I am solving a differential equation related to fluid mechanics, a rigid air bubble rising towards the surface of a liquid. Doing all the calculations, I have come to this differential equation, which I have not been able to solve, by far.
$$\frac{d^2y}{dt^2}=A(P_{atm}-\rho g y)^{2/3}\cdot \left(\frac{dy}{dt}\right)^2+By+C \tag{1}$$ Where $A,B,C, P_{atm}, \rho$ and $g$ are all constants.
I have created an alternative differential equation by substituting $v=y'$, as it seems like a legitimate method, however I am not sure how I will solve this, as this equation also seems unsolvable:
$$v\frac{dv}{dy}=A(P_{atm}-\rho g y)^{2/3}\cdot v^2+By+C$$
$$\frac{dv}{dy}=A(P_{atm}-\rho g y)^{2/3}\cdot v+\frac{By+C}{v}\ \tag{2}$$
In conclusion, I would like some guidance to obtain a solution for $y(t)$. I am not sure whether if I solved $(2)$, if it would still be possible to get $y(t)$ from $v(y)$.
The idea to substitute $v = \frac{\text{d} y}{\text{d} t}$ is a good one. To start with your last equation, we can multiply on both sides with $v$ to obtain \begin{equation} v \frac{\text{d} v}{\text{d} y} = A (P- \rho g\, y)^{\frac{2}{3}} v^2 + B y + C. \tag{1} \end{equation} Now, we recognise the left hand side as the derivative of $\frac{1}{2} v^2$. Therefore, writing $w = v^2$, we can write $(1)$ in terms of $w$ as \begin{equation} \frac{\text{d} w}{\text{d} y} = 2A (P- \rho g\, y)^{\frac{2}{3}} w + 2B y + 2C, \tag{2} \end{equation} which is a linear first order ODE for $w$. Now, I would suggest that you rescale $w$ and $y$ to simplify $(2)$ a bit. A good choice would be to introduce \begin{equation} \eta = \frac{P - \rho g y}{(\frac{\rho g}{2A})^{\frac{3}{5}}} \;\Leftrightarrow \; y = \frac{P - (\frac{\rho g}{2A})^{\frac{3}{5}}\eta}{\rho g} \end{equation} and \begin{equation} z = \frac{\rho g A}{B} (\frac{2A}{\rho g})^{\frac{1}{5}} w\;\Leftrightarrow\;w = \frac{B}{\rho g A} (\frac{\rho g}{2A})^{\frac{1}{5}} z \end{equation} to reduce $(2)$ to \begin{equation} \frac{\text{d} z}{\text{d} \eta} = - \eta^{\frac{2}{3}} z + \eta - D, \tag{3} \end{equation} where \begin{equation} D = \frac{B P +C \rho g}{B (\frac{\rho g}{2A})^{\frac{3}{5}}}. \end{equation} You can solve $(3)$ explicitly in terms of the exponential integral.