Analytical solution to a nonlinear ODE $\dot x = cy(1-xy) - \beta cx$ and $\dot y = xc(1-xy)$

Question

Is it possible to find an analytical solution to the following system of ODEs: $\dot x = cy(1-xy) - bcx$ and $\dot y = cx(1-xy)$? I can find fixed points and do stability analysis but I am looking for an analytical solution if possible. Solutions in special cases will also potentially be helpful.

Answer 1

Partial answer (but way too long for a comment):

• Trivial case $c=0$: In this case the solution is given by $$ x=x_0\\ y=y_0 $$

If $c\neq0$ a simple rescaling $t\to c\cdot t'$ of the variable allows us to remove $c$ from the equations and therefore we can without loss of generality only consider the case of $c=1$.

• Case $\beta=0$: In this case we can simply calculate (similar to the comment from Nap D. Lover but we only need $\beta=0$) $$ x\dot x-y\dot y=0\\ \Leftrightarrow\frac{\partial}{\partial t} x^2-\frac{\partial}{\partial t} y^2=0 \\ \Leftrightarrow x^2-y^2=C $$

• General case: if we consider $\frac{dx}{dy}=1-\frac{\beta}{1-xy}$, I was not able to solve this by hand and the software Maple produces a solution only in terms of the higly loaded WhittakerM/W functions $$ \frac{(1+\beta)WhittakerM(-\frac{2+3\beta}{4\beta}, \frac{1}{4}, \frac{(x-y)^2}{2\beta})+(yx-y^2-1)WhittakerM(\frac{\beta-2}{4\beta}, \frac{1}{4}, \frac{(x-y)^2}{2\beta})}{(1+\beta)WhittakerW(-\frac{2+3\beta}{4\beta}, \frac{1}{4}, \frac{(x-y)^2}{2\beta})+2\beta(yx-y^2-1)WhittakerW(\frac{\beta-2}{4\beta}, \frac{1}{4}, \frac{(x-y)^2}{2\beta})}=C $$

For some special cases of $\beta$ this will simplify to more standard functions such as $\beta=\pm 1$.