I am solving two the first order ODEs ($'=\dfrac{d}{dz}$, all other variables are known constants):
$$p_0 p_0'=-\dfrac{32 \beta}{R^4}$$
$$(p_0p_1)'=-\dfrac{2-\sigma_v}{\sigma_v}\dfrac{8}{R}p_0'$$
I have next conditions
$p_0|_{z=0}=p_{0i}$ (I can choose value)
$p_0|_{z=1}=1$
$p_1|_{z=0}=0$
$p_1|_{z=1}=0$
It is necessary to find $p_0'|_{z=0}$ and $p_1'|_{z=0}$ with shooting method (literature says like that), according to already mentioned $p_0|_{z=1}=1$ and $p_1|_{z=1}=0$. How to connect this two conditions and shoot $p_0'|_{z=0}$ for already known $p_0|_{z=1}=1$?
Are that conditions $p_0'|_{z=0}$ and $p_1'|_{z=0}$ necessary, because these are the first order equations, is there only one initial condition enough?
Instead of missing conditions, I also need to solve numerically two ODEs, from the beginning of text, with Runge Kutta method. How to connect that solving with shooting? Is it possible to find missing condition with bvp4c, as shooting method, and after that solve equation with ode45?
Yes, for a boundary value problem you better use a BVP solver. For an order 1 system in 2 components you can only ever hope to satisfy 2 arbitrary boundary conditions, thus if you chose to fix values to $p_0(0)$ and $p_0(1)$, the values of $p_1$ are dependents, there is no guarantee that they will be even close to zero. However, as the equation for $p_0$ is independent of $p_1$, you can always only fix one boundary condition for $p_0$, the other has to be for $p_1$ and you could solve the problem in 2 stages, first find the other boundary value of $p_0$ and then solve the coupled problem from the side where the condition for $p_1$ is given.