Solve $\dot v + \frac{\alpha}{m}v + \frac{\beta}{m} v^2 = g$ $\alpha, \beta$ are constants.
I know this is a Riccati equation, and I've tried to find a particular solution, with no success.
Also, $v = v(t)$, but $t$ is not part of the equation explicitly, so maybe that could help somehow?
I don't know what else to try, I'd appreciate a push in the right direction.
Darsen is of course correct for the case where $\dot v =0 \; .$ That being said, my first observation is that this equation is integrable. Observe that $$\dot v  = g - \frac{\alpha}{m}v - \frac{\beta}{m} v^2$$
$$\text{implies}$$
$$\frac{\dot v}{g - \frac{\alpha}{m}v - \frac{\beta}{m} v^2}  = 1 $$
whenever $\dot v \neq 0$ thus we can find an expression for $v$ using
$$\int\frac{dv}{g - \frac{\alpha}{m}v - \frac{\beta}{m} v^2}  = \int dt \; .$$
Use some good integral tables and you will find a solution where $v$ is easy to isolate. Let me know if you need any more guidance.
Edit:
pujfei is also correct in his answer and you'll find that the integral above is a flavor of the $\arctan $ function.
Result
Given the original equation:
$$
v'(t)+\frac{\alpha  }{m}v(t)+\frac{ \beta }{m}v(t)^2=g
$$
The general solution is:
\begin{equation}
v(t)= \frac{-\alpha +\sqrt{-\alpha ^2-4 \beta  g m} \tan \left(\frac{1}{2} \left(-\frac{t \sqrt{-\alpha ^2-4 \beta  g m}}{m}-c_1 \sqrt{-\alpha ^2-4 \beta  g m}\right)\right)}{2 \beta }
\end{equation}
Or:
$$
\begin{align*}
v(t)&amp;=\frac{-\alpha+\sqrt{-\alpha^2-4bgm}\tan{\big[\frac{1}{2}\big(-\frac{\sqrt{-\alpha^2-4{\beta}gm}}{m}{\cdot}t-\sqrt{-\alpha^2-4{\beta}gm}{\cdot}c_1\big)\big]}}{2{\beta}}\\
&amp;=\frac{1}{2{\beta}}\bigg(-\alpha+\sqrt{-\alpha^2-4{\beta}gm}\tan{\big[\frac{1}{2}\big(-c_1\sqrt{-\alpha^2-4{\beta}gm}-\frac{\sqrt{-\alpha^2-4{\beta}gm}}{m}{\cdot}t\big)\big]}\bigg)\\
&amp;=\frac{1}{2{\beta}}\bigg(-\alpha+\sqrt{\Delta}\tan{\big[\frac{1}{2}\big(-c_1\sqrt{\Delta}-\frac{\sqrt{\Delta}}{m}{\cdot}t\big)\big]}\bigg)
\end{align*}
$$
In which the discriminant for this kind of differential equations is apparently:
$$
\Delta=-\alpha^2-4{\beta}gm
$$
Proof
The solution is achieved by applying the following transformation to a general form of Riccati Equation.
A Riccati Equation is an equation that looks like this:
$$y'(x)=q_0(x)+q_1(x)y(x)+q_2(x)y^2(x),q_2(x)\neq0\tag{1}$$
It can always be reduced to a second order linear ordinary differential equation (ODE).
With:
$$y'=q_0(x)+q_1(x)y+q_2(x)y^2\tag{2}$$
Let $v=yq_2$, we'll have:
$$v'=v^2+R(x)v+S(x)\tag{3}$$
where
$$
\left\{
\begin{align}
S&amp;=q_0q_2\\
R&amp;=q_1+\frac{q^{'}_2}{q_2}
\end{align}\tag{4}
\right.
$$
To continue, substitute:
$$v=-\frac{u'}{u}\tag{5}$$
We'll achieve a linear 2nd order ODE:
$$u''-Ru'+Su=0\tag{6}$$
Whose general solution is:
$$
\begin{equation}
u(x)= c_1 e^{\frac{1}{2} x \left(R-\sqrt{R^2-4 S}\right)}+c_2 e^{\frac{1}{2} x \left(R+\sqrt{R^2-4 S}\right)}
\end{equation}\tag{7}
$$
Thus we have:
$$
\begin{align}
y&amp;=-\frac{u'}{q_2u}\\
&amp;=\frac{1}{2q_2}\left(-R+\sqrt{R^2-4 S} \left(-1+\frac{2 c_1}{c_2 e^{x \sqrt{R^2-4 S}}+c_1}\right)\right)
\end{align}\tag{8}
$$
For the instance in this question:
$$
\left\{
\begin{align}
q_0&amp;=g\\
q_1&amp;=-\frac{\alpha}{m}\\
q_2&amp;=-\frac{\beta}{m}\\
R&amp;=q_1=-\frac{\alpha}{m}\\
S&amp;=q_2q_0=-\frac{g\beta}{m}\\
R^2-4S&amp;=\frac{\alpha^2+4{\beta}mg}{m^2}
\end{align}\tag{9}
\right.
$$