Does this system of non-linear differential equations have an analytical solution? I have this system of linear differential equations:
\begin{align*}
\frac{dA}{dt} & = k_1 - k_2AB - k_3A\\
\frac{dB}{dt} & = -k_2AB - k_4B\\
\frac{dC}{dt} & = k_2AB - k_5C\\
\frac{dD}{dt} & = k_6C
\end{align*}
Where $k_{i}$ are constants for $i\in\{1,2,3,4,5,6\}$.
Will an analytic solution to these equations exist? Is there an easy way to tell?
Thank you in advance
 A: \begin{align*}
A'=\frac{dA}{dt} & = k_1 - k_2AB - k_3A \tag 1\\
B'=\frac{dB}{dt} & = -k_2AB - k_4B \tag 2\\
C'=\frac{dC}{dt} & = k_2AB - k_5C \tag 3\\
D'=\frac{dD}{dt} & = k_6C \tag 4
\end{align*}
The equations $(1)$ and $(2)$ are independent from $(3)$ and $(4)$. 
From $(2)$ :
$$A=-\frac{1}{k_2} \left(\frac{B'}{B}+k_4 \right) \tag 5$$
$$A'=-\frac{1}{k_2} \left(\frac{B''}{B}-\frac{(B')^2}{B^2} \right)$$
Putting them into $(1)$ :
$$-\frac{1}{k_2} \left(\frac{B''}{B}-\frac{(B')^2}{B^2} \right)= k_1 +(B'+k_4B)  +\frac{k_3}{k_2} \left(\frac{B'}{B}+k_3k_4 \right)$$
$$BB'' -B'^2 +(k_2B +k_3)B B' +(k_1k_2+k_3k_4)B^2+k_2k_4B^3=0$$
This is a second order non-linear ODE, with only one unknown function. Of course, they are an infinity of solutions $B(t)$. But most likely in the general case the solutions cannot be expressed on a closed form.
It is certain that closed form exist in case of some particular values of the coefficients. In those particular cases, $B(t)$ which is obtained can be put into equation (5) leading to $A(t)$. Then, putting $A$ and $B$ into (3) leads to a first order linear ODE which theoretically can be solved. May be not in practice, if the functions involved become too complicated. So, in some cases, $C(t)$ can be obtained. Then, $D(t)=k_6\int C(t)dt$.$ 
A: PARTIAL SOLUTION (case $k_3=k_4=k_5$)
Upon eliminating the $k_2AB$ terms, the system in $A,B,C$ becomes
\begin{align*}
\dot{A}-\dot{B}&=k_1-k_3(A-B)\\
\dot{B}+\dot{C}&=-k_3(B+C)\\
\dot{A}+\dot{C}&=k_1-k_3(A+C)
\end{align*}
which may be integrated directly to yield
\begin{align*}
A-B&=\frac{k_1-\bigg(k_1-k_3(A_0-B_0)\bigg)\exp(-k_3t)}{k_3}&=f_1\\
B+C&=(B_0+C_0)\exp(-k_3t)&=f_2\\
A+C&=\frac{k_1-\bigg(k_1-k_3(A_0+C_0)\bigg)\exp(-k_3t)}{k_3}&=f_3
\end{align*}
where $A_0=A(0)$ and so on. It follows that $B_0+C_0=0$ and hence $B(t)+C(t)=0$.
In matrix form,
$\begin{pmatrix}
A\\
B\\
C\\
\end{pmatrix}=
\begin{pmatrix}
1&-1&0\\
0&1&1\\
1&0&1\\
\end{pmatrix}^{-1}
\begin{pmatrix}
f_1\\
0\\
f_3\\
\end{pmatrix}$
and it is clear that there exists no unique solution because the $3\times 3$ matrix is singular.
A: Suppose $k_3=k_4$ and $k_2=-k_3^2/k_1$.  Then one solution to the first two equations is
\begin{align}
A(t)&=\frac{k_1}{k_3}\left(1-\frac{1}{k_3(t-T)}\right),\\
B(t)&=-\frac{k_1}{k_3^2(t-T)},
\end{align}
where $T$ is a constant.  The other two are then straightforward to integrate.
