $g_pf+g_v=0$ condition, it only have properties of tangent to the solution of differential equation.
Example u=2.5pv and g=-p, with condition
$u_pf+u_v=g$ obtain
$f=-1.4\frac{p}{v}$ as well
$u(p_0,v_0)+g(p_0,v_0)=u(p_2,v_0+h)$
$2.5p_0v_0-p_0(h)=2.5(p_2)(v_0+h)$
$p_2=P_0[1-1.4\frac{p}{v}-0.6(\frac{h}{v_0})^2+...]$
 A: Method - Solving PDE $xu_x+(x+t)u_t=1$ with $u(1,t)=t$ 
Say initial data $g_0(z)=g(\mathbf x_0(z))$ is prescribed on a curve (that is not characteristic, i.e. not parallel to $(f,1)$ ) $ \{(p,v)  : (p,v) = \mathbf  x_0(z), z\in \mathbb R \}$. Solve first the equations for $\mathbf X(z,s), G(z,s)$,
$$ \partial_s \mathbf X(z,s) = \binom{f(\mathbf X(z,s))}{1}, \quad \mathbf X(z,0) = \mathbf x_0(z)\\ \partial_s G(z,s) = 0, \quad G(z,0) = g_0(z) $$
Clearly $G(z,s) = g_0(z)$, and $\mathbf X_2(z,s) = \mathbf x_0(z) + s.$ Sometimes you get lucky and $\mathbf X_1(z,s) = \mathbf x_0(z) + \int_0^s f(\mathbf X(z,s)) ds $ has an easy solution. But the solution exists for nice enough $f$ by Picard's theorem.
If you're even luckier you can write down the formula for the inverse $(z,s)=\mathbf X^{-1}(p,v)$ of $\mathbf X$. But the inverse function theorem guarantees it exists, since $\nabla X$ has full rank. (this is where we need the initial data curve to not be parallel to $(f,1)$)
Now note that we can define the solution $g$ by
$$ g(\mathbf X(z,s)) = G(z,s) \iff g(p,v):= G(\mathbf X^{-1}(p,v)) =g_0(\mathbf X^{-1}_1(p,v)). $$
OK, now we know $g$. We can repeat the same thing for $u$. We need initial data for $u$, also prescribed on $\mathbf x_0$ for simplicity. i.e. $u(\mathbf x_0) = u_0$. Define
$$ \partial_s U(z,s) =g(\mathbf X(z,s)) = G(z,s) ,\quad  U(z,0) = u_0(z)$$
from the above we note
$$ U(z,s) = u_0(\mathcal z)+ G(z,0)s.$$
Then we can define 
$$ u(\mathbf X(z,s)) =U(z,s) \iff u(p,v) := U(\mathbf X^{-1}(p,v)) = u_0( \mathbf X^{-1}_1(p,v)) + G(\mathbf X^{-1}_1(p,v)) \mathbf X^{-1}_2(p,v).$$
TL;DR your two equations in some weird coordinates that depend on $f$ are simply $ U'=G, G'=0$ so $G=C_1$ and $U = C_1s + C_2$, but the constants $C_1,C_2$ are functions and you need to invert the coordinates so thats a little annoying
A: The third equation stipulates that $p$ is a function of $v$ only. Therefore, the second equation may be written as a total derivative
$$
\frac{\text d }{\text d v} g(p(v),v) = g_p p_v + g_v = 0\, ,
$$
which means that $g = g_0$ is constant along the curves $p_v = f(p,v)$. Thus, the first equation may be written as follows
$$
\frac{\text d }{\text d v} u(p(v),v) = u_p p_v + u_v = g_0 \, ,
$$
leading to $u = g_0 (v-v_0) + u_0$ (the parameter $v_0$ is such that the boundary data is known for $v=v_0$). Now, expressing the solution analytically over the plane $(p, v) \in \Bbb R^2$ relies on the solution of $p_v = f(p,v)$, which is not straightforward in general...
For instance, in the case where $f(p,v)=k$ is constant, we have $p=k(v-v_0)+p_0$. Hence, the solution is known along a family of lines in the $p$-$v$ plane. The boundary data is function of $v_0 = v-({p-p_0})/{k}$, so that the solutions $g=g_0(v_0)$ and $u=g_0(v_0)(v-v_0)+u_0(v_0)$ rewrite as
$$ g(p,v)=g_0\big(v-\tfrac{p-p_0}{k}\big)
$$
$$
u(p,v)=g_0\big(v-\tfrac{p-p_0}{k}\big)(v-v_0) + u_0\big(v-\tfrac{p-p_0}{k}\big) .
$$
As $f(p,v)$ becomes more complicated, computations become more involved.
