For the Muskat’s Material Balance equation:
\[\newcommand{\numD}{{\dfrac {S_o}{B_o B_g} \dfrac {dR_s}{dP} + \dfrac {S_o}{B_o} \dfrac {k_g}{k_o} \dfrac {\mu_o}{\mu_g} \dfrac {dB_o}{dP} + (1 - S_o - S_w) \dfrac {1}{B_g} \dfrac {dB_g}{dP} }}\]
\[\dfrac {dS_o}{dP} = \dfrac {\numD} {1 + \dfrac {k_g}{k_o} \dfrac {\mu_o}{\mu_g} }\]
All the terms on the right side are function of pressure (\(P\)) and saturation (\(S\)), the equation could be reduced to:
\[ \dfrac{dS}{dP} = f(P, S)\]
A first-order ordinary diferential equation (ODE).
Given the ODE: \[ \dfrac{dS}{dP} = (2P + S)\]
Find the first saturation values if the initial conditions are \(P_o\) = 0, and \(S_o\) = 1.