Is there an analytical solution to this differential equation : $y''+\alpha yy'+\beta y'+\gamma y^{3}=0?$ I have this set of differential equations:
\begin{align*}
\frac{dD(t)}{dt}&=-\frac{k}{2}D(t)^2-KI(t)D(t)\\
\frac{dI(t)}{dt}&=\frac{k}{2}D(t)^2\\
\frac{dF(t)}{dt}&=KI(t)D(t)
\end{align*}
After some calculations, I obtained a differential equation of the form :
$$D''(t)+\alpha D(t)D'(t)+\beta D'(t)+\gamma D(t)^{3}=0$$
I wonder if an analytical solution exists ? I found a solution for $$D''(t)+\alpha D(t) D'(t)+\gamma D(t)^3=0$$
but the term $\beta D'(t)$ is giving me trouble.
Perhaps there exists also a simpler way to solve this set of differential equations?
Thanks for your answer.
 A: Starting from your first two ODE's
\begin{align}
D'=-\frac{k}{2}D^2-KID,\ \ \ \ \
I'=\frac{k}{2}D^2,
\end{align}
divide the first by $D$ and take a derivative,
\begin{align}
\frac{D''}{D}-&\left(\frac{D'}{D}\right)^2+\frac{k}{2}D'+KI'=0,\\
\frac{D''}{D}-&\left(\frac{D'}{D}\right)^2+\frac{k}{2}D'+\frac{kK}{2}D^2=0.
\end{align}
Let
\begin{align}
D'=\frac{1}{p},\ \ \ \ \ 
D''=-\frac{1}{p^3}\frac{\mathrm dp}{\mathrm dD},
\end{align}
so that the ODE becomes
\begin{align}
-\frac{1}{Dp^3}\frac{\mathrm dp}{\mathrm dD}-\frac{1}{(pD)^2}+\frac{k}{2p}+\frac{kK}{2}D^2=0.
\end{align}
Now substitute
\begin{align}
p=\frac{1}{Du},\ \ \ \ \ 
\frac{\mathrm dp}{\mathrm dD}=-\frac{1}{D^2u}-\frac{1}{Du^2}\frac{\mathrm du}{\mathrm dD},
\end{align}
so then \begin{align}
u\frac{\mathrm du}{\mathrm dD}+\frac{k}{2}u+\frac{kK}{2}D=0.
\end{align}
The final substitution of $D=-2\xi/k$ brings the equation to
\begin{align}
u\frac{\mathrm du}{\mathrm d\xi}=u+\frac{2K}{k}\xi.
\end{align}
The solution to this is presented in the book Handbook of Exact Solutions for Ordinary Differential Equations by Polyanin & Zaitsev, (I'm unsure how this was derived), in parametric form
\begin{align}
\xi=c_1 \exp\left(-\int \frac{\tau \mathrm d\tau}{\tau^2-\tau-2K/k}\right), \ \ \ \ \ u=\tau\xi,
\end{align}
Where $c_1$ is the constant of integration. After undoing the substitutions you'll get that
\begin{align}
D=-\frac{2c_1}{k}\exp\left(-\int \frac{\tau \mathrm d\tau}{\tau^2-\tau-2K/k}\right), \ \ \ \ \ \tau=-\frac{2}{k}\frac{D'}{D^2}.
\end{align}
I believe the author intentionally left the integral as is, since I don't think you can find $\tau(D)$ analytically (but the form in the book is a bit more general, a constant tacked onto the Abel equation, which only slightly complicates the parametric solution), here is the algebraic equation to solve:
\begin{align}
\left(\frac{kD}{2c_1}\right)^2(k\tau^2-k\tau-2K)\left(\frac{2k\tau-\sqrt{k(8K+k)}-k}{2k\tau+\sqrt{k(8K+k)}-k}\right)^{\frac{k}{\sqrt{k(8K+k)}}}=\tau^2,
\end{align}
which I won't be typing into Wolfram Alpha. If you can solve this, you'd be able to separate and integrate to find $t(D)$ and possibly $D(t)$:
\begin{align}
\int \frac{\mathrm dD}{\tau(D)D^2}=c_2-\frac{k}{2}t.
\end{align}
I believe you could do this numerically, but at that point you could just numerically solve the original system.
----Edit----
Substituting the second ODE into the first yields the equation
\begin{align}
\frac{1}{K}\frac{I''}{I}+\left(\frac{k}{2K^2}-\frac{1}{K}\right)\left(\frac{I'}{I}\right)^2+I'=0.
\end{align}
Substituting $I'=\Lambda\rightarrow I''=\Lambda\mathrm d\Lambda/\mathrm dI$, yields
\begin{align}
\frac{\mathrm d\Lambda}{\mathrm dI}+\left(\frac{k}{2K}-1\right)\frac{\Lambda}{I}+I=0. \tag{$\dagger$}
\end{align}
Multiplying by the integrating factor $I^{k/2K-1}$ gives
\begin{align}
\frac{\mathrm d}{\mathrm dI}\left(I^{k/2K-1}\Lambda\right)+I^{k/2K}=0\\
I^{k/2K-1}\Lambda+\frac{2K}{k+2K}I^{k/2K+1}=c_1\\
\Lambda=I'=c_1I^{1-k/2K}-\frac{2K}{k+2K}I^2\\
\int \left(c_1I^{1-k/2K}-\frac{2K}{k+2K}I^2\right)^{-1}\mathrm dI=t+c_2.
\end{align}
The solution to this integral is a Hypergeometric function,
\begin{align}
\frac{I^{2-k/2K}}{c_1}\ _2F_1\left(1, \frac{k-4K}{k-6K};\frac{2k-10K}{k-6K};\frac{2K}{c_1(k+2K)}I^{3-k/2K}\right)=t+c_2,
\end{align}
which you can only find the inverse of numerically. For $k$ being integer multiples of $K$ the integral is nicer (simpler non-elementary functions), but I could only find one case where the result was invertible via elementary function, $k=2K$, which yields the solution
\begin{align}
   I(t)=c_1\tanh\left[\frac{c_1}{2}t+c_2\right].
\end{align}
But even then, the solution for $D(t)$ still requires a Hypergeometric function, and I can't imagine you'd be able to analytically find $F(t)$.
I apologize for the wall of math. All of this was to say it's not worth the effort! I believe it's time to turn to numerical methods.
