How to solve diffrential equations of Chemical Kinetics 3rd order reaction [Problem setting]

(ii)  A chemical reaction is described by
$$
A+B+C \to^{k_1} D\\
D \to_{k_2} A+B+C
$$
If $x_1, x_2, x_3, x_4$ are the concentrations of A, B, C and D respectively, then write down the governing equations.
Hence deduce the equations for
$$
A+A+A \to^{k_1} A_3 \\
A_3 \to_{k_2} A + A + A \\
$$
Solve these equations and illustrate the solutions in a graph of the concentration of $A$ against $A_3$.
D.N.Burghers, M.S.Borrie, "Modelling with Differential Equations",  1990, p.134 Chapter6 Exercises 3.(ii)

[Solved halfway]
If $x$ is the concentration of $A$, $y$ of $A_3$,
\begin{eqnarray*}
-\frac{1}{3}\frac{dx}{dt} 
&=& k_{1}x^3 - k_{2}y\\
-\frac{dy}{dt} &=& k_2y - k_1x^{3}
\end{eqnarray*}
where $k_1, k_2$ are the reaction rates.
rearrange,
\begin{eqnarray}
\left\{
\begin{array}{ll}
\frac{dx}{dt} 
&=& 3k_{2}y - 3k_{1}x^3 & (1) \\
\frac{dy}{dt} &=& k_1x^{3} - k_2y & (2)
\end{array}
\right.
\end{eqnarray}
rearrange eq(1),
$$
\frac{dx}{dt} = 3k_2y - 3k_{1}x^3\\
y = \frac{1}{3k_2}(3k_{1}x^3 + \frac{dx}{dt})\\
\frac{dy}{dt} = \frac{1}{3k_2}(3k_{1}3x^2\frac{dx}{dt} + \frac{d^2x}{dt^2})
= \frac{3k_{1}x^2}{k_2}\frac{dx}{dt} + \frac{1}{3k_2}\frac{d^2x}{dt^2}
$$
substitute to (2) from it, derive differential equation.
\begin{eqnarray*}
\frac{dy}{dt} &=& k_1 x^3 - k_2y\\
\frac{3k_{1}x^2}{k_2}\frac{dx}{dt} + \frac{1}{3k_2}\frac{d^2x}{dt^2} &=& k_1 x^3 - k_2\frac{1}{3k_2}(3k_{1}x^3 + \frac{dx}{dt})\\
\frac{3k_{1}x^2}{k_2}\frac{dx}{dt} + \frac{1}{3k_2}\frac{d^2x}{dt^2} &=& k_1 x^3 - k_{1}x^3 -\frac{1}{3} \frac{dx}{dt}\\
9k_{1}x^2 \frac{dx}{dt} + \frac{d^2x}{dt^2} &=&  -k_2 \frac{dx}{dt}\\
\frac{d^2x}{dt^2} + (9k_{1}x^2 + k_2)\frac{dx}{dt} &=& 0\\
\end{eqnarray*}
replace $p=dx/dt$,
$$
\frac{d^2 x}{dt^2} 
= \frac{dp}{dt}
= \frac{dp}{dx}\frac{dx}{dt}
= \frac{dp}{dx} p
$$
substitute $p$,
$$
\frac{d^2x}{dt^2} + (9k_{1}x^2 + k_2)\frac{dx}{dt} = 0\\
\frac{dp}{dx} p + (9k_{1}x^2 + k_2)p = 0\\
$$
if $p \neq 0$,
$$
\frac{dp}{dx} p + (9k_{1}x^2 + k_2)p = 0\\
\frac{dp}{dx} + (9k_{1}x^2 + k_2) = 0\\
\frac{dp}{dx} = -(9k_{1}x^2 + k_2)\\
p = -3k_1x^3 -k_2x + c_0
$$
now have a first order differential equation and can separate the variables to give,
$$
\int \frac{dx}{3k_1x^3 + k_2x - c_0} = - \int dt\\
\frac{1}{3k_1} \int \frac{dx}{x^3 + \frac{k_2}{3k_1}x - \frac{c_0}{3k_1} } = -t+ c_1\\
$$
How can I solve the rest?

From general principles of mass conservation, or the construction of the system by mass exchange terms that balance, you can easily see that
$$
3x+y=C=3x_0+y_0
$$
is a constant. Thus the system lives on the line $y=C-3x$. From the right side of the first equation we see that equilibrium points satisfy $k_1x^3-k_2y=0$. There is only one intersection point of the rising cubic $y=\frac{k_1}{k_2}x^3$ with the falling line $y=C-3x$, thus exactly one stationary point for the system.
Eliminating $y$ from the first ODE, one finds
$$
\dot x = -3(k_1x^3-k_2y)=-3(k_1x^3+3k_2x-k_2C)
$$
so that the $x$-derivative of the right side is negative for all $x>0$. This means that the stationary point is attracting or stable, all "physical" solutions converge to that point.
