Tried to propose this statement before but it looks like i have problem in writing the correct mathematical symbol especially dealing v and p making it very confusing. Please help to edit the symbol if the idea sound comprehensible.
Given $u(v,p)$,$g(v,p)$ and $f(v,p)$ satisfy the relationship such that
$f=\frac{dp}{dv}=\frac{g-u_v}{u_p}$ with initial condition $p(v_0)=p_0$.
While $p_1$ is a solution that satisfy below equation
$u(p_1,v_0+h)-u(p_0,v_0)=g(p_0,v_0)h$
The solution of $p_1$ that satisfy the above equation will also be the exact solution of the DE $\frac{dp}{dv}=f(v,p)$ at point $p(v_0+h)$ if $g(v,p)$ satisfy the condition $g_pf+g_v=0$
Example
$u(v,p)=pv^{2.4}$ and $g(v,p)=pv^{1.4}$
then
$\frac{dp}{dv}=\frac{g-u_v}{u_p}=\frac{pv^{1.4}-2.4pv^{1.4}}{v^{1.4}}=-1.4\frac{p}{v}$
Solve this DE and will obtain $pv^{1.4}$=constant, consider $p(v_0)=p_0$ hence the value of p at point $(v_0+h)$ will be
$p(v_0+h)=\frac{p_0v_0^{1.4}}{(v_0+h)^{1.4}}$
while
the value obtain directly from this formulae
$u(p_1,v_0+h)-u(p_0,v_0)=g(p_0,v_0)h$
$p_0v_0^{2.4}+hp_0v_0^{1.4}=p_1(v_0+h)^{2.4}$
$p_0v_0^{1.4}(v_0+h)=p_1(v_0+h)^{2.4}$
$p_1=\frac{p_0v_0^{1.4}}{(v_0+h)^{1.4}}$
Can try with u(v,p) and g(v,p) with other function such that it yield the same DE $\frac{dp}{dv}=-1.4\frac{p}{v}$ but it wont give the same result by direct application of this formulae $u(p_1,v_0+h)-u(p_0,v_0)=g(p_0,v_0)h$
ie $u(v,p)=pv^a$ and $g(v,p)=(-1.4+a)pv^{a-1}$
It produce the same $f(v,p)$ with all value of a but only $a=2.4$ give the above result
Is the proof below able to verify both sequence yield the same value of $p_n$ when n tend to infinity? This equation is actually a one step numerical integration which is approximate using an Euler method but a=2.4 give the most accurate result since it is the exact solution. Does this apply the same for all u,g and f if it satisfy above conditions?
Example 2
u(v,p)=pv and g(v,p)=p
then $\frac{dp}{dv}=\frac{g-u_v}{u_p}=\frac{p-p}{v}=0$ assume $p(v_0)=p_0$
while from this formula $u(p_1,v_0+h)-u(p_0,v_0)=g(p_0,v_0)h$
$p_0v_0+p_0(h)=p_1(v_0+h)$
$p_1=p_0$