I've been working with the following system of differential equations:
$x' = a_1 - b_1x - cx(x+y)$
$y' = a_2 - b_2y - cy(x+y)$
With initial conditions $x(0)=y(0)=0$.
I was wondering if anyone here by any chance happens to know an analytical solution to this, since I haven't been able to find one anywhere else.
Because of its non-linear nature (assuming that $c \neq 0$, because otherwise everything is trivial) it's not going to be particularly amenable to analytic solutions, but you can remove the dependence between $x$ and $y$ to apply univariate ODE solution methods (although in practice there are a bunch of different cases depending on the specific values of the coefficients). For example:
First, look at what happens when we separately add and subtract the DEs from each other:
$$\begin{eqnarray} x' + y' & = & a_1 + a_2 - b_1 x - b_2 y - cx (x + y) - cy (x + y) \\ & = & (a_1 + a_2) - \frac{b_1 + b_2}{2} (x + y) - \frac{b_1 - b_2}{2} (x - y) - c(x + y)^2 \\ x' - y' & = & a_1 - a_2 - b_1 x + b_2 y - cx (x + y) + cy (x + y) \\ & = & a_1 - a_2 - \frac{b_1 - b_2}{2} (x + y) - \frac{b_1 + b_2}{2} (x - y) - c(x - y)(x + y) \end{eqnarray}$$
I'm now going to make the substitutions $u = x + y$, $v = x - y$, $A_1 = a_1 + a_2$, $A_2 = a_1 - a_2$, $B_1 = \frac{b_1 + b_2}{2}$, $B_2 = \frac{b_1 - b_2}{2}$, giving:
$$\begin{eqnarray} u' & = & A_1 - B_1 u - B_2 v - c u^2 \\ v' & = & A_2 - B_2 u - B_1 v - c uv \end{eqnarray}$$
Now, if $B_2 = 0$ (i.e. if $b_1 = b_2$), then the first equation is just a separable first-order ODE in $u$ that you can solve for an implicit expression of $u$, and for certain values of the coefficients you might even be able to explicitly solve for $u$. From there, you can substitute the form of $u$ into the second equation to similarly solve for $v$.
Then if $B_2 \neq 0$, we can do some more manipulation:
$$\begin{eqnarray} B_2 v & = & A_1 - B_1 u - c u^2 - u' \\ B_2 v' & = & - B_1 u' - 2 c u u' - u'' \\ B_2 (A_2 - B_2 u - B_1 v - c uv) & = & - B_1 u' - 2 c u u' - u'' \\ B_2 v (B_1 + cu) & = & u'' + 2cuu' + B_1 u' - B_2^2 u + A_2 B_2 \\ (A_1 - B_1 u - c u^2 - u')(B_1 + cu) & = & u'' + 2cuu' + B_1 u' - B_2^2 u + A_2 B_2 \end{eqnarray}$$
and I'll let you try manipulating that last one, but eventually you'll get something like $u'' + p(u) u' + q(u) = 0$, where $p$ and $q$ are polynomials, which again is going to have a bunch of special cases that may be more or less approachable (e.g. if $p = 0$ or $q = 0$).