How to solve the Lienard equation $v'' = av -bvv'$? For a Physics project that I'm doing in school, I have the following differential equation:
$$v'' = av -bvv'$$
where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time. 
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.
It's really important that I solve it, and I'm not able to find the solution anywhere. 
This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks
 A: Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.
A: First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.
Writing $v' = w(v)$ with a new function $w$, $v'' = w \frac{dw}{dv}$, we obtain the first-order separable ODE
\begin{equation}
w \frac{dw}{dv} = a v - b v w = v (a - b w).
\end{equation}
Assuming that $w \not \equiv \frac{a}{b}$ and using separation of variables we obtain
\begin{equation}
\frac{w}{a-bw} \, \mathrm{d}w = v \, \mathrm{d}v \quad \Leftrightarrow \quad \ln\left|\frac{b}{a}w-1\right| + \frac{b}{a} w - 1 = - \frac{b^2}{2 a} v^2 + C_1, \quad C_1 \in \mathbb{R}.
\end{equation}
Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.
Finally, the solution of the separable ODE $\frac{dv}{dt} = w(v)$ yields an implicit representation
\begin{equation}
\int \frac{1}{w(v)} \, \mathrm{d}v = t
\end{equation}
for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.
Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = \frac{a}{b}$, is given by the linear function $v(t) = \frac{a}{b} t + v_0$. This is the special case of a constant $w \equiv \frac{a}{b}$, for which we don't need to do the separation of variables.
